AP Physics C: Electricity and Magnetism · ⁨AP Fisika C: Listrik dan Magnetisme⁩

Tips

AP Physics C: Electricity and Magnetism is the calculus-based electromagnetism course: charges, fields and Gauss's law; electric potential; conductors and capacitors; circuits; magnetic fields; and electromagnetic induction. It is usually taken after Mechanics and assumes that level of calculus.

Gauss's law and Ampère's law decide how this course feels. Both are really about choosing a surface or loop that makes the integral trivial — a skill you build by doing many, not by reading the law. Symmetry is the whole game.

Induction questions punish sign errors more than anything else, so fix one convention for flux direction and use it every time.

Notes cover Gauss's law through to Maxwell's equations, unit by unit, with the vector calculus written out. The library carries the released FRQs; the hardest marks in E&M go to choosing the right symmetry before integrating, so the examples start there.

Train all vocabulary · ⁨Latih semua kosakata⁩

Past Papers · ⁨Soal-Soil Masa Lalu⁩

Electric Charges, Fields, and Gauss's Law · ⁨Muatan Listrik, Medan, dan Hukum Gauss⁩

Watch lesson · ⁨Tonton pelajaran⁩

Open this topic on its own page · ⁨Buka topik ini di halaman sendiri⁩

8.1

Electric Charge and Electric Force · ⁨Muatan Listrik dan Gaya Listrik⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 8.1.A: Describe the electric force that results from the interactions between charged objects or systems.

8.1.A.1 Charge is a fundamental property of all matter.
- 8.1.A.1.i Charge is a scalar quantity and is described as positive or negative.
- 8.1.A.1.ii The magnitude of the charge of a single electron or proton, the elementary charge $e$, can be considered to be the smallest indivisible amount of charge.
- 8.1.A.1.iii The charge of an electron is $-e$ and the charge of a proton is $+e$, and a neutron has no electric charge.
- 8.1.A.1.iv A point charge is a model in which the physical size of a charged object or system is negligible in the context of the situation being analyzed.
8.1.A.2 Coulomb's law describes the electrostatic force between two charged objects as directly proportional to the magnitude of each of the charges and inversely proportional to the square of the distance between the objects.
- Relevant equation: $\left| \vec{F}_E \right| = \dfrac{1}{4\pi\varepsilon_0} \dfrac{|q_1 q_2|}{r^2} = k \dfrac{|q_1 q_2|}{r^2}$
8.1.A.3 The direction of the electrostatic force depends on the signs of the charges of the interacting objects and is along the line of separation between the objects.
- 8.1.A.3.i Two objects with charges of the same sign exert repulsive forces on each other.
- 8.1.A.3.ii Two objects with charges of opposite signs exert attractive forces on each other.
8.1.A.4 Electric forces are responsible for some of the macroscopic properties of objects in everyday experiences. However, the large number of particle interactions that occur make it more convenient to treat everyday forces in terms of nonfundamental forces called contact forces, such as normal force, friction, and tension.

Learning Objective 8.1.B: Describe the electric and gravitational forces that result from interactions between charged objects with mass.

8.1.B.1 Electrostatic forces can be attractive or repulsive, while gravitational forces are always attractive.
8.1.B.2 For any two objects that have mass and electric charge, the magnitude of the gravitational force is usually much smaller than the magnitude of the electrostatic force.
8.1.B.3 Gravitational forces dominate at larger scales even though they are weaker than electrostatic forces, because systems at large scales tend to be electrically neutral.

Learning Objective 8.1.C: Describe the electric permittivity of a material or medium.

8.1.C.1 Electric permittivity is a measurement of the degree to which a material or medium is polarized in the presence of an electric field.
8.1.C.2 Electric polarization can be modeled as the induced rearrangement of electrons by an external electric field, resulting in a separation of positive and negative charges within a material or medium.
8.1.C.3 Free space has a constant value of electric permittivity, $\varepsilon_0$, that appears in physical relationships.
8.1.C.4 The permittivity of matter has a value different from that of free space that arises from the matter's composition and arrangement.
- 8.1.C.4.i In a given material, electric permittivity is determined by the ease with which electrons can change configurations within the material.
- 8.1.C.4.ii Conductors are made from electrically conducting materials in which charge carriers move easily; insulators are made from electrically nonconducting materials in which charge carriers cannot move easily.

Boundary statement: AP Physics C: Electricity & Magnetism only expects students to make calculations of the electric force between four or fewer interacting charged objects or systems. The analysis of the resulting electric force from more charges is allowed in situations of high symmetry. Note that students are expected to calculate the electric fields of charge distributions, as described in Topics 8.4 and 8.6.

Bahasa Indonesia

Tujuan Pembelajaran 8.1.A: Jelaskan gaya listrik yang dihasilkan dari interaksi antara benda atau sistem yang bermuatan.

8.1.A.1 Muatan adalah sifat fundamental dari semua materi.
- 8.1.A.1.i Muatan adalah besaran skalar dan dijelaskan sebagai positif atau negatif.
- 8.1.A.1.ii Besar muatan elektron tunggal atau proton, yaitu muatan elementer $e$, dapat dianggap sebagai jumlah muatan terkecil yang tak terbagi.
- 8.1.A.1.iii Muatan elektron adalah $-e$ dan muatan proton adalah $+e$, sedangkan neutron tidak memiliki muatan listrik.
- 8.1.A.1.iv Muatan titik adalah model di mana ukuran fisik benda atau sistem bermuatan diabaikan dalam konteks situasi yang sedang dianalisis.
8.1.A.2 Hukum Coulomb mendeskripsikan gaya elektrostatik antara dua benda bermuatan yang sebanding langsung dengan besar masing-masing muatan dan berbanding terbalik dengan kuadrat jarak antar benda.
- Persamaan relevan: $\left| \vec{F}_E \right| = \dfrac{1}{4\pi\varepsilon_0} \dfrac{|q_1 q_2|}{r^2} = k \dfrac{|q_1 q_2|}{r^2}$
8.1.A.3 Arah gaya elektrostatik bergantung pada tanda muatan dari benda-benda yang berinteraksi dan berada sepanjang garis pemisah antara benda-benda tersebut.
- 8.1.A.3.i Dua benda dengan muatan yang sama tanda saling memberikan gaya tolak-menolak.
- 8.1.A.3.ii Dua benda dengan muatan yang berbeda tanda saling memberikan gaya tarik-menarik.
8.1.A.4 Gaya listrik bertanggung jawab atas beberapa sifat makroskopis objek dalam pengalaman sehari-hari. Namun, karena banyaknya interaksi partikel yang terjadi, lebih nyaman untuk memperlakukan gaya-gaya sehari-hari dalam istilah gaya non-fundamental yang disebut gaya kontak, seperti gaya normal, gesekan, dan tegangan.

Tujuan Pembelajaran 8.1.B: Jelaskan gaya listrik dan gravitasi yang dihasilkan dari interaksi antara benda bermuatan dengan massa.

8.1.B.1 Gaya elektrostatik bisa bersifat tarik-menarik atau tolak-menolak, sementara gaya gravitasi selalu tarik-menarik.
8.1.B.2 Untuk setiap dua benda yang memiliki massa dan muatan listrik, besar gaya gravitasi biasanya jauh lebih kecil daripada besar gaya elektrostatik.
8.1.B.3 Gaya gravitasi mendominasi pada skala yang lebih besar meskipun lebih lemah daripada gaya elektrostatik, karena sistem pada skala besar cenderung netral secara listrik.

Tujuan Pembelajaran 8.1.C: Jelaskan permitivitas listrik suatu material atau medium.

8.1.C.1 Permitivitas listrik adalah pengukuran seberapa jauh suatu material atau medium terpolarisasi ketika berada dalam medan listrik.
8.1.C.2 Polaritas listrik dapat dimodelkan sebagai penataan ulang elektron yang diinduksi oleh medan listrik eksternal, yang menghasilkan pemisahan muatan positif dan negatif di dalam material atau medium.
8.1.C.3 Ruang hampa memiliki nilai konstan permitivitas listrik, $\varepsilon_0$, yang muncul dalam hubungan fisika.
8.1.C.4 Permitivitas materi memiliki nilai berbeda dari ruang hampa yang timbul dari komposisi dan susunan materinya.
- 8.1.C.4.i Dalam suatu material tertentu, permitivitas listrik ditentukan oleh kemudahan elektron mengubah konfigurasi di dalam material tersebut.
- 8.1.C.4.ii Konduktor dibuat dari bahan yang menghantarkan listrik di mana pembawa muatan bergerak dengan mudah; isolator dibuat dari bahan yang tidak menghantarkan listrik di mana pembawa muatan tidak dapat bergerak dengan mudah.

Pernyataan batas: AP Physics C: Listrik & Magnetisme hanya mengharapkan siswa melakukan perhitungan gaya listrik antara empat atau kurang objek/sistem bermuatan yang berinteraksi. Analisis gaya listrik yang dihasilkan dari lebih banyak muatan diperbolehkan dalam situasi simetri tinggi. Perhatikan bahwa siswa diharapkan menghitung medan listrik dari distribusi muatan, sebagaimana dijelaskan dalam Topik 8.4 dan 8.6.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

Electric charge 电荷 comes in positive and negative; like charges repel, opposites attract. Charge is quantized 量子化 – every charge is a whole-number multiple of the elementary charge 基本电荷 $e=1.6\times10^{-19}\ \text{C}$ – and obeys conservation of charge 电荷守恒: charge is never created or destroyed, only moved. The force between two point charges 点电荷 is Coulomb's law 库仑定律:

$$\vec{F}=\frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}\,\hat{r},$$

an inverse-square 平方反比 force along the line joining them, with $\dfrac{1}{4\pi\varepsilon_0}=k=9.0\times10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$. For several charges (AP uses four or fewer, unless symmetry helps), add the force vectors on the charge you care about – superposition 叠加.

Charge is a fundamental property 基本属性 of matter, and a point charge is a model that ignores the size of a charged object. The constant $\varepsilon_0$ is the permittivity of free space 真空介电常数 – a material property of the vacuum. The permittivity of matter differs from $\varepsilon_0$, set by how easily the material polarizes, which is why a dielectric between charges weakens the force. Note too that although the electric force is far stronger than gravity, gravity dominates at astronomical scales because large bodies are nearly neutral.

Worked example. Two $+2.0\ \mu\text{C}$ charges sit $0.30\ \text{m}$ apart: $F=\dfrac{kq_1q_2}{r^2}=\dfrac{9.0\times10^{9}(2.0\times10^{-6})^2}{(0.30)^2}=0.40\ \text{N}$, repulsive.

Worked example (vectors). Charges $+3.0\ \mu\text{C}$ and $-3.0\ \mu\text{C}$ sit at two corners of a right angle, each $0.30\ \text{m}$ from a $+1.0\ \mu\text{C}$ charge at the corner. Each exerts $F=\dfrac{(9.0\times10^{9})(3.0\times10^{-6})(1.0\times10^{-6})}{0.09}=0.30\ \text{N}$ – one a push, one a pull, at right angles to each other. The net force is $F_{\text{net}}=\sqrt{0.30^2+0.30^2}=0.42\ \text{N}$, pointing between them. Never add magnitudes blindly: components first.

Bahasa Indonesia

Muatan listrik ada dalam bentuk positif dan negatif; muatan sejenis tolak-menolak, berlawanan tarik-menarik. Muatan bersifat terkuantisasi – setiap muatan adalah kelipatan bilangan bulat dari muatan elementer $e=1.6\times10^{-19}\ \text{C}$ – dan mematuhi kekekalan muatan: muatan tidak pernah tercipta atau musnah, hanya berpindah. Gaya antara dua muatan titik dijelaskan oleh hukum Coulomb:

$$\vec{F}=\frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}\,\hat{r},$$

sebuah gaya kuadrat terbalik sepanjang garis yang menghubungkannya, dengan $\dfrac{1}{4\pi\varepsilon_0}=k=9.0\times10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$. Untuk beberapa muatan (AP menggunakan empat atau kurang, kecuali simetri membantu), tambahkan vektor gaya pada muatan yang Anda perhatikan – superposisi.

Muatan merupakan sifat fundamental materi, dan muatan titik adalah model yang mengabaikan ukuran benda bermuatan. Konstanta $\varepsilon_0$ adalah permitivitas ruang hampa – sifat material dari vakum. Permitivitas materi berbeda dari $\varepsilon_0$, ditentukan oleh seberapa mudah bahan tersebut terpolarisasi, sehingga dielektrik di antara muatan melemahkan gaya. Perhatikan pula bahwa meskipun gaya listrik jauh lebih kuat daripada gravitasi, gravitasi mendominasi pada skala astronomis karena benda-benda besar hampir netral.

Contoh kerja. Dua $+2.0\ \mu\text{C}$ muatan terletak $0.30\ \text{m}$ terpisah: $F=\dfrac{kq_1q_2}{r^2}=\dfrac{9.0\times10^{9}(2.0\times10^{-6})^2}{(0.30)^2}=0.40\ \text{N}$, tolak-menolak.

Contoh kerja (vektor). Muatan $+3.0\ \mu\text{C}$ dan $-3.0\ \mu\text{C}$ terletak di dua sudut siku-siku, masing-masing $0.30\ \text{m}$ dari sebuah muatan $+1.0\ \mu\text{C}$ di sudut tersebut. Masing-masing memberikan gaya $F=\dfrac{(9.0\times10^{9})(3.0\times10^{-6})(1.0\times10^{-6})}{0.09}=0.30\ \text{N}$ – satu mendorong, satu menarik, tegak lurus satu sama lain. Resultan gayanya adalah $F_{\text{net}}=\sqrt{0.30^2+0.30^2}=0.42\ \text{N}$, mengarah di antara keduanya. Jangan pernah menjumlahkan besaran secara buta: hitung komponen dulu.

Explore · ⁨Jelajahi⁩

Explore the field of a source charge · ⁨Jelajahi medan dari muatan sumber⁩

Change the charge's sign and size. Lines point away from positive and toward negative, and their $1/r^2$ crowding near the charge mirrors why Coulomb's force $F = k\,|q_1 q_2|/r^2$ weakens with distance. · ⁨Ubah tanda dan besar muatan. Garis menunjukkan menjauhi positif dan menuju negatif, serta $1/r^2$ kerapatan dekat muatan mencerminkan mengapa gaya Coulomb $F = k\,|q_1 q_2|/r^2$ melemah seiring jarak.⁩

8.2

Electric Charge and the Process of Charging · ⁨Muatan Listrik dan Proses Pemuatan⁩

Syllabus · ⁨Silabus⁩

Learning Objective	Essential Knowledge
8.2.A Describe the behavior of a system using conservation of charge.	8.2.A.1 The net charge or charge distribution of a system can change in response to the presence of, or changes in, the net charge or charge distribution of other systems. 8.2.A.1.i The net charge of a system can change due to friction or contact between systems. 8.2.A.1.ii Induced charge separation occurs when the electrostatic force between two systems alters the distribution of charges within the systems, resulting in the polarization of one or both systems. 8.2.A.1.iii Induced charge separation can occur in neutral systems. 8.2.A.2 Any change to a system's net charge is due to a transfer of charge between the system and its surroundings. 8.2.A.2.i The charging of a system typically involves the transfer of electrons to and from the system. 8.2.A.2.ii The net charge of a system will be constant unless there is a transfer of charge to or from the system. 8.2.A.3 Grounding involves electrically connecting a charged object to a much larger and approximately neutral system (e.g., Earth).

Learning Objective

Essential Knowledge

8.2.A
Describe the behavior of a system using conservation of charge.

8.2.A.1 The net charge or charge distribution of a system can change in response to the presence of, or changes in, the net charge or charge distribution of other systems.
- 8.2.A.1.i The net charge of a system can change due to friction or contact between systems.
- 8.2.A.1.ii Induced charge separation occurs when the electrostatic force between two systems alters the distribution of charges within the systems, resulting in the polarization of one or both systems.
- 8.2.A.1.iii Induced charge separation can occur in neutral systems.
8.2.A.2 Any change to a system's net charge is due to a transfer of charge between the system and its surroundings.
- 8.2.A.2.i The charging of a system typically involves the transfer of electrons to and from the system.
- 8.2.A.2.ii The net charge of a system will be constant unless there is a transfer of charge to or from the system.
8.2.A.3 Grounding involves electrically connecting a charged object to a much larger and approximately neutral system (e.g., Earth).

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

A conductor 导体 lets charge move freely; an insulator 绝缘体 holds it in place. Objects gain net charge three ways:

Charging by friction 摩擦起电: rubbing transfers electrons from one surface to the other.
Charging by conduction 传导起电: touching a charged object shares charge of the same sign.
Charging by induction 感应起电: a nearby charge pushes the conductor's charge apart; ground 接地 the far side, remove the ground wire first, and the conductor is left with the opposite sign – all without contact.

An insulator cannot pass charge, but it can develop polarization 极化: its molecules stretch or turn so one face is slightly positive and the other slightly negative. That is why a charged comb picks up neutral paper scraps. An electroscope 验电器 shows net charge by its leaves repelling; on any isolated conductor the excess charge sits entirely on the outer surface.

Bahasa Indonesia

Konduktor memungkinkan muatan bergerak bebas; isolator menahan muatan di tempatnya. Benda memperoleh muatan bersih melalui tiga cara:

Pemuatan dengan gesekan: menggosok mentransfer elektron dari satu permukaan ke permukaan lainnya.
Pemuatan dengan konduksi: menyentuh benda bermuatan membagikan muatan dengan tanda yang sama.
Pemuatan dengan induksi: muatan terdekat mendorong muatan konduktor menjauh; bumikan sisi yang jauh, lepaskan kabel bumi pertama, dan konduktor akan tersisa dengan tanda berlawanan – semuanya tanpa kontak langsung.

Isolator tidak dapat mengalirkan muatan, tetapi dapat mengalami polarisasi: molekul-molekulnya meregang atau berputar sehingga satu sisi sedikit positif dan sisi lainnya sedikit negatif. Itulah mengapa sisir bermuatan dapat mengangkat serpihan kertas netral. Elektroskop menunjukkan muatan bersih melalui daun-daunnya yang tolak-menolak; pada konduktor terisolasi mana pun, kelebihan muatan berada sepenuhnya di permukaan luar.

Explore · ⁨Jelajahi⁩

Explore charging by rubbing · ⁨Jelajahi pengisian dengan menggosok⁩

Rubbing transfers electrons onto the object, leaving it negative. A charged object then attracts a neutral wall or hair (by polarization) but repels another like-charged object — the same electron transfer behind friction, conduction, and induction. · ⁨Menggosok mentransfer elektron ke objek, membuatnya bermuatan negatif. Objek bermuatan kemudian menarik dinding atau rambut netral (melalui polarisasi) tetapi menolak objek bermuatan sejenis lainnya — transfer elektron yang sama di balik gesekan, konduksi, dan induksi.⁩

8.3

Electric Fields · ⁨Medan Listrik⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 8.3.A: Describe the electric field produced by a charged object or configuration of point charges.

8.3.A.1 Electric fields may originate from charged objects.
8.3.A.2 The electric field at a given point is the ratio of the electric force exerted on a test charge at the point to the charge of the test charge.
- Relevant equation: $\vec{E} = \dfrac{\vec{F}_E}{q}$
- 8.3.A.2.i A test charge is a point charge of small enough magnitude such that its presence does not significantly affect an electric field in its vicinity.
- 8.3.A.2.ii An electric field points away from isolated positive charges and toward isolated negative charges.
- 8.3.A.2.iii The electric force exerted on a positive test charge by an electric field is in the same direction as the electric field.
8.3.A.3 The electric field is a vector quantity and can be represented in space using vector field maps.
- 8.3.A.3.i The net electric field at a given location is the vector sum of individual electric fields created by nearby charged objects.
- 8.3.A.3.ii Electric field maps use vectors to depict the magnitude and direction of the electric field at many locations within a given region.
- 8.3.A.3.iii Electric field line diagrams are simplified models of electric field maps and can be used to determine the relative magnitude and direction of the electric field at any position in the diagram.

Learning Objective 8.3.B: Describe the electric field generated by charged conductors or insulators.

8.3.B.1 While in electrostatic equilibrium, the excess charge of a conductor is distributed on the surface of the conductor, and the electric field within the conductor is zero.
- 8.3.B.1.i At the surface of a charged conductor, the electric field is perpendicular to the surface.
- 8.3.B.1.ii The electric field outside an isolated sphere with spherically symmetric charge distribution is the same as the electric field due to a point charge with the same net charge as the sphere located at the center of the sphere.
8.3.B.2 While in electrostatic equilibrium, the excess charge of an insulator is distributed throughout the interior of the insulator as well as at the surface, and the electric field within the insulator may have a nonzero value.

Bahasa Indonesia

Tujuan Pembelajaran 8.3.A: Jelaskan medan listrik yang dihasilkan oleh objek bermuatan atau konfigurasi muatan titik.

8.3.A.1 Medan listrik dapat berasal dari objek-objek bermuatan.
8.3.A.2 Medan listrik pada suatu titik adalah rasio gaya listrik yang diberikan pada muatan uji di titik tersebut terhadap muatan dari muatan uji itu sendiri.
- Persamaan relevan: $\vec{E} = \dfrac{\vec{F}_E}{q}$
- 8.3.A.2.i Muatan uji adalah muatan titik dengan besaran yang cukup kecil sehingga keberadaannya tidak secara signifikan mempengaruhi medan listrik di sekitarnya.
- 8.3.A.2.ii Medan listrik mengarah menjauhi muatan positif terisolasi dan menuju muatan negatif terisolasi.
- 8.3.A.2.iii Gaya listrik yang diberikan pada muatan uji positif oleh sebuah medan listrik berada dalam arah yang sama dengan medan listrik tersebut.
8.3.A.3 Medan listrik adalah besaran vektor dan dapat direpresentasikan dalam ruang menggunakan peta medan vektor.
- 8.3.A.3.i Medan listrik total pada suatu lokasi tertentu adalah hasil penjumlahan vektor dari medan-medan listrik individual yang diciptakan oleh objek-objek bermuatan di dekatnya.
- 8.3.A.3.ii Peta medan listrik menggunakan vektor untuk menggambarkan besaran dan arah medan listrik di banyak lokasi dalam suatu wilayah tertentu.
- 8.3.A.3.iii Diagram garis medan listrik adalah model sederhana dari peta medan listrik dan dapat digunakan untuk menentukan besaran relatif dan arah medan listrik pada posisi apa pun dalam diagram.

Tujuan Pembelajaran 8.3.B: Jelaskan medan listrik yang dihasilkan oleh konduktor atau isolator bermuatan.

8.3.B.1 Saat berada dalam kesetimbangan elektrostatik, kelebihan muatan pada konduktor terdistribusi pada permukaan konduktor, dan medan listrik di dalam konduktor adalah nol.
- 8.3.B.1.i Di permukaan konduktor bermuatan, medan listrik tegak lurus terhadap permukaan.
- 8.3.B.1.ii Medan listrik di luar bola terisolasi dengan distribusi muatan simetri bola sama dengan medan listrik akibat muatan titik dengan muatan bersih yang sama dengan bola tersebut yang terletak di pusat bola.
8.3.B.2 Saat berada dalam kesetimbangan elektrostatik, kelebihan muatan pada isolator terdistribusi di seluruh bagian dalam isolator serta di permukaannya, dan medan listrik di dalam isolator mungkin memiliki nilai tak nol.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

The electric field of a dipole

The electric field 电场 at a point is the force per unit charge that a small positive test charge 检验电荷 would feel there:

$$\vec{E}=\frac{\vec{F}}{q},\qquad \vec{E}=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\,\hat{r}\ \text{(point charge)}.$$

A charge placed in a field feels $\vec{F}=q\vec{E}$ – positive charges along the field, negative charges against it. Fields from several sources add as vectors (superposition again). Field lines 电场线 make the field visible: they point away from positive charge and toward negative, their density shows the strength, and they never cross.

In a uniform 均匀 field a charge feels a constant force, so it accelerates uniformly – launched sideways, it follows a parabola, just like a projectile in gravity.

Bahasa Indonesia

Medan listrik dari sebuah dipol

Kumpulan Tesla dengan cincin logam di atasnya, memancarkan percikan ungu panjang ke udara — Kumpulan Tesla memancarkan percikan: ia menaikkan tegangan sedemikian rupa sehingga medan listrik mengionisasi udara di sekitarnya, dan muatan melompat sebagai pelepasan terlihat

Medan listrik pada suatu titik adalah gaya per satuan muatan yang akan dirasakan oleh muatan uji positif kecil di sana:

$$\vec{E}=\frac{\vec{F}}{q},\qquad \vec{E}=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\,\hat{r}\ \text{(point charge)}.$$

Muatan yang ditempatkan dalam medan merasakan $\vec{F}=q\vec{E}$ – muatan positif searah medan, muatan negatif berlawanan arahnya. Medan dari beberapa sumber ditambahkan sebagai vektor (superposisi kembali). Garis-garis medan membuat medan terlihat: mereka mengarah menjauhi muatan positif dan menuju muatan negatif, kepadatannya menunjukkan kekuatan, dan mereka tidak pernah bersilangan.

Pola garis medan listrik untuk pelat sejajar, dipol, dan muatan titik

Garis-garis medan listrik dipol mengarah dari muatan positif ke muatan negatif

Dalam medan seragam, muatan merasakan gaya konstan, sehingga dipercepat secara seragam – diluncurkan menyamping, ia mengikuti lintasan parabola, sama seperti proyektil dalam gravitasi.

Muatan yang menyeberangi medan seragam mengikuti lintasan parabola

Seseorang menyentuh generator Van de Graaff dengan rambut berdiri tegak — Muatan menyebar di atas kubah dan ke rambut; muatan sejenis tolak-menolak, sehingga helai-helai saling mendorong

Explore · ⁨Jelajahi⁩

Explore field lines of a point charge · ⁨Jelajahi garis medan muatan titik⁩

Set the charge positive or negative. Field lines start on positive and end on negative charge, never cross, and crowd together where $\vec{E}$ is strongest — the line density is proportional to the field magnitude. · ⁨Atur muatan positif atau negatif. Garis medan dimulai pada positif dan berakhir pada negatif, tidak pernah bersilangan, dan mengkerat di tempat $\vec{E}$ terkuat — kepadatan garis sebanding dengan besaran medan.⁩

Vocabulary · ⁨Kosa kata⁩ Train · ⁨Latih⁩

English	Bahasa Indonesia
Electric charge/ɪˈlektrɪk tʃɑːdʒ/	Muatan listrik
quantized/ˈkwɒntaɪzd/	terkuantisasi
elementary charge/ˌelɪˈmentəri tʃɑːdʒ/	muatan elementer
conservation of charge/ˌkɒnsəˈveɪʃn ɒv tʃɑːdʒ/	kekekalan muatan
point charges/pɔɪnt ˈtʃɑːdʒɪz/	Muatan titik
Coulomb's law/ˈkuːlɒmz lɔː/	Hukum Coulomb
inverse-square/ɪnˈvɜːs skweə/	kuadrat terbalik
superposition/ˌsuːpəpəˈzɪʃn/	superposisi
fundamental property/ˌfʌndəˈmentl ˈprɒpəti/	Sifat fundamental
permittivity of free space/ˌpɜːmɪˈtɪvəti ɒv friː speɪs/	permitivitas ruang hampa
conductor/kənˈdʌktə/	konduktor
insulator/ˈɪnsjuːleɪtə/	isolator
Charging by friction/ˈtʃɑːdʒɪŋ baɪ ˈfrɪkʃn/	Pengisian dengan gesekan
Charging by conduction/ˈtʃɑːdʒɪŋ baɪ kənˈdʌkʃn/	Pengisian dengan konduksi
Charging by induction/ˈtʃɑːdʒɪŋ baɪ ɪnˈdʌkʃn/	Pengisian dengan induksi
ground/ɡraʊnd/	landasan
polarization/ˌpəʊləraɪˈzeɪʃn/	Polarisasi
electroscope/ɪˌlektrəʊˈskəʊp/	Elektroskop
electric field/ɪˈlektrɪk fiːld/	medan listrik
test charge/test tʃɑːdʒ/	muatan uji
Field lines/fiːld laɪnz/	Garis medan
uniform/ˈjuːnɪfɔːm/	seragam
continuous charge distribution/kənˈtɪnjuːəs tʃɑːdʒ ˌdɪstrɪˈbjuːʃn/	Distribusi muatan kontinu
linear charge density/ˈlɪnɪə tʃɑːdʒ ˈdensɪti/	Kepadatan muatan linear
surface charge density/ˈsɜːfɪs tʃɑːdʒ ˈdensɪti/	densitas muatan permukaan
volume charge density/ˈvɒljuːm tʃɑːdʒ ˈdensɪti/	Kepadatan muatan volume
symmetry/ˈsɪmətri/	simetri

8.4

Electric Fields of Charge Distributions · ⁨Medan Listrik dari Distribusi Muatan⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 8.4.A: Describe the electric field resulting from a given charge distribution.

8.4.A.1 Expressions for the electric field of specified charge distributions can be found using integration and the principle of superposition.
- Relevant equation: $\vec{E} = \dfrac{1}{4\pi\varepsilon_0} \displaystyle\int \dfrac{dq}{r^2} \hat{r}$
8.4.A.2 Symmetry considerations of certain charge distributions can simplify analysis of the electric field resulting from those charge distributions.

Boundary statement: AP Physics C: Electricity & Magnetism only expects students to use calculus to find the electric field resulting from the following charge distributions and locations: an infinitely long, uniformly charged wire or cylinder at a distance from its central axis, a thin ring of charge at a location along the axis of the ring, a semicircular arc or part of a semicircular arc at its center, and a finite wire or line charge at a point collinear with the line charge or at a location along its perpendicular bisector.

Bahasa Indonesia

Tujuan Pembelajaran 8.4.A: Jelaskan medan listrik yang timbul dari distribusi muatan tertentu.

8.4.A.1 Ekspresi untuk medan listrik dari distribusi muatan tertentu dapat ditemukan menggunakan integrasi dan prinsip superposisi.
- Persamaan relevan: $\vec{E} = \dfrac{1}{4\pi\varepsilon_0} \displaystyle\int \dfrac{dq}{r^2} \hat{r}$
8.4.A.2 Pertimbangan simetri dari distribusi muatan tertentu dapat menyederhanakan analisis medan listrik yang timbul dari distribusi muatan tersebut.

Pernyataan Batas: AP Physics C: Listrik & Magnetisme hanya mengharapkan siswa menggunakan kalkulus untuk menemukan medan listrik yang timbul dari distribusi muatan dan lokasi berikut: kawat atau silinder bermuatan seragam yang sangat panjang pada jarak dari sumbu tengahnya, cincin tipis bermuatan pada lokasi sepanjang sumbu cincin, busur setengah lingkaran atau sebagian busur setengah lingkaran di pusatnya, dan kawat atau garis muatan terbatas pada titik yang segaris dengan garis muatan atau pada lokasi sepanjang garis bagi tegaknya.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

For a continuous charge distribution 连续电荷分布, cut the object into pieces $dq$ and integrate their point-charge fields:

$$\vec{E}=\frac{1}{4\pi\varepsilon_0}\int \frac{dq}{r^2}\,\hat{r}.$$

Write $dq$ using the right density: $\lambda\,dl$ with the linear charge density 线电荷密度 (a line or ring), $\sigma\,dA$ with the surface charge density 面电荷密度, or $\rho\,dV$ with the volume charge density 体电荷密度. Then use symmetry 对称性 to cancel components before integrating – that sentence is usually a scored step. AP's calculus cases: a finite line (collinear point or perpendicular bisector), an infinite wire or cylinder, a ring on its axis, and an arc at its centre.

Worked example (ring, on axis). A ring of radius $R$ carries charge $Q$. At a point $z$ along the axis, each element is a distance $\sqrt{R^2+z^2}$ away, and by symmetry the sideways components cancel, leaving only the axial part ($\cos\alpha=z/\sqrt{R^2+z^2}$):

$$E_z=\frac{1}{4\pi\varepsilon_0}\int\frac{dq}{R^2+z^2}\cdot\frac{z}{\sqrt{R^2+z^2}}=\frac{1}{4\pi\varepsilon_0}\frac{Qz}{(R^2+z^2)^{3/2}}.$$

Check the limits: $E=0$ at the centre ($z=0$), and far away ($z\gg R$) it becomes $kQ/z^2$ – a point charge, as it must.

Bahasa Indonesia

Untuk distribusi muatan kontinu, potong benda menjadi bagian-bagian $dq$ dan integralkan medan muatan titiknya:

$$\vec{E}=\frac{1}{4\pi\varepsilon_0}\int \frac{dq}{r^2}\,\hat{r}.$$

Tulis $dq$ menggunakan densitas yang tepat: $\lambda\,dl$ dengan densitas muatan linear (garis atau cincin), $\sigma\,dA$ dengan densitas muatan permukaan, atau $\rho\,dV$ dengan densitas muatan volume. Kemudian gunakan simetri untuk membatalkan komponen sebelum mengintegralkan – kalimat ini biasanya merupakan langkah bernilai skor. Kasus kalkulus AP: garis terbatas (titik segaris atau sumbu tegak lurus), kawat atau silinder tak hingga, cincin pada sumbunya, dan busur di pusatnya.

Contoh terpadu (cincin, pada sumbu). Sebuah cincin berjari-jari $R$ membawa muatan $Q$. Pada titik $z$ sepanjang sumbu, setiap elemen berada pada jarak $\sqrt{R^2+z^2}$, dan oleh karena simetri komponen menyampingnya saling meniadakan, hanya menyisakan komponen aksial ($\cos\alpha=z/\sqrt{R^2+z^2}$):

$$E_z=\frac{1}{4\pi\varepsilon_0}\int\frac{dq}{R^2+z^2}\cdot\frac{z}{\sqrt{R^2+z^2}}=\frac{1}{4\pi\varepsilon_0}\frac{Qz}{(R^2+z^2)^{3/2}}.$$

Periksa batasnya: $E=0$ di pusat ($z=0$), dan jauh di kejauhan ($z\gg R$) menjadi $kQ/z^2$ – sebuah muatan titik, sebagaimana mestinya.

8.5

Electric Flux · ⁨Fluks Listrik⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 8.5.A: Describe the electric flux through an arbitrary area or geometric shape.

8.5.A.1 Flux describes the amount of a given quantity that passes through a given area.
8.5.A.2 For an electric field $\vec{E}$ that is constant across an area $\vec{A}$, the electric flux through the area is defined as $\Phi_E = \vec{E} \bullet \vec{A}$.
- 8.5.A.2.i The direction of the area vector is defined as perpendicular to the plane of the surface and outward from a closed surface.
- 8.5.A.2.ii The sign of flux is given by the dot product of the electric field vector and the area vector.
8.5.A.3 The total electric flux passing through a surface is defined by the surface integral of the electric field over the surface.
- Relevant equation: $\Phi_E = \displaystyle\int \vec{E} \cdot d\vec{A}$

Bahasa Indonesia

Tujuan Pembelajaran 8.5.A: Jelaskan fluks listrik melalui area sembarang atau bentuk geometris.

8.5.A.1 Fluks menggambarkan jumlah besaran tertentu yang melewati area tertentu.
8.5.A.2 Untuk medan listrik $\vec{E}$ yang konstan melintasi area $\vec{A}$, fluks listrik melalui area didefinisikan sebagai $\Phi_E = \vec{E} \bullet \vec{A}$.
- 8.5.A.2.i Arah vektor area didefinisikan sebagai tegak lurus terhadap bidang permukaan dan menjauhi permukaan tertutup.
- 8.5.A.2.ii Tanda fluks ditentukan oleh perkalian titik (dot product) dari vektor medan listrik dan vektor area.
8.5.A.3 Total fluks listrik yang melewati suatu permukaan didefinisikan oleh integral permukaan dari medan listrik di atas permukaan tersebut.
- Persamaan relevan: $\Phi_E = \displaystyle\int \vec{E} \cdot d\vec{A}$

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

Electric flux & Gauss's law

Electric flux 电通量 measures how much field passes through a surface:

$$\Phi_E=\int \vec{E}\cdot d\vec{A}.$$

For a uniform field through a flat area, $\Phi_E=EA\cos\theta$, where the area vector 面积矢量 is perpendicular to the surface. Flux is positive where field lines exit a closed surface and negative where they enter – only the perpendicular component of $\vec{E}$ counts.

Bahasa Indonesia

Fluks listrik & Hukum Gauss

Fluks listrik mengukur berapa banyak medan yang menembus suatu permukaan:

$$\Phi_E=\int \vec{E}\cdot d\vec{A}.$$

Untuk medan seragam melalui area datar, $\Phi_E=EA\cos\theta$, di mana vektor luas tegak lurus terhadap permukaan. Fluks positif di mana garis medan keluar dari permukaan tertutup dan negatif di mana mereka masuk – hanya komponen tegak lurus dari $\vec{E}$ yang dihitung.

Fluks listrik melalui permukaan bergantung pada sudut antara medan dan normal permukaan

Vocabulary · ⁨Kosa kata⁩ Train · ⁨Latih⁩

English	Bahasa Indonesia
Electric flux/ɪˈlektrɪk flʌks/	Fluks listrik
area vector/ˈeərɪə ˈvektə/	vektor luas
Gauss's law/ˈɡaʊsɪz lɔː/	Hukum Gauss

8.6

Gauss's Law · ⁨Hukum Gauss⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 8.6.A: Describe the properties of a charge distribution by applying Gauss's law.

8.6.A.1 Gauss's law relates electric flux through a Gaussian surface to the charge enclosed by that surface.
- Relevant equations: $\Phi_E = \dfrac{q_{\text{enc}}}{\varepsilon_0}$ and $\displaystyle\oint \vec{E} \cdot d\vec{A} = \dfrac{q_{\text{enc}}}{\varepsilon_0}$
8.6.A.2 A Gaussian surface is a three-dimensional, closed surface.
8.6.A.3 The total electric flux through a Gaussian surface is independent of the size of the Gaussian surface if the amount of enclosed charge remains constant.
8.6.A.4 Gaussian surfaces are typically constructed such that the electric field generated by the enclosed charge is either perpendicular or parallel to different regions of the Gaussian surface, resulting in a simplified surface integral.
8.6.A.5 If a function of charge density is given for a charge distribution, the total charge can be determined by integrating the charge density over the length (one dimension), area (two dimensions), or volume (three dimensions) of the charge distribution. For example: $Q_{\text{total}} = \displaystyle\int \rho(\vec{r})\, dV$
8.6.A.6 Maxwell's equations are the collection of equations that fully describe electromagnetism. Gauss's law is Maxwell's first equation.
- Relevant equation: $\displaystyle\oint \vec{E} \cdot d\vec{A} = \dfrac{q_{\text{enc}}}{\varepsilon_0}$

Boundary statement: AP Physics C: Electricity & Magnetism only expects students to quantitatively apply Gauss's law to point charges and charge distributions that have spherical, cylindrical, or planar symmetry.

Bahasa Indonesia

Tujuan Pembelajaran 8.6.A: Jelaskan sifat distribusi muatan dengan menerapkan hukum Gauss.

8.6.A.1 Hukum Gauss menghubungkan fluks listrik melalui permukaan Gaussian dengan muatan yang terjebak di dalam permukaan tersebut.
- Persamaan Relevan: $\Phi_E = \dfrac{q_{\text{enc}}}{\varepsilon_0}$ dan $\displaystyle\oint \vec{E} \cdot d\vec{A} = \dfrac{q_{\text{enc}}}{\varepsilon_0}$
8.6.A.2 Permukaan Gaussian adalah permukaan tiga dimensi yang tertutup.
8.6.A.3 Total fluks listrik melalui permukaan Gaussian tidak bergantung pada ukuran permukaan Gaussian jika jumlah muatan yang terjebak tetap konstan.
8.6.A.4 Permukaan Gaussian biasanya dibangun sedemikian rupa sehingga medan listrik yang dihasilkan oleh muatan yang terjebakeither tegak lurus atau sejajar dengan berbagai region permukaan Gaussian, menghasilkan integral permukaan yang disederhanakan.
8.6.A.5 Jika fungsi kerapatan muatan diberikan untuk distribusi muatan, total muatan dapat ditentukan dengan mengintegrasikan kerapatan muatan sepanjang panjang (satu dimensi), luas (dua dimensi), atau volume (tiga dimensi) dari distribusi muatan tersebut. Sebagai contoh: $Q_{\text{total}} = \displaystyle\int \rho(\vec{r})\, dV$
8.6.A.6 Persamaan Maxwell adalah kumpulan persamaan yang secara lengkap mendeskripsikan elektromagnetisme. Hukum Gauss adalah persamaan pertama Maxwell.
- Persamaan relevan: $\displaystyle\oint \vec{E} \cdot d\vec{A} = \dfrac{q_{\text{enc}}}{\varepsilon_0}$

Pernyataan Batasan: AP Physics C: Listrik & Magnetisme hanya mengharapkan siswa untuk menerapkan hukum Gauss secara kuantitatif terhadap muatan titik dan distribusi muatan yang memiliki simetri bola, silinder, atau planar.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

Gauss's law 高斯定律 – the first of Maxwell's equations – relates the flux through any closed surface to the charge inside it:

$$\oint \vec{E}\cdot d\vec{A}=\frac{Q_{\text{enc}}}{\varepsilon_0}.$$

It is true for any closed surface, but it solves for $E$ only when symmetry lets you choose a Gaussian surface 高斯面 on which $E$ is constant and perpendicular (or parallel, contributing zero). The three symmetries AP tests: spherical symmetry 球对称 (concentric sphere), cylindrical symmetry 柱对称 (coaxial cylinder), and planar symmetry 平面对称 (a straddling "pillbox").

Worked example (sphere). Outside a sphere of total charge $Q$, a spherical Gaussian surface of radius $r$ gives $E(4\pi r^2)=Q/\varepsilon_0$, so $E=\dfrac{kQ}{r^2}$ – identical to a point charge at the centre. Inside a uniformly charged solid sphere of radius $R$, the surface encloses only $Q_{\text{enc}}=Q\,r^3/R^3$, so $E=\dfrac{kQr}{R^3}$: zero at the centre, growing linearly to the surface.

Worked example (wire). For an infinite wire with charge per length $\lambda$, take a coaxial cylinder of radius $r$ and length $L$. The ends contribute nothing ($\vec{E}\perp d\vec{A}$), so $E(2\pi rL)=\dfrac{\lambda L}{\varepsilon_0}$ and $E=\dfrac{\lambda}{2\pi\varepsilon_0 r}$. The same pillbox method gives an infinite sheet's field, $E=\dfrac{\sigma}{2\varepsilon_0}$, uniform on both sides.

Exam skill. A full-credit Gauss's-law answer has four parts: name the surface, state the symmetry argument (why $E$ is constant and perpendicular on it), count $Q_{\text{enc}}$, then solve. Skipping the symmetry sentence loses the reasoning point – and remember Gauss's law also explains why $E=0$ inside any conductor in equilibrium.

Bahasa Indonesia

Hukum Gauss – persamaan Maxwell pertama – menghubungkan fluks melalui permukaan tertutup apa pun dengan muatan di dalamnya:

$$\oint \vec{E}\cdot d\vec{A}=\frac{Q_{\text{enc}}}{\varepsilon_0}.$$

Hal ini berlaku untuk setiap permukaan tertutup, tetapi hanya menyelesaikan $E$ ketika simetri memungkinkan Anda memilih permukaan Gaussian di mana $E$ konstan dan tegak lurus (atau sejajar, berkontribusi nol). Tiga jenis simetri yang diuji AP: simetri bola (bola konsentris), simetri silinder (silinder seaksis), dan simetri planar ("kotak pil" yang melintasi bidang).

Permukaan Gaussian dipilih untuk sesuai dengan simetri muatan

Contoh terpecahkan (bola). Di luar bola dengan total muatan $Q$, permukaan Gaussian berbentuk bola dengan jari-jari $r$ menghasilkan $E(4\pi r^2)=Q/\varepsilon_0$, sehingga $E=\dfrac{kQ}{r^2}$ – identik dengan muatan titik di pusat. Di dalam bola pejal bermuatan seragam dengan jari-jari $R$, permukaan tersebut hanya membungkus $Q_{\text{enc}}=Q\,r^3/R^3$, sehingga $E=\dfrac{kQr}{R^3}$: nol di pusat, meningkat secara linear menuju permukaan.

Contoh terpecahkan (kawat). Untuk kawat tak hingga dengan muatan per panjang $\lambda$, ambil silinder seaksis dengan jari-jari $r$ dan panjang $L$. Bagian ujung tidak memberikan kontribusi ($\vec{E}\perp d\vec{A}$), sehingga $E(2\pi rL)=\dfrac{\lambda L}{\varepsilon_0}$ dan $E=\dfrac{\lambda}{2\pi\varepsilon_0 r}$. Metode kotak pil yang sama memberikan medan lembaran tak hingga, $E=\dfrac{\sigma}{2\varepsilon_0}$, seragam di kedua sisinya.

Keterampilan ujian. Jawaban hukum Gauss yang mendapat nilai penuh memiliki empat bagian: sebut nama permukaan, nyatakan argumen simetri (mengapa $E$ konstan dan tegak lurus padanya), hitung $Q_{\text{enc}}$, lalu selesaikan. Melewatkan kalimat simetri akan menghilangkan poin penalaran – dan ingat bahwa hukum Gauss juga menjelaskan mengapa $E=0$ di dalam konduktor apa pun dalam kesetimbangan.

Vocabulary · ⁨Kosa kata⁩ Train · ⁨Latih⁩

English	Bahasa Indonesia
Gaussian surface/ˈɡaʊʃn ˈsɜːfɪs/	permukaan Gauss
spherical symmetry/sˈferɪkl ˈsɪmətri/	Simetri bola
cylindrical symmetry/sɪˈlɪndrɪkl ˈsɪmətri/	Simetri silinder
planar symmetry/ˈpleɪnə ˈsɪmətri/	Simetri planar

8.6

Exam tips · ⁨Tips ujian⁩

English

Use Coulomb's law $F=\tfrac{1}{4\pi\varepsilon_0}\tfrac{q_1 q_2}{r^2}$ and superpose forces as vectors (components, not magnitudes).
For a continuous charge, integrate $d\vec E$ over $dq=\lambda\,dl,\ \sigma\,dA,\ \rho\,dV$; use symmetry to cancel components.
The field points away from positive and toward negative charge; draw field lines correctly.
Distinguish the force on a charge ($\vec F=q\vec E$) from the field the charge creates.
Keep the constant $k=\tfrac{1}{4\pi\varepsilon_0}$ straight and check units.

Bahasa Indonesia

Gunakan hukum Coulomb $F=\tfrac{1}{4\pi\varepsilon_0}\tfrac{q_1 q_2}{r^2}$ dan jumlahkan gaya sebagai vektor (komponen, bukan besaran).
Untuk muatan kontinu, integralkan $d\vec E$ terhadap $dq=\lambda\,dl,\ \sigma\,dA,\ \rho\,dV$; gunakan simetri untuk membatalkan komponen.
Medan mengarah jauh dari muatan positif dan menuju muatan negatif; gambar garis medan dengan benar.
Bedakan gaya pada suatu muatan ($\vec F=q\vec E$) dari medan yang dihasilkan oleh muatan tersebut.
Pertahankan konstanta $k=\tfrac{1}{4\pi\varepsilon_0}$ tetap dan periksa satuan.

Electric Potential · ⁨Potensial Listrik⁩

Watch lesson · ⁨Tonton pelajaran⁩

Open this topic on its own page · ⁨Buka topik ini di halaman sendiri⁩

9.1

Electric Potential Energy · ⁨Energi Potensial Listrik⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 9.1.A: Describe the electric potential energy of a system.

9.1.A.1 The electric potential energy of a system of two point charges equals the amount of work required for an external force to bring the point charges to their current positions from infinitely far away.
9.1.A.2 The general form for the electric potential energy between two charged objects is given by the equation
- Equation: $U_E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r} = k\dfrac{q_1 q_2}{r}$.
9.1.A.3 The total electric potential energy of a system can be determined by finding the sum of the electric potential energies of the individual interactions between each pair of charged objects in the system.

Bahasa Indonesia

Tujuan Pembelajaran 9.1.A: Deskripsikan energi potensial listrik dari suatu sistem.

9.1.A.1 Energi potensial listrik dari sistem dua muatan titik sama dengan jumlah kerja yang diperlukan oleh gaya eksternal untuk mendekatkan muatan titik ke posisi saat ini dari jarak tak hingga.
9.1.A.2 Bentuk umum untuk energi potensial listrik antara dua benda bermuatan diberikan oleh persamaan
- Persamaan: $U_E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r} = k\dfrac{q_1 q_2}{r}$.
9.1.A.3 Total energi potensial listrik dari suatu sistem dapat ditentukan dengan menemukan jumlah dari energi potensial listrik dari interaksi individu antara setiap pasangan benda bermuatan dalam sistem tersebut.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

When you push two like charges together, you do work 功 against the electric force. That work is stored by the pair as electric potential energy 电势能:

$$U_E=\frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r}=k\frac{q_1 q_2}{r}.$$

$U_E$ is defined as the work an external force must do to bring the charges from infinitely far away to a distance $r$ apart. The signs carry the physics:

Like charges: $U_E>0$. An outside push was needed to bring them together. If released, they fly apart, and the stored energy becomes kinetic energy 动能.
Opposite charges: $U_E<0$. The pair is bound. You must supply $|U_E|$ of work to pull them apart to infinity.

The electric force is a conservative force 保守力: the work it does depends only on the start and end points, not the path, and equals $-\Delta U_E$.

For a system 系统 of several charges, add the potential energy of every distinct pair:

$$U_{\text{total}}=k\!\left(\frac{q_1q_2}{r_{12}}+\frac{q_1q_3}{r_{13}}+\frac{q_2q_3}{r_{23}}\right).$$

Worked example. Three $+2.0\ \mu\text{C}$ charges sit at the corners of an equilateral triangle with sides $0.30\ \text{m}$. Each pair stores $U=\dfrac{kq^2}{r}=\dfrac{(9.0\times10^{9})(2.0\times10^{-6})^2}{0.30}=0.12\ \text{J}$. There are three pairs, so $U_{\text{total}}=3\times0.12\ \text{J}=0.36\ \text{J}$. This is the work needed to assemble the triangle from infinity – and the kinetic energy the charges would share if all three were released.

Bahasa Indonesia

Ketika Anda mendorong dua muatan sejenis bersama, Anda melakukan usaha melawan gaya listrik. Usaha itu disimpan oleh pasangan tersebut sebagai energi potensial listrik:

$$U_E=\frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r}=k\frac{q_1 q_2}{r}.$$

$U_E$ didefinisikan sebagai usaha yang harus dilakukan gaya eksternal untuk membawa muatan-muatan tersebut dari jarak tak hingga hingga berjarak $r$. Tanda-tandanya membawa makna fisika:

Muatan sejenis: $U_E>0$. Dorongan dari luar diperlukan untuk mendekatkan mereka. Jika dilepaskan, mereka terbang menjauh, dan energi yang tersimpan berubah menjadi energi kinetik.
Muatan berlawanan: $U_E<0$. Pasangan tersebut terikat. Anda harus menyediakan $|U_E|$ usaha untuk memisahkan mereka hingga tak hingga.

Gaya listrik adalah gaya konservatif: usaha yang dilakukannya hanya bergantung pada titik awal dan akhir, bukan lintasan, dan sama dengan $-\Delta U_E$.

Untuk sistem beberapa muatan, tambahkan energi potensial dari setiap pasangan unik:

$$U_{\text{total}}=k\!\left(\frac{q_1q_2}{r_{12}}+\frac{q_1q_3}{r_{13}}+\frac{q_2q_3}{r_{23}}\right).$$

Contoh terpecahkan. Tiga muatan $+2.0\ \mu\text{C}$ berada di sudut-sudut segitiga sama sisi dengan sisi $0.30\ \text{m}$. Setiap pasangan menyimpan $U=\dfrac{kq^2}{r}=\dfrac{(9.0\times10^{9})(2.0\times10^{-6})^2}{0.30}=0.12\ \text{J}$. Ada tiga pasangan, sehingga $U_{\text{total}}=3\times0.12\ \text{J}=0.36\ \text{J}$. Ini adalah usaha yang dibutuhkan untuk merakit segitiga dari tak hingga – dan energi kinetik yang akan dibagi oleh muatan-muatan tersebut jika ketiganya dilepaskan.

9.2

Electric Potential · ⁨Potensial Listrik⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 9.2.A: Describe the electric potential due to a configuration of charged objects.

9.2.A.1 Electric potential describes the electric potential energy per unit charge at a point in space.
9.2.A.2 Expressions for the electric potential of charge distributions can be found using integration and the principle of superposition.
- Equation: $V = \dfrac{1}{4\pi\varepsilon_0}\displaystyle\int \dfrac{dq}{r}$
- 9.2.A.2.i The electric potential for single point charge is
  - Equation: $V = \dfrac{q}{4\pi\varepsilon_0 r}$.
- 9.2.A.2.ii The electric potential due to multiple point charges can be determined by the principle of scalar superposition of the electric potential due to each of the point charges.
  - Equation: $V = \dfrac{1}{4\pi\varepsilon_0}\displaystyle\sum_i \dfrac{q_i}{r_i}$
9.2.A.3 The electric potential difference between two points is the change in electric potential energy per unit charge when a test charge is moved between the two points.
- Equation: $\Delta V = \dfrac{\Delta U_E}{q}$
9.2.A.4 Electric potential difference may also result from chemical processes that cause positive and negative charges to separate, such as in a battery.

Boundary statement: AP Physics C: Electricity & Magnetism only expects students to use calculus to find the electric potential resulting from the following charge distributions and locations: an infinitely long, uniformly charged wire or cylinder at a distance from its central axis, a thin ring of charge at a location along the axis of the ring, a semicircular arc or part of a semicircular arc at its center, and a finite wire or line charge at a point collinear with the line charge or at a location along its perpendicular bisector.

Learning Objective 9.2.B: Describe the relationship between electric potential and electric field.

9.2.B.1 The value of an electric field component in any direction at a given location is equal to the negative of the spatial rate of change in electric potential at that location.
- Equation: $E_x = -\dfrac{dV}{dx}$
9.2.B.2 The change in electric potential between two points can be determined by integrating the dot product of the electric field and the displacement along the path connecting the points.
- Equation: $\Delta V = V_b - V_a = -\displaystyle\int_a^b \vec{E}\cdot d\vec{r}$
9.2.B.3 Electric field vector maps and equipotential lines are tools to describe the field produced by a charge or configuration of charges and can be used to predict the motion of charged objects in the field.
- 9.2.B.3.i Equipotential lines represent lines of equal electric potential. These lines are also referred to as isolines of electric potential.
- 9.2.B.3.ii Isolines are perpendicular to electric field vectors. An isoline map of electric potential can be constructed from an electric field vector map, and an electric field map may be constructed from an isoline map.
- 9.2.B.3.iii An electric field vector points in the direction of decreasing potential.
- 9.2.B.3.iv There is no component of an electric field along an isoline.

Bahasa Indonesia

Tujuan Pembelajaran 9.2.A: Deskripsikan potensial listrik akibat konfigurasi benda-benda bermuatan.

9.2.A.1 Potensial listrik menggambarkan energi potensial listrik per satuan muatan pada suatu titik di ruang.
9.2.A.2 Ekspresi untuk potensial listrik dari distribusi muatan dapat ditemukan menggunakan integrasi dan prinsip superposisi.
- Persamaan: $V = \dfrac{1}{4\pi\varepsilon_0}\displaystyle\int \dfrac{dq}{r}$
- 9.2.A.2.i Potensial listrik untuk muatan titik tunggal adalah
  - Persamaan: $V = \dfrac{q}{4\pi\varepsilon_0 r}$.
- 9.2.A.2.ii Potensial listrik akibat beberapa muatan titik dapat ditentukan melalui prinsip superposisi skalar dari potensial listrik masing-masing muatan titik.
  - Persamaan: $V = \dfrac{1}{4\pi\varepsilon_0}\displaystyle\sum_i \dfrac{q_i}{r_i}$
9.2.A.3 Selisih potensial listrik antara dua titik adalah perubahan energi potensial listrik per satuan muatan ketika muatan uji dipindahkan antara kedua titik tersebut.
- Persamaan: $\Delta V = \dfrac{\Delta U_E}{q}$
9.2.A.4 Selisih potensial listrik juga dapat dihasilkan dari proses kimia yang menyebabkan pemisahan muatan positif dan negatif, seperti dalam baterai.

Pernyataan Batasan: AP Physics C: Listrik & Magnetisme hanya mengharapkan siswa menggunakan kalkulus untuk menemukan potensial listrik yang dihasilkan dari distribusi dan lokasi muatan berikut: kawat atau silinder bermuatan seragam yang sangat panjang pada jarak dari sumbu tengahnya, cincin tipis bermuatan pada lokasi sepanjang sumbu cincin, busur setengah lingkaran atau bagian dari busur setengah lingkaran di pusatnya, dan kawat terbatas atau garis muatan pada titik yang segaris dengan garis muatan atau pada lokasi sepanjang garis bagi tegak lurusnya.

Tujuan Pembelajaran 9.2.B: Deskripsikan hubungan antara potensial listrik dan medan listrik.

9.2.B.1 Nilai komponen medan listrik dalam arah apa pun pada lokasi tertentu sama dengan negatif dari laju perubahan spasial potensial listrik di lokasi tersebut.
- Persamaan: $E_x = -\dfrac{dV}{dx}$
9.2.B.2 Perubahan potensial listrik antara dua titik dapat ditentukan dengan mengintegrasikan hasil kali titik dari medan listrik dan perpindahan sepanjang lintasan yang menghubungkan titik-titik tersebut.
- Persamaan: $\Delta V = V_b - V_a = -\displaystyle\int_a^b \vec{E}\cdot d\vec{r}$
9.2.B.3 Peta vektor medan listrik dan garis ekuipotensial adalah alat untuk mendeskripsikan medan yang dihasilkan oleh muatan atau konfigurasi muatan dan dapat digunakan untuk memprediksi gerakan benda bermuatan dalam medan tersebut.
- 9.2.B.3.i Garis ekuipotensial mewakili garis dengan potensial listrik yang sama. Garis-garis ini juga disebut isolines potensial listrik.
- 9.2.B.3.ii Isolines tegak lurus terhadap vektor medan listrik. Peta isolines potensial listrik dapat dibangun dari peta vektor medan listrik, dan peta medan listrik dapat dibangun dari peta isolines.
- 9.2.B.3.iii Vektor medan listrik mengarah ke arah penurunan potensial.
- 9.2.B.3.iv Tidak ada komponen medan listrik sepanjang isoline.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

Electric potential 电势 is electric potential energy per unit charge at a point in space. It is measured in volts ($1\ \text{V}=1\ \text{J/C}$):

$$V=\frac{U_E}{q},\qquad V=\frac{1}{4\pi\varepsilon_0}\frac{q}{r}\ \text{(point charge)}.$$

Potential is a scalar 标量: it has a sign but no direction. This makes $V$ far easier to work with than the field $\vec{E}$. For several point charges 点电荷, use scalar superposition 叠加 – add the potentials with their signs:

$$V=\frac{1}{4\pi\varepsilon_0}\sum_i \frac{q_i}{r_i}.$$

Worked example. The potential $0.10\ \text{m}$ from a $+3.0\ \text{nC}$ point charge is $V=\dfrac{kq}{r}=\dfrac{9.0\times10^{9}(3.0\times10^{-9})}{0.10}=270\ \text{V}$. Now add a $-3.0\ \text{nC}$ charge $0.30\ \text{m}$ from the same point: it contributes $\dfrac{9.0\times10^{9}(-3.0\times10^{-9})}{0.30}=-90\ \text{V}$, so the total is $270-90=180\ \text{V}$. Just a signed sum – no vector components to resolve.

Continuous charge distributions

For a continuous charge distribution 连续电荷分布, cut it into small pieces $dq$, treat each piece as a point charge, and integrate 积分:

$$V=\frac{1}{4\pi\varepsilon_0}\int\frac{dq}{r}.$$

This is a scalar integral, so it is usually much easier than the field integral. AP expects you to handle four shapes with calculus: a thin ring (at a point on its axis), an arc (at its centre), a finite line or wire (collinear, or on its perpendicular bisector), and an infinitely long wire or cylinder (at a distance from its axis).

Worked example (ring of charge). A thin ring of radius $R$ carries total charge $Q$. For a point $P$ on the axis a distance $z$ from the centre, every element $dq$ sits at the same distance $r=\sqrt{R^2+z^2}$ from $P$. The distance is constant, so it comes out of the integral:

$$V=\frac{1}{4\pi\varepsilon_0}\int\frac{dq}{\sqrt{R^2+z^2}}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{\sqrt{R^2+z^2}}.$$

On an FRQ, say that $r$ is the same for every element – that statement is the marked step. The same idea gives $V=kQ/R$ at the centre of any arc, whatever its angle.

From potential to field

Potential and field contain the same information, linked by a derivative in each direction and by a path integral:

$$E_x=-\frac{dV}{dx},\qquad \Delta V=V_b-V_a=-\int_a^b \vec{E}\cdot d\vec{r}.$$

The field is the negative gradient 梯度 of the potential: $\vec{E}$ points from high potential to low potential, "downhill". Where $V$ changes quickly, the field is strong.

Worked example. Along the $x$-axis, $V(x)=3x^{2}-2x$ (volts, with $x$ in metres). Then $E_x=-\dfrac{dV}{dx}=2-6x\ \text{V/m}$. At $x=0.50\ \text{m}$, $E_x=-1.0\ \text{V/m}$: the field there points in the $-x$ direction.

The potential difference 电势差 between two points is the change in potential energy per unit charge moved between them: $\Delta V=\Delta U_E/q$. A battery makes a potential difference chemically: reactions inside separate positive from negative charge and hold the terminals a fixed $\Delta V$ apart.

Equipotential maps

An equipotential 等势面 line (also called an isoline 等值线) joins points that share the same potential. Four rules let you read any map:

Isolines are perpendicular 垂直 to field lines 电场线 everywhere they cross.
$\vec{E}$ points from high $V$ to low $V$ – never along an isoline. Moving a charge along an isoline takes no work.
Closely spaced isolines mean a strong field: $E\approx-\Delta V/\Delta x$.
You can sketch the field map from an isoline map, and the isoline map from a field map.

Exam skill. Given a map with isolines every $10\ \text{V}$ spaced about $2\ \text{cm}$ apart, estimate $E\approx\dfrac{10}{0.02}=500\ \text{V/m}$, pointing from the higher-value line to the lower one. The work an external agent does moving a charge $q$ slowly from $A$ to $B$ is $W=q(V_B-V_A)$ – the path taken does not matter.

Bahasa Indonesia

Potensial listrik adalah energi potensial listrik per satuan muatan pada suatu titik di ruang. Potensial ini diukur dalam volt ($1\ \text{V}=1\ \text{J/C}$):

$$V=\frac{U_E}{q},\qquad V=\frac{1}{4\pi\varepsilon_0}\frac{q}{r}\ \text{(point charge)}.$$

Potensial adalah skalar: ia memiliki tanda tetapi tidak memiliki arah. Hal ini membuat $V$ jauh lebih mudah dihitung daripada medan $\vec{E}$. Untuk beberapa muatan titik, gunakan superposisi skalar – tambahkan potensial beserta tandanya:

$$V=\frac{1}{4\pi\varepsilon_0}\sum_i \frac{q_i}{r_i}.$$

Contoh terpecahkan. Potensial $0.10\ \text{m}$ dari muatan titik $+3.0\ \text{nC}$ adalah $V=\dfrac{kq}{r}=\dfrac{9.0\times10^{9}(3.0\times10^{-9})}{0.10}=270\ \text{V}$. Sekarang tambahkan muatan $-3.0\ \text{nC}$ yang berjarak $0.30\ \text{m}$ dari titik yang sama: hal ini berkontribusi $\dfrac{9.0\times10^{9}(-3.0\times10^{-9})}{0.30}=-90\ \text{V}$, sehingga totalnya adalah $270-90=180\ \text{V}$. Hanya penjumlahan bertanda – tidak ada komponen vektor yang perlu diuraikan.

Distribusi muatan kontinu

Untuk distribusi muatan kontinu, potong menjadi bagian-bagian kecil $dq$, anggap setiap bagian sebagai muatan titik, dan integralkan:

$$V=\frac{1}{4\pi\varepsilon_0}\int\frac{dq}{r}.$$

Ini adalah integral skalar, jadi biasanya jauh lebih mudah daripada integral medan. AP mengharapkan Anda menangani empat bentuk dengan kalkulus: cincin tipis (pada titik di sumbunya), busur (di pusatnya), garis atau kawat terbatas (segaris, atau pada garis bagi tegak lurusnya), dan kawat atau silinder tak hingga (pada jarak dari sumbunya).

Cincin bermuatan: setiap elemen dq berada pada jarak yang sama dari titik sumbu

Contoh terpecahkan (cincin bermuatan). Cincin tipis berjari-jari $R$ membawa total muatan $Q$. Untuk titik $P$ pada sumbu pada jarak $z$ dari pusat, setiap elemen $dq$ berada pada jarak yang sama $r=\sqrt{R^2+z^2}$ dari $P$. Jaraknya konstan, sehingga dapat dikeluarkan dari integral:

$$V=\frac{1}{4\pi\varepsilon_0}\int\frac{dq}{\sqrt{R^2+z^2}}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{\sqrt{R^2+z^2}}.$$

Pada soal FRQ, nyatakan bahwa $r$ sama untuk setiap elemen – pernyataan ini adalah langkah bernilai. Gagasan yang sama menghasilkan $V=kQ/R$ di pusat busur apa pun, terlepas dari sudutnya.

Dari potensial ke medan

Potensial dan medan mengandung informasi yang sama, terhubung melalui turunan di setiap arah dan melalui integral lintasan:

$$E_x=-\frac{dV}{dx},\qquad \Delta V=V_b-V_a=-\int_a^b \vec{E}\cdot d\vec{r}.$$

Medan adalah gradien negatif dari potensial: $\vec{E}$ mengarah dari potensial tinggi ke rendah, "turunan". Di mana $V$ berubah cepat, medannya kuat.

Contoh terpecahkan. Sepanjang sumbu-$x$, $V(x)=3x^{2}-2x$ (volt, dengan $x$ dalam meter). Maka $E_x=-\dfrac{dV}{dx}=2-6x\ \text{V/m}$. Pada $x=0.50\ \text{m}$, $E_x=-1.0\ \text{V/m}$: medan di sana mengarah ke arah $-x$.

Dalam medan seragam, potensial turun secara teratur seiring jarak

Selisih potensial antara dua titik adalah perubahan energi potensial per satuan muatan yang dipindahkan di antaranya: $\Delta V=\Delta U_E/q$. Baterai menciptakan selisih potensial secara kimia: reaksi di dalamnya memisahkan muatan positif dari negatif dan menahan terminal-terminalnya berjarak tetap $\Delta V$.

Potensial dekat muatan titik berubah sebanding dengan 1/r — Potensial dekat muatan titik bervariasi sebagai 1/r

Peta ekuipotensial

Garis ekuipotensial (juga disebut isoline) menghubungkan titik-titik yang memiliki potensial yang sama. Empat aturan memungkinkan Anda membaca peta apa pun:

Isoline tegak lurus terhadap garis medan di setiap tempat mereka berpotongan.
$\vec{E}$ mengarah dari $V$ tinggi ke $V$ rendah – tidak pernah sepanjang garis ekuipotensial. Memindahkan muatan sepanjang garis ekuipotensial tidak memerlukan usaha.
Isoline yang rapat menandakan medan yang kuat: $E\approx-\Delta V/\Delta x$.
Anda dapat menggambar peta medan dari peta isoline, dan peta isoline dari peta medan.

Garis ekuipotensial di sekitar dipol berpotongan dengan garis medan secara tegak lurus

Keterampilan ujian. Diberikan peta dengan isoline setiap $10\ \text{V}$ dengan jarak sekitar $2\ \text{cm}$, perkirakan $E\approx\dfrac{10}{0.02}=500\ \text{V/m}$, yang menunjuk dari garis bernilai lebih tinggi ke yang lebih rendah. Usaha yang dilakukan agen eksternal memindahkan muatan $q$ perlahan dari $A$ ke $B$ adalah $W=q(V_B-V_A)$ – lintasan yang diambil tidak penting.

Saluran listrik bertegangan tinggi: energi potensial listrik dikonversi dan ditransmisikan sebagai arus pada tegangan tinggi

Explore · ⁨Jelajahi⁩

Field lines and the potential around a charge · ⁨Garis medan dan potensial di sekitar muatan⁩

Electric potential is the energy per unit charge. It falls off with distance from a positive charge; the field lines point from high potential to low. · ⁨Potensial listrik adalah energi per satuan muatan. Menurun seiring jarak dari muatan positif; garis medan menunjukkan dari potensial tinggi ke rendah.⁩

Vocabulary · ⁨Kosa kata⁩ Train · ⁨Latih⁩

English	Bahasa Indonesia
work/wɜːk/	usaha
electric potential energy/ɪˈlektrɪk pəˈtenʃl ˈenədʒi/	energi potensial listrik
kinetic energy/kɪˈnetɪk ˈenədʒi/	energi kinetik
conservative force/kənˈsɜːvətɪv fɔːs/	gaya konservatif
system/ˈsɪstəm/	sistem
Electric potential/ɪˈlektrɪk pəˈtenʃl/	Potensi elektrik
scalar/ˈskeɪlə/	skalar
point charges/pɔɪnt ˈtʃɑːdʒɪz/	Muatan titik
superposition/ˌsuːpəpəˈzɪʃn/	superposisi
continuous charge distribution/kənˈtɪnjuːəs tʃɑːdʒ ˌdɪstrɪˈbjuːʃn/	Distribusi muatan kontinu
integrate/ˈɪntɪɡreɪt/	integralkan
gradient/ˈɡreɪdɪənt/	gradien
potential difference/pəˈtenʃl ˈdɪfrəns/	beda potensial
equipotential/ˌiːkwɪpəˈtenʃl/	equipotensial
isoline/ˈaɪsəliːn/	Isoline
perpendicular/ˌpɜːpənˈdɪkjʊlə/	tegak lurus
field lines/fiːld laɪnz/	garis medan

9.3

Conservation of Electric Energy · ⁨Kekekalan Energi Listrik⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 9.3.A: Describe changes in a system due to a difference in electric potential between two locations.

9.3.A.1 When a charged object moves between two locations with different electric potentials, the resulting change in the electric potential energy of the object-field system is given by the following equation.
- Equation: $\Delta U_E = q\Delta V$
9.3.A.2 The movement of a charged object between two points with different electric potentials results in a change in kinetic energy of the object consistent with the conservation of energy.

Bahasa Indonesia

Tujuan Pembelajaran 9.3.A: Deskripsikan perubahan dalam sistem akibat perbedaan potensial listrik antara dua lokasi.

9.3.A.1 Ketika benda bermuatan bergerak antara dua lokasi dengan potensial listrik yang berbeda, perubahan energi potensial listrik dari sistem benda-medan yang timbul diberikan oleh persamaan berikut.
- Persamaan: $\Delta U_E = q\Delta V$
9.3.A.2 Pergerakan benda bermuatan antara dua titik dengan potensial listrik yang berbeda menghasilkan perubahan energi kinetik dari benda tersebut yang sesuai dengan kekekalan energi.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

When a charge $q$ moves between two points that differ in potential by $\Delta V$, the potential energy of the charge–field system changes by

$$\Delta U_E=q\,\Delta V.$$

If only the electric force acts, total energy is conserved, so the kinetic energy changes by the opposite amount:

$$\Delta K=-\Delta U_E=-q\,\Delta V.$$

Watch the two signs together: a positive charge speeds up when it moves to lower potential, while a negative charge speeds up when it moves to higher potential. Both are just the system trading potential energy for kinetic energy. This is how particle accelerators give charged particles their energy, and it underlies energy analysis in circuits.

Worked example. A proton ($q=1.6\times10^{-19}\ \text{C}$, $m=1.67\times10^{-27}\ \text{kg}$) starts at rest and is accelerated through a potential drop of $500\ \text{V}$ (it moves to lower potential, so $\Delta V=-500\ \text{V}$ and $\Delta K=-q\,\Delta V=+8.0\times10^{-17}\ \text{J}$). Setting $\Delta K=\tfrac12 mv^2$:

$$v=\sqrt{\frac{2(8.0\times10^{-17})}{1.67\times10^{-27}}}=3.1\times10^{5}\ \text{m/s}.$$

Exam skill. Choose energy methods, not kinematics, whenever the field is non-uniform or the path curves: only the end-point potentials matter. A typical FRQ chain is $q\,\Delta V \to \Delta K \to v$, with one line of justification: "the electric force is conservative, so energy is conserved."

Bahasa Indonesia

Ketika sebuah muatan $q$ bergerak antara dua titik yang berbeda dalam potensial sebesar $\Delta V$, energi potensial sistem muatan-medan berubah sebesar

$$\Delta U_E=q\,\Delta V.$$

Jika hanya gaya listrik yang bekerja, total energi kekal, sehingga energi kinetik berubah sebesar jumlah yang berlawanan:

$$\Delta K=-\Delta U_E=-q\,\Delta V.$$

Perhatikan kedua tanda bersama-sama: muatan positif bergerak semakin cepat ketika bergerak ke potensial yang lebih rendah, sementara muatan negatif bergerak semakin cepat ketika bergerak ke potensial yang lebih tinggi. Keduanya hanyalah sistem yang menukar energi potensial menjadi energi kinetik. Inilah bagaimana akselerator partikel memberikan energi pada partikel bermuatan, dan ini mendasari analisis energi dalam rangkaian.

Contoh terpadu. Sebuah proton ($q=1.6\times10^{-19}\ \text{C}$, $m=1.67\times10^{-27}\ \text{kg}$) mulai dari keadaan diam dan dipercepat melalui penurunan potensial $500\ \text{V}$ (berpindah ke potensial yang lebih rendah, sehingga $\Delta V=-500\ \text{V}$ dan $\Delta K=-q\,\Delta V=+8.0\times10^{-17}\ \text{J}$). Menetapkan $\Delta K=\tfrac12 mv^2$:

$$v=\sqrt{\frac{2(8.0\times10^{-17})}{1.67\times10^{-27}}}=3.1\times10^{5}\ \text{m/s}.$$

Keterampilan ujian. Pilih metode energi, bukan kinematika, setiap kali medan tidak seragam atau lintasan melengkung: hanya potensial titik akhir yang relevan. Rantai FRQ典型 adalah $q\,\Delta V \to \Delta K \to v$, dengan satu baris justifikasi: "gaya listrik bersifat konservatif, sehingga energi kekal."

9.3

Exam tips · ⁨Tips ujian⁩

English

Relate field and potential by $V=-\int \vec E\cdot d\vec l$ and $\vec E=-\nabla V$ (in 1-D, $E_x=-\tfrac{dV}{dx}$).
Potential is a scalar — add contributions with signs, no vector components needed.
Potential energy of a pair is $U=\tfrac{1}{4\pi\varepsilon_0}\tfrac{q_1 q_2}{r}$; use energy conservation for a charge's speed.
Know that no work is done moving along an equipotential, which is perpendicular to $\vec E$.
Choose the zero of potential (usually infinity) and state it.

Bahasa Indonesia

Hubungkan medan dan potensial melalui $V=-\int \vec E\cdot d\vec l$ dan $\vec E=-\nabla V$ (dalam dimensi-1, $E_x=-\tfrac{dV}{dx}$).
Potensial adalah skalar – tambahkan kontribusinya dengan tanda, tanpa perlu komponen vektor.
Energi potensial pasangan adalah $U=\tfrac{1}{4\pi\varepsilon_0}\tfrac{q_1 q_2}{r}$; gunakan kekekalan energi untuk kecepatan muatan.
Tahu bahwa tidak ada usaha yang dilakukan saat bergerak sepanjang ekuipotensial, yang tegak lurus terhadap $\vec E$.
Pilih nol potensial (biasanya tak hingga) dan nyatakan itu.

Conductors and Capacitors · ⁨Konduktor dan Kapasitor⁩

Watch lesson · ⁨Tonton pelajaran⁩

Open this topic on its own page · ⁨Buka topik ini di halaman sendiri⁩

10.1

Electrostatics with Conductors · ⁨Elektrostatis dengan Konduktor⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 10.1.A: Describe the charge distribution within a conductor.

10.1.A.1 An ideal conductor is a material in which electrons are able to move freely.
10.1.A.2 When a conductor is in electrostatic equilibrium, mutual repulsion of excess charge carriers results in those charge carriers residing entirely on the surface of the conductor.
- 10.1.A.2.i In a conductor with a negative net charge, excess electrons reside on the surface of the conductor.
- 10.1.A.2.ii In a conductor with a positive net charge, the surface becomes deficient in electrons, and can be modeled as if positive charge carriers reside on the surface of the conductor.
10.1.A.3 Excess charges will move to the surface of a conductor to create a state of electrostatic equilibrium within the conductor.
- 10.1.A.3.i The time interval over which charges reach electrostatic equilibrium within a conductor is so short as to be negligible.
- 10.1.A.3.ii When a conductor reaches electrostatic equilibrium, all points on the surface of the conductor have the same electric potential, and the conductor becomes an equipotential surface.
- 10.1.A.3.iii The charge density on the surface of a conductor will be greater where there are points or edges compared to planar areas.
10.1.A.4 All excess charges reside on the surface of a conductor, which means there is no net charge in the interior of the conductor, and the electric field is zero within the conductor.
10.1.A.5 The electric field is perpendicular to the outer surface of a conductor.
10.1.A.6 A conductor can be polarized in the presence of an external electric field. This is a consequence of the conductor remaining an equipotential surface.
10.1.A.7 Electrostatic shielding is the process of surrounding an area with a closed, conducting shell to create a region inside the conductor that is free from external electric fields.

Bahasa Indonesia

Tujuan Pembelajaran 10.1.A: Jelaskan distribusi muatan di dalam konduktor.

10.1.A.1 Konduktor ideal adalah bahan di mana elektron dapat bergerak bebas.
10.1.A.2 Ketika konduktor berada dalam kesetimbangan elektrostatis, tolak-menolak timbal balik pembawa muatan berlebih menyebabkan pembawa muatan tersebut menempati seluruh permukaan konduktor.
- 10.1.A.2.i Pada konduktor dengan muatan bersih negatif, elektron berlebih menempati permukaan konduktor.
- 10.1.A.2.ii Pada konduktor dengan muatan bersih positif, permukaannya kekurangan elektron, dan dapat dimodelkan seolah-olah pembawa muatan positif menempati permukaan konduktor.
10.1.A.3 Muatan berlebih akan bergerak ke permukaan konduktor untuk menciptakan keadaan kesetimbangan elektrostatis di dalam konduktor.
- 10.1.A.3.i Selang waktu di mana muatan mencapai kesetimbangan elektrostatik dalam sebuah konduktor sangat singkat sehingga dapat diabaikan.
- 10.1.A.3.ii Ketika sebuah konduktor mencapai kesetimbangan elektrostatik, semua titik pada permukaan konduktor memiliki potensial listrik yang sama, dan konduktor tersebut menjadi permukaan ekuipotensial.
- 10.1.A.3.iii Kepadatan muatan pada permukaan konduktor akan lebih besar di daerah yang memiliki titik atau tepi dibandingkan dengan area datar.
10.1.A.4 Semua muatan berlebih berada pada permukaan konduktor, yang berarti tidak ada muatan bersih di bagian dalam konduktor, dan medan listriknya nol di dalam konduktor.
10.1.A.5 Medan listrik tegak lurus terhadap permukaan luar konduktor.
10.1.A.6 Konduktor dapat terpolarisasi di hadapan medan listrik eksternal. Ini merupakan akibat dari konduktor yang tetap mempertahankan status sebagai permukaan ekuipotensial.
10.1.A.7 Peredaman elektrostatik adalah proses mengelilingi suatu area dengan kulit konduktor tertutup untuk menciptakan wilayah di dalam konduktor yang bebas dari medan listrik eksternal.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

In an ideal conductor 导体, electrons move freely. Put extra charge on one and the charges repel each other until, almost instantly, they settle into electrostatic equilibrium 静电平衡. In that state the conductor has four properties you must be able to state and use:

The electric field inside the conductor is zero. If it were not, free electrons would still be moving.
All excess charge sits on the surface. (Negative net charge = extra electrons on the surface; positive = a shortage of electrons there.)
The field just outside is perpendicular 垂直 to the surface, with magnitude $E=\sigma/\varepsilon_0$. Any parallel component would push charge sideways along the surface.
The whole conductor is one equipotential surface 等势面: every point, inside and on the surface, is at the same potential.

Charge density is largest where the surface curves sharply – at points and edges – so the outside field is strongest there. In an external field a conductor polarizes: charge shifts on its surface so that the interior stays field-free and the body stays an equipotential. Surrounding a region with a closed conducting shell keeps outside fields out entirely – electrostatic shielding 静电屏蔽, the idea behind the Faraday cage 法拉第笼.

Bahasa Indonesia

Dalam konduktor ideal, elektron bergerak bebas. Letakkan muatan berlebih di salah satu sisi dan muatan-muatan tersebut saling tolak hingga, hampir seketika, mereka menetap dalam keseimbangan elektrostatis. Dalam keadaan ini konduktor memiliki empat sifat yang harus Anda bisa nyatakan dan gunakan:

Medan listrik di dalam konduktor adalah nol. Jika tidak, elektron bebas masih akan bergerak.
Semua muatan berlebih berada di permukaan. (Muatan bersih negatif = elektron berlebih di permukaan; positif = kekurangan elektron di sana.)
Medan tepat di luar adalah tegak lurus terhadap permukaan, dengan besarannya $E=\sigma/\varepsilon_0$. Komponen sejajar apa pun akan mendorong muatan menyamping sepanjang permukaan.
Seluruh konduktor adalah satu permukaan ekuipotensial: setiap titik, di dalam dan di permukaan, berada pada potensial yang sama.

Kepadatan muatan terbesar di mana permukaan melengkung tajam – di titik dan tepi – sehingga medan di luarnya paling kuat di sana. Dalam medan eksternal, konduktor terpolarisasi: muatan bergeser di permukaannya sehingga bagian dalamnya tetap bebas medan dan badannya tetap ekuipotensial. Mengelilingi suatu wilayah dengan cangkang konduktor tertutup sepenuhnya menghalangi medan di luar – pelindung elektrostatis, ide di balik kandang Faraday.

10.2

Redistribution of Charge Between Conductors · ⁨Redistribusi Muatan Antara Konduktor⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 10.2.A: Describe the movement of charge and the resulting interactions when conductors physically contact each other.

10.2.A.1 When conductors are in electrical contact, charges will be redistributed such that the surfaces of each conductor are at the same electric potential.
10.2.A.2 Ground is an idealized reference point that has zero electric potential and can absorb or provide an infinite amount of charge without changing its electric potential.
10.2.A.3 Charge can be induced on a conductor by grounding the conductor in the presence of an external electric field.

Bahasa Indonesia

Tujuan Pembelajaran 10.2.A: Jelaskan pergerakan muatan dan interaksi yang terjadi ketika konduktor bersentuhan secara fisik.

10.2.A.1 Ketika konduktor terhubung secara listrik, muatan akan terdistribusi ulang sedemikian rupa sehingga permukaan setiap konduktor berada pada potensial listrik yang sama.
10.2.A.2 Tanah (ground) adalah titik referensi ideal yang memiliki nol potensial listrik dan dapat menyerap atau menyediakan jumlah muatan tak terhingga tanpa mengubah potensial listriknya.
10.2.A.3 Muatan dapat diinduksi pada konduktor dengan cara menghubungkan konduktor tersebut ke tanah (grounding) di hadapan medan listrik eksternal.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

When two conductors touch (or are wired together), charge flows between them until both surfaces reach the same potential – that is the stopping condition, not "equal charges". A larger sphere holds more charge at the same potential ($V=kQ/R$), so it takes the larger share.

Worked example. A small sphere of radius $R$ carrying $+6.0\ \mu\text{C}$ touches a distant sphere of radius $2R$, then they separate. Equal potentials require $\dfrac{kq_1}{R}=\dfrac{kq_2}{2R}$, so $q_2=2q_1$. With $q_1+q_2=6.0\ \mu\text{C}$: $q_1=2.0\ \mu\text{C}$ and $q_2=4.0\ \mu\text{C}$.

Ground 接地 is an idealized reference at zero potential that can absorb or supply any amount of charge. Grounding a conductor while an external charge is nearby leaves the conductor with a net induced charge 感应电荷: the external field pushes charge of one sign to ground, and cutting the ground wire before removing the external charge traps the rest.

Bahasa Indonesia

Ketika dua konduktor bersentuhan (atau dihubungkan dengan kawat), muatan mengalir di antara keduanya hingga kedua permukaan mencapai potensial yang sama – itulah kondisi berhenti, bukan "muatan yang sama". Bola yang lebih besar menyimpan lebih banyak muatan pada potensial yang sama ($V=kQ/R$), sehingga ia mengambil porsi yang lebih besar.

Contoh terpadu. Sebuah bola kecil berjari-jari $R$ membawa muatan $+6.0\ \mu\text{C}$ menyentuh bola jauh berjari-jari $2R$, kemudian keduanya terpisah. Potensial yang sama mensyaratkan $\dfrac{kq_1}{R}=\dfrac{kq_2}{2R}$, sehingga $q_2=2q_1$. Dengan $q_1+q_2=6.0\ \mu\text{C}$: $q_1=2.0\ \mu\text{C}$ dan $q_2=4.0\ \mu\text{C}$.

Tanah adalah referensi ideal pada nol potensial yang dapat menyerap atau menyediakan muatan dalam jumlah berapa pun. Menguatkan (grounding) konduktor sementara muatan eksternal berada di dekatnya meninggalkan konduktor dengan muatan induksi netto: medan eksternal mendorong muatan dari satu tanda ke tanah, dan memutus kawat tanah sebelum menghapus muatan eksternal menjebak sisanya.

Vocabulary · ⁨Kosa kata⁩ Train · ⁨Latih⁩

English	Bahasa Indonesia
conductor/kənˈdʌktə/	konduktor
electrostatic equilibrium/ɪˌlektrəʊˈstætɪk ˌiːkwɪˈlɪbrɪəm/	kesetimbangan elektrostatik
perpendicular/ˌpɜːpənˈdɪkjʊlə/	tegak lurus
equipotential surface/ˌiːkwɪpəˈtenʃl ˈsɜːfɪs/	Permukaan ekuipotensial
electrostatic shielding/ɪˌlektrəʊˈstætɪk ˈʃiːldɪŋ/	Peredaman elektrostatik
Faraday cage/ˈfærədeɪ keɪdʒ/	sangkar Faraday
ground/ɡraʊnd/	landasan
induced charge/ɪnˈdjuːst tʃɑːdʒ/	Muatan terinduksi

10.3

Capacitors · ⁨Kapasitor⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 10.3.A: Describe the physical properties of a parallel-plate capacitor.

10.3.A.1 A parallel-plate capacitor consists of two separated parallel conducting surfaces that can hold equal amounts of charge with opposite signs.
10.3.A.2 Capacitance relates the magnitude of the charge stored on each plate to the electric potential difference created by the separation of those charges.
- Equation: $C = \dfrac{Q}{\Delta V}$
- 10.3.A.2.i The capacitance of a capacitor depends only on the physical properties of the capacitor, such as the capacitor's shape and the material used to separate the plates.
- 10.3.A.2.ii The capacitance of a parallel-plate capacitor is proportional to the area of one of its plates and inversely proportional to the distance between its plates. The constant of proportionality is the product of the dielectric constant, $\kappa$, of the material between the plates and the electric permittivity of free space, $\varepsilon_0$.
  - Equation: $C = \dfrac{\kappa \varepsilon_0 A}{d}$
10.3.A.3 The electric field between two charged parallel plates with uniformly distributed electric charge, such as in a parallel-plate capacitor, is constant in both magnitude and direction, except near the edges of the plates.
- 10.3.A.3.i The magnitude of the electric field between two charged parallel plates, where the plate separation is much smaller than the dimensions of the plates, can be determined by applying Gauss's law and the principle of superposition.
  - Equation: $E = \dfrac{Q}{\varepsilon_0 A}$
- 10.3.A.3.ii The electric field is proportional to the surface charge density on either plate of the capacitor.
- 10.3.A.3.iii A charged particle between two oppositely charged parallel plates undergoes constant acceleration, and therefore its motion shares characteristics with the projectile motion of an object with mass in the gravitational field near Earth's surface.
10.3.A.4 The electric potential energy stored in a capacitor is equal to the work done by an external force to separate that amount of charge on the capacitor.
10.3.A.5 The electric potential energy stored in a capacitor is described by the equation $U_C = \dfrac{1}{2} Q \Delta V$.

Boundary statement: While other shapes are also able to separate charges, AP Physics C: Electricity & Magnetism only expects the quantitative analysis and description of parallel-plate capacitors, concentric spherical capacitors, and coaxial cylindrical capacitors.

Bahasa Indonesia

Tujuan Pembelajaran 10.3.A: Jelaskan sifat fisik kapasitor pelat sejajar.

10.3.A.1 Kapasitor pelat sejajar terdiri dari dua permukaan konduktor sejajar yang terpisah dan mampu menampung muatan yang sama besar dengan tanda berlawanan.
10.3.A.2 Kapasitas membandingkan besaran muatan yang tersimpan pada setiap pelat dengan beda potensial listrik yang dihasilkan oleh pemisahan muatan tersebut.
- Persamaan: $C = \dfrac{Q}{\Delta V}$
- 10.3.A.2.i Kapasitas kapasitor hanya bergantung pada sifat fisik kapasitor itu sendiri, seperti bentuk kapasitor dan bahan yang digunakan untuk memisahkan pelat-pelatnya.
- 10.3.A.2.ii Kapasitas kapasitor pelat sejajar berbanding lurus dengan luas salah satu pelatnya dan berbanding terbalik dengan jarak antar pelatnya. Konstanta perbandingan ini adalah hasil kali konstanta dielektrik, $\kappa$, dari material di antara pelat dan permitivitas ruang hampa, $\varepsilon_0$.
  - Persamaan: $C = \dfrac{\kappa \varepsilon_0 A}{d}$
10.3.A.3 Medan listrik di antara dua pelat sejajar bermuatan dengan distribusi muatan listrik yang seragam, seperti pada kapasitor pelat sejajar, bersifat konstan baik dalam besaran maupun arah, kecuali di dekat tepi pelat.
- 10.3.A.3.i Besaran medan listrik di antara dua pelat sejajar bermuatan, di mana jarak antar pelat jauh lebih kecil daripada dimensi pelat, dapat ditentukan dengan menerapkan hukum Gauss dan prinsip superposisi.
  - Persamaan: $E = \dfrac{Q}{\varepsilon_0 A}$
- 10.3.A.3.ii Medan listrik berbanding lurus dengan kepadatan muatan permukaan pada salah satu pelat kapasitor.
- 10.3.A.3.iii Partikel bermuatan di antara dua pelat sejajar bermuatan berlawanan mengalami percepatan konstan, sehingga gerakannya memiliki karakteristik serupa dengan gerak peluru benda bermassa di dalam medan gravitasi dekat permukaan Bumi.
10.3.A.4 Energi potensial listrik yang tersimpan dalam kapasitor sama dengan usaha yang dilakukan oleh gaya eksternal untuk memisahkan sejumlah muatan tersebut pada kapasitor.
10.3.A.5 Energi potensial listrik yang tersimpan dalam kapasitor dijelaskan oleh persamaan $U_C = \dfrac{1}{2} Q \Delta V$.

Pernyataan Batasan: Meskipun bentuk lain juga mampu memisahkan muatan, AP Physics C: Listrik & Magnetisme hanya mengharapkan analisis kuantitatif dan deskripsi kapasitor pelat sejajar, kapasitor bola konsentris, dan kapasitor silinder koaksial.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

Charging a capacitor (RC)

A capacitor 电容器 stores charge on two conductors separated by a gap: $+Q$ on one plate, $-Q$ on the other. Its capacitance 电容 relates the stored charge to the potential difference between the plates:

$$C=\frac{Q}{\Delta V}.$$

Capacitance depends only on geometry and the material in the gap – never on $Q$ or $\Delta V$ themselves.

Worked derivation (parallel plates). For a parallel-plate capacitor 平行板电容器 with plate area $A$ and small gap $d$: Gauss's law plus superposition 叠加 gives a uniform 均匀 field between the plates, $E=\dfrac{\sigma}{\varepsilon_0}=\dfrac{Q}{\varepsilon_0 A}$ (each plate alone contributes $\sigma/2\varepsilon_0$; between the plates the two add, outside they cancel). A uniform field means $\Delta V=Ed=\dfrac{Qd}{\varepsilon_0 A}$, so

$$C=\frac{Q}{\Delta V}=\frac{\varepsilon_0 A}{d}.$$

This $E\to\Delta V\to C$ chain is a standard FRQ derivation – learn it as three steps, and quote each one. The same method handles the other two shapes AP expects: concentric spheres ($C=4\pi\varepsilon_0\dfrac{ab}{b-a}$) and a coaxial cylinder of length $L$, where $E=\dfrac{\lambda}{2\pi\varepsilon_0 r}$ gives $\Delta V=\dfrac{\lambda}{2\pi\varepsilon_0}\ln\dfrac{b}{a}$ and so $C=\dfrac{2\pi\varepsilon_0 L}{\ln(b/a)}$.

Worked example. Plates of area $A=0.020\ \text{m}^2$ and gap $d=1.0\ \text{mm}$: $C=\dfrac{8.85\times10^{-12}(0.020)}{1.0\times10^{-3}}=1.8\times10^{-10}\ \text{F}$. Charged to $100\ \text{V}$ it holds $Q=CV=1.8\times10^{-8}\ \text{C}$.

Because the field between the plates is uniform, a charged particle there feels a constant force, so it moves with constant acceleration – exactly like projectile motion 抛体运动 in gravity: constant speed across, uniform acceleration towards a plate.

Worked example. An electron enters midway between plates at $v_0=2.0\times10^{7}\ \text{m/s}$, parallel to them. The field is $E=1.0\times10^{3}\ \text{N/C}$ and the plates are $4.0\ \text{cm}$ long. Acceleration: $a=\dfrac{eE}{m}=\dfrac{(1.6\times10^{-19})(1.0\times10^{3})}{9.11\times10^{-31}}=1.8\times10^{14}\ \text{m/s}^2$. Time between the plates: $t=\dfrac{0.040}{2.0\times10^{7}}=2.0\times10^{-9}\ \text{s}$. Deflection: $y=\tfrac12at^{2}=\tfrac12(1.8\times10^{14})(2.0\times10^{-9})^{2}\approx3.5\times10^{-4}\ \text{m}$ – about $0.35\ \text{mm}$ towards the positive plate.

Storing charge takes work 功: an external force must move each bit of charge against the field of the charge already there. The total work ends up as stored potential energy,

$$U_C=\tfrac12\,Q\,\Delta V=\tfrac12 C(\Delta V)^2=\frac{Q^2}{2C}.$$

The factor $\tfrac12$ is the area of the triangle under the $Q$–$\Delta V$ line: the first charge moved across costs almost nothing, the last costs the full $\Delta V$.

Capacitors combine oppositely to resistors: in parallel 并联 the capacitances add ($C_{\text{eq}}=C_1+C_2$, same $\Delta V$, charges add), while in series 串联 the reciprocals add ($\tfrac{1}{C_{\text{eq}}}=\tfrac{1}{C_1}+\tfrac{1}{C_2}$, same $Q$, voltages add).

Worked example. A $2\ \mu\text{F}$ and a $4\ \mu\text{F}$ capacitor in series: $\tfrac{1}{C}=\tfrac12+\tfrac14$, so $C=\tfrac43\ \mu\text{F}$. The same pair in parallel: $6\ \mu\text{F}$.

Bahasa Indonesia

Mengisi kapasitor (RC)

Sebuah kapasitor menyimpan muatan pada dua konduktor yang dipisahkan oleh celah: $+Q$ pada pelat satu, $-Q$ pada pelat lainnya. Kapasitansinya menghubungkan muatan yang tersimpan dengan beda potensial antar pelat:

$$C=\frac{Q}{\Delta V}.$$

Kapasitansi hanya bergantung pada geometri dan material di celah – tidak pernah pada $Q$ atau $\Delta V$ itu sendiri.

Turunan contoh terpecahkan (pelat paralel). Untuk kapasitor pelat paralel dengan luas pelat $A$ dan celah kecil $d$: Hukum Gauss ditambah superposisi menghasilkan medan seragam di antara pelat, $E=\dfrac{\sigma}{\varepsilon_0}=\dfrac{Q}{\varepsilon_0 A}$ (setiap pelat saja berkontribusi $\sigma/2\varepsilon_0$; di antara pelat keduanya menambah, di luar mereka saling menghilangkan). Medan seragam berarti $\Delta V=Ed=\dfrac{Qd}{\varepsilon_0 A}$, jadi

$$C=\frac{Q}{\Delta V}=\frac{\varepsilon_0 A}{d}.$$

Rantai $E\to\Delta V\to C$ ini adalah turunan FRQ standar – pelajari sebagai tiga langkah, dan kutipkan masing-masing. Metode yang sama menangani dua bentuk lain yang diharapkan AP: bola konsentris ($C=4\pi\varepsilon_0\dfrac{ab}{b-a}$) dan silinder koaksial sepanjang $L$, di mana $E=\dfrac{\lambda}{2\pi\varepsilon_0 r}$ menghasilkan $\Delta V=\dfrac{\lambda}{2\pi\varepsilon_0}\ln\dfrac{b}{a}$ dan seterusnya $C=\dfrac{2\pi\varepsilon_0 L}{\ln(b/a)}$.

Contoh terpecahkan. Pelat dengan luas $A=0.020\ \text{m}^2$ dan celah $d=1.0\ \text{mm}$: $C=\dfrac{8.85\times10^{-12}(0.020)}{1.0\times10^{-3}}=1.8\times10^{-10}\ \text{F}$. Dimuat hingga $100\ \text{V}$ menyimpan $Q=CV=1.8\times10^{-8}\ \text{C}$.

Karena medan antara pelat seragam, partikel bermuatan di sana merasakan gaya konstan, sehingga bergerak dengan percepatan konstan – persis seperti gerak proyektil dalam gravitasi: kecepatan tetap secara horizontal, percepatan seragam menuju pelat.

*Partikel bermuatan di antara pelat mengikuti parabola, seperti proyektil

Contoh terpecahkan. Elektron memasuki di tengah antara pelat pada $v_0=2.0\times10^{7}\ \text{m/s}$, sejajar dengannya. Medan adalah $E=1.0\times10^{3}\ \text{N/C}$ dan pelat panjangnya $4.0\ \text{cm}$. Percepatan: $a=\dfrac{eE}{m}=\dfrac{(1.6\times10^{-19})(1.0\times10^{3})}{9.11\times10^{-31}}=1.8\times10^{14}\ \text{m/s}^2$. Waktu di antara pelat: $t=\dfrac{0.040}{2.0\times10^{7}}=2.0\times10^{-9}\ \text{s}$. Defleksi: $y=\tfrac12at^{2}=\tfrac12(1.8\times10^{14})(2.0\times10^{-9})^{2}\approx3.5\times10^{-4}\ \text{m}$ – sekitar $0.35\ \text{mm}$ menuju pelat positif.

Menyimpan muatan memerlukan usaha: gaya eksternal harus memindahkan setiap bagian muatan melawan medan dari muatan yang sudah ada. Total usaha berakhir sebagai energi potensial tersimpan,

$$U_C=\tfrac12\,Q\,\Delta V=\tfrac12 C(\Delta V)^2=\frac{Q^2}{2C}.$$

*Energi tersimpan dalam kapasitor adalah area di bawah garis muatan-tegangannya

Faktor $\tfrac12$ adalah area segitiga di bawah garis $Q$–$\Delta V$: muatan pertama yang dipindahkan hampir tidak biaya, yang terakhir biaya penuh $\Delta V$.

Kapasitor digabungkan berlawanan dengan resistor: dalam paralel kapasitansinya bertambah ($C_{\text{eq}}=C_1+C_2$, tegangan $\Delta V$ sama, muatan bertambah), sedangkan dalam seri kebalikannya bertambah ($\tfrac{1}{C_{\text{eq}}}=\tfrac{1}{C_1}+\tfrac{1}{C_2}$, muatan $Q$ sama, tegangan bertambah).

*Kapasitor dalam seri membawa muatan yang sama, dan beda potensialnya bertambah

Contoh terpecahkan. Kapasitor $2\ \mu\text{F}$ dan $4\ \mu\text{F}$ dalam seri: $\tfrac{1}{C}=\tfrac12+\tfrac14$, sehingga $C=\tfrac43\ \mu\text{F}$. Pasangan yang sama dalam paralel: $6\ \mu\text{F}$.

*Berbagai kapasitor: perangkat yang menyimpan muatan dan energi dalam medan listrik antar konduktor

*Kapasitor elektrolitik pada papan sirkuit: kapasitansi menyimpan energi sebagai U = ½CV²

Explore · ⁨Jelajahi⁩

Charge and discharge a capacitor · ⁨Muati dan kosongkan kapasitor⁩

A capacitor stores charge on two plates, filling and emptying exponentially with time constant $\tau=RC$. Bigger $R$ or $C$ slows it down. · ⁨Kapasitor menyimpan muatan pada dua pelat, mengisi dan mengosongkan secara eksponensial dengan konstanta waktu $\tau=RC$. $R$ atau $C$ yang lebih besar memperlambatnya.⁩

Vocabulary · ⁨Kosa kata⁩ Train · ⁨Latih⁩

English	Bahasa Indonesia
capacitor/kəˈpæsɪtə/	kapasitor
capacitance/kəˈpæsɪtəns/	kapasitansi
parallel-plate capacitor/ˈpærəlel pleɪt kəˈpæsɪtə/	Kapasitor pelat sejajar
superposition/ˌsuːpəpəˈzɪʃn/	superposisi
uniform/ˈjuːnɪfɔːm/	seragam
projectile motion/prəˈdʒektaɪl ˈməʊʃn/	gerak peluru
work/wɜːk/	usaha
series/ˈsɪəriːz/	seri

10.4

Dielectrics · ⁨Dielektrik⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 10.4.A: Describe how a dielectric inserted between the plates of a capacitor changes the properties of the capacitor.

10.4.A.1 In a dielectric material, electric charges are not as free to move as they are in a conductor. Instead, the material becomes polarized in the presence of an external electric field.
10.4.A.2 The dielectric constant of a material relates the electric permittivity of that material to the permittivity of free space.
- Equation: $\kappa = \dfrac{\varepsilon}{\varepsilon_0}$
10.4.A.3 The electric field created by a polarized dielectric is opposite in direction to the external field.
10.4.A.4 The electric field between the plates of an isolated parallel-plate capacitor decreases when a dielectric is placed between the plates.
- Equation: $\kappa = \dfrac{E_0}{E}$
10.4.A.5 The insertion of a dielectric into a capacitor may change the capacitance of the capacitor.
- Equation: $C = \kappa C_0$

Bahasa Indonesia

Tujuan Pembelajaran 10.4.A: Jelaskan bagaimana penyisipan dielektrik di antara pelat kapasitor mengubah sifat kapasitor.

10.4.A.1 Dalam bahan dielektrik, muatan listrik tidak sebebas bergerak dibandingkan dengan konduktor. Sebaliknya, material tersebut terpolarisasi di hadapan medan listrik eksternal.
10.4.A.2 Konstanta dielektrik dari suatu material membandingkan permitivitas material tersebut dengan permitivitas ruang hampa.
- Persamaan: $\kappa = \dfrac{\varepsilon}{\varepsilon_0}$
10.4.A.3 Medan listrik yang dihasilkan oleh dielektrik yang terpolarisasi berarah berlawanan dengan medan eksternal.
10.4.A.4 Medan listrik di antara pelat kapasitor pelat sejajar terisolasi menurun ketika dielektrik diletakkan di antara pelat-pelatnya.
- Persamaan: $\kappa = \dfrac{E_0}{E}$
10.4.A.5 Penyisipan dielektrik ke dalam kapasitor dapat mengubah kapasitas kapasitor tersebut.
- Persamaan: $C = \kappa C_0$

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

A dielectric 电介质 is an insulating material. Its charges cannot travel, but in an external field each molecule stretches or turns slightly – the material becomes polarized 极化. The lined-up molecules create their own small field opposite to the applied one, so the net field inside the material drops:

$$E=\frac{E_0}{\kappa},\qquad \kappa=\frac{\varepsilon}{\varepsilon_0}\ \ (\kappa>1),$$

where $\kappa$ is the dielectric constant 介电常数, the ratio of the material's permittivity to the permittivity of free space 真空介电常数.

Filling a capacitor with dielectric multiplies its capacitance:

$$C=\kappa\,C_0\qquad\left(\text{parallel plates: } C=\frac{\kappa\varepsilon_0 A}{d}\right).$$

What happens next depends on what stays fixed – a favourite exam trap:

	battery stays connected ($\Delta V$ fixed)	battery removed first ($Q$ fixed)
charge $Q$	rises to $\kappa Q_0$	unchanged
voltage $\Delta V$	unchanged	drops to $\Delta V_0/\kappa$
field $E$	unchanged	drops to $E_0/\kappa$
energy $U$	rises to $\kappa U_0$	drops to $U_0/\kappa$

Worked example. A $100\ \text{pF}$ capacitor is charged to $12\ \text{V}$, disconnected, then filled with a $\kappa=3$ dielectric. $Q$ is trapped, so $\Delta V$ falls to $4\ \text{V}$ and the stored energy falls to one-third – the missing energy went into pulling the dielectric in. Reconnected to the $12\ \text{V}$ battery instead, the capacitor would hold three times the charge and three times the energy.

Exam skill. Always begin dielectric questions by writing down which quantity is held fixed ($Q$ or $\Delta V$), then let $C=\kappa C_0$ drive everything else through $Q=C\Delta V$ and $U=\tfrac12 C(\Delta V)^2$.

Bahasa Indonesia

Dielektrik adalah bahan isolator. Muatannya tidak dapat berpindah, tetapi dalam medan eksternal setiap molekul meregang atau berputar sedikit – material menjadi terpolarisasi. Molekul yang sejajar menciptakan medan kecil mereka sendiri berlawanan dengan medan yang diterapkan, sehingga medan total di dalam material menurun:

$$E=\frac{E_0}{\kappa},\qquad \kappa=\frac{\varepsilon}{\varepsilon_0}\ \ (\kappa>1),$$

di mana $\kappa$ adalah konstanta dielektrik, rasio permitivitas material terhadap permitivitas ruang hampa.

*Dielektrik terpolarisasi menciptakan medan internal yang berlawanan dengan medan yang diterapkan

Mengisi kapasitor dengan dielektrik mengalikan kapasitansinya:

$$C=\kappa\,C_0\qquad\left(\text{parallel plates: } C=\frac{\kappa\varepsilon_0 A}{d}\right).$$

Apa yang terjadi selanjutnya tergantung pada apa yang tetap konstan – jebakan ujian favorit:

	baterai tetap terhubung ($\Delta V$ tetap)	baterai dilepas dulu ($Q$ tetap)
muatan $Q$	naik ke $\kappa Q_0$	tak berubah
tegangan $\Delta V$	tak berubah	turun ke $\Delta V_0/\kappa$
medan $E$	tak berubah	turun ke $E_0/\kappa$
energi $U$	naik ke $\kappa U_0$	turun ke $U_0/\kappa$

Contoh terpecahkan. Kapasitor $100\ \text{pF}$ dimuat hingga $12\ \text{V}$, dilepas, lalu diisi dengan dielektrik $\kappa=3$. $Q$ terperangkap, sehingga $\Delta V$ turun ke $4\ \text{V}$ dan energi tersimpan turun menjadi sepertiga – energi yang hilang masuk untuk menarik dielektrik ke dalam. Jika disambungkan kembali ke baterai $12\ \text{V}$, kapasitor akan menyimpan tiga kali lipat muatan dan tiga kali lipat energi.

Keterampilan ujian. Selalu mulai pertanyaan dielektrik dengan menulis besaran mana yang dijaga tetap ($Q$ atau $\Delta V$), lalu biarkan $C=\kappa C_0$ mendorong semua hal lainnya melalui $Q=C\Delta V$ dan $U=\tfrac12 C(\Delta V)^2$.

Explore · ⁨Jelajahi⁩

Add a dielectric · ⁨Tambahkan dielektrik⁩

A dielectric between the plates raises the capacitance, so the capacitor holds more charge at the same voltage. Watch the charge build faster. · ⁨Dielektrik di antara pelat meningkatkan kapasitansi, sehingga kapasitor menyimpan lebih banyak muatan pada tegangan yang sama. Saksikan muatan membangun lebih cepat.⁩

Vocabulary · ⁨Kosa kata⁩ Train · ⁨Latih⁩

English	Bahasa Indonesia
parallel/ˈpærəlel/	paralel
dielectric/ˌdaɪɪˈlektrɪk/	dielektrik
polarized/ˈpəʊləraɪzd/	Terpolarisasi
dielectric constant/ˌdaɪɪˈlektrɪk ˈkɒnstənt/	konstanta dielektrik
permittivity of free space/ˌpɜːmɪˈtɪvəti ɒv friː speɪs/	permitivitas ruang hampa

10.4

Exam tips · ⁨Tips ujian⁩

English

In electrostatic equilibrium a conductor has $\vec E=0$ inside and all excess charge on the surface.
Apply Gauss's law $\oint \vec E\cdot d\vec A=\tfrac{q_{enc}}{\varepsilon_0}$ with a symmetry-matched Gaussian surface (sphere, cylinder, pillbox).
The whole conductor is one equipotential, and the surface field is perpendicular to it.
For a capacitor use $C=\tfrac{Q}{V}$, energy $U=\tfrac12 CV^2$, and how a dielectric raises $C$.
Pick the Gaussian surface so $\vec E$ is constant and parallel (or zero) on each part.

Bahasa Indonesia

Dalam kesetimbangan elektrostatik konduktor memiliki $\vec E=0$ di dalamnya dan semua kelebihan muatan di permukaan.
Terapkan hukum Gauss $\oint \vec E\cdot d\vec A=\tfrac{q_{enc}}{\varepsilon_0}$ dengan permukaan Gaussian yang sesuai simetrinya (bola, silinder, kotak pil).
Seluruh konduktor adalah satu equipotensial, dan medan permukaannya tegak lurus terhadapnya.
Untuk kapasitor gunakan $C=\tfrac{Q}{V}$, energi $U=\tfrac12 CV^2$, dan bagaimana dielektrik meningkatkan $C$.
Pilih permukaan Gaussian agar $\vec E$ konstan dan sejajar (atau nol) di setiap bagiannya.

Electric Circuits · ⁨Rangkaian Listrik⁩

Watch lesson · ⁨Tonton pelajaran⁩

Open this topic on its own page · ⁨Buka topik ini di halaman sendiri⁩

11.1

Electric Current · ⁨Arus Listrik⁩

Syllabus · ⁨Silabus⁩

Learning Objective	Essential Knowledge
11.1.A Describe the movement of electric charges through a medium.	11.1.A.1 Current is the rate at which charge passes through a cross-sectional area of a wire. Equation: $I = \dfrac{dq}{dt}$ 11.1.A.1.i Current within a conductor consists of charge carriers traveling through the conductor with an average drift velocity. Equation: $I = nqv_d A$ 11.1.A.1.ii Electric charge moves in a circuit in response to an electric potential difference, sometimes referred to as electromotive force, or $\mathrm{emf}$ ($\mathcal{E}$). 11.1.A.1.iii If the current is zero in a section of wire, the net motion of charge carriers in the wire is also zero, although individual charge carriers will not have zero speed. 11.1.A.2 Current density is the flow of charge per unit area. Equation: $I = \int \vec{J} \cdot d\vec{A}$ 11.1.A.2.i Current density is related to the motion of the charge carriers within a conductor. Equation: $\vec{J} = nq\vec{v}_d$ 11.1.A.2.ii Current density is a vector quantity. 11.1.A.2.iii A potential difference across a conductor creates an electric field within the conductor that is proportional to the resistivity of the conductor and the current density. Equation: $\vec{E} = \rho\vec{J}$ 11.1.A.3 If a function of current density is given, the total current can be determined by integrating the current density over the area. Equation: $I_{\text{tot}} = \int \vec{J}(r) \cdot d\vec{A}$ 11.1.A.4 Although current is a scalar quantity, it does have a direction. Because its direction is relative to the current carrier and not space, current does not obey the laws of vector addition and has no vector components. 11.1.A.4.i The direction of conventional current is chosen to be the direction in which positive charge would move. 11.1.A.4.ii In common circuits, the current is actually due to the movement of electrons (negative charge carriers).

Learning Objective

Essential Knowledge

11.1.A
Describe the movement of electric charges through a medium.

11.1.A.1 Current is the rate at which charge passes through a cross-sectional area of a wire.
- Equation: $I = \dfrac{dq}{dt}$
- 11.1.A.1.i Current within a conductor consists of charge carriers traveling through the conductor with an average drift velocity.
  - Equation: $I = nqv_d A$
- 11.1.A.1.ii Electric charge moves in a circuit in response to an electric potential difference, sometimes referred to as electromotive force, or $\mathrm{emf}$ ($\mathcal{E}$).
- 11.1.A.1.iii If the current is zero in a section of wire, the net motion of charge carriers in the wire is also zero, although individual charge carriers will not have zero speed.
11.1.A.2 Current density is the flow of charge per unit area.
- Equation: $I = \int \vec{J} \cdot d\vec{A}$
- 11.1.A.2.i Current density is related to the motion of the charge carriers within a conductor.
  - Equation: $\vec{J} = nq\vec{v}_d$
- 11.1.A.2.ii Current density is a vector quantity.
- 11.1.A.2.iii A potential difference across a conductor creates an electric field within the conductor that is proportional to the resistivity of the conductor and the current density.
  - Equation: $\vec{E} = \rho\vec{J}$
11.1.A.3 If a function of current density is given, the total current can be determined by integrating the current density over the area.
- Equation: $I_{\text{tot}} = \int \vec{J}(r) \cdot d\vec{A}$
11.1.A.4 Although current is a scalar quantity, it does have a direction. Because its direction is relative to the current carrier and not space, current does not obey the laws of vector addition and has no vector components.
- 11.1.A.4.i The direction of conventional current is chosen to be the direction in which positive charge would move.
- 11.1.A.4.ii In common circuits, the current is actually due to the movement of electrons (negative charge carriers).

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

Electric current 电流 is the rate at which charge passes a cross-section of wire, $I=\dfrac{dq}{dt}$, measured in amperes 安培. Conventional current points the way positive charge would move. Microscopically, a current is a slow drift of many carriers:

$$I=nqv_dA,$$

with $n$ the number of carriers per volume, $q$ the charge each carries, $v_d$ the drift velocity 漂移速度, and $A$ the cross-sectional area 横截面积.

Worked example. A copper wire with $A=1.0\times10^{-6}\ \text{m}^2$ and $n=8.5\times10^{28}\ \text{m}^{-3}$ carries $1.7\ \text{A}$. Then $v_d=\dfrac{I}{nqA}=\dfrac{1.7}{(8.5\times10^{28})(1.6\times10^{-19})(1.0\times10^{-6})}\approx1.3\times10^{-4}\ \text{m/s}$ – the carriers drift slower than a snail, even though the signal travels near light speed.

Current density 电流密度 is charge flow per unit area, $\vec{J}=nq\vec{v}_d$, linked to the field driving it by $\vec{E}=\rho\vec{J}$. In general $I=\int\vec{J}\cdot d\vec{A}$; if $J(r)$ varies across the wire, integrate it over the cross-section to get the total current. One care point: current has a direction along its wire, but it is a scalar 标量 – currents do not add as vectors, and there are no "components of current".

Bahasa Indonesia

Arus listrik adalah laju muatan yang melewati penampang kawat, $I=\dfrac{dq}{dt}$, diukur dalam amperes. Arus konvensional menunjuk arah positif muatan akan bergerak. Secara mikroskopis, arus adalah aliran lambat banyak pembawa:

$$I=nqv_dA,$$

dengan $n$ jumlah pembawa per volume, $q$ muatan yang dibawanya, $v_d$ kecepatan hanyut, dan $A$ luas penampang.

Pembawa muatan mengalir lambat melalui konduktor untuk menghasilkan arus — Pembawa muatan melambat merayap melalui konduktor untuk menghasilkan arus

Contoh terpecahkan. Kawat tembaga dengan $A=1.0\times10^{-6}\ \text{m}^2$ dan $n=8.5\times10^{28}\ \text{m}^{-3}$ membawa $1.7\ \text{A}$. Maka $v_d=\dfrac{I}{nqA}=\dfrac{1.7}{(8.5\times10^{28})(1.6\times10^{-19})(1.0\times10^{-6})}\approx1.3\times10^{-4}\ \text{m/s}$ – pembawa bergerak lebih lambat dari siput, meskipun sinyal berjalan mendekati kecepatan cahaya.

Kepadatan arus adalah aliran muatan per satuan area, $\vec{J}=nq\vec{v}_d$, terkait dengan medan yang menggerakkannya melalui $\vec{E}=\rho\vec{J}$. Secara umum $I=\int\vec{J}\cdot d\vec{A}$; jika $J(r)$ bervariasi di seluruh kawat, integralkan di atas penampang untuk mendapatkan arus total. Satu catatan: arus memiliki arah sepanjang kawatnya, tetapi ia adalah skalar – arus tidak dijumlahkan sebagai vektor, dan tidak ada "komponen arus".

Osiloskop: tegangan terhadap waktu menunjukkan bagaimana arus dan muatan berevolusi dalam sirkuit — Osiloskop: tegangan terhadap waktu mengungkap bagaimana arus dan muatan berevolusi dalam sebuah sirkuit

Vocabulary · ⁨Kosa kata⁩ Train · ⁨Latih⁩

English	Bahasa Indonesia
Electric current/ɪˈlektrɪk ˈkʌrənt/	Arus listrik
amperes/ˈæmpeəz/	ampere
drift velocity/drɪft vəˈlɒsɪti/	kecepatan rambat
cross-sectional area/krɒs ˈsekʃənl ˈeərɪə/	luas penampang
Current density/ˈkʌrənt ˈdensɪti/	Kepadatan arus
scalar/ˈskeɪlə/	skalar
circuit diagram/ˈsɜːkɪt ˈdaɪəɡræm/	Diagram rangkaian
series/ˈsɪəriːz/	seri

11.2

Electric Circuits · ⁨Sirkuit Listrik⁩

Syllabus · ⁨Silabus⁩

Learning Objective	Essential Knowledge
11.2.A Describe the behavior of a circuit.	11.2.A.1 A circuit is composed of electrical loops, which can include wires, batteries, resistors, lightbulbs, capacitors, inductors, switches, ammeters, and voltmeters. 11.2.A.2 A closed electrical loop is a closed path through which charges may flow. 11.2.A.2.i A closed circuit is one in which charges would be able to flow. 11.2.A.2.ii An open circuit is one in which charges would not be able to flow. 11.2.A.2.iii A short circuit is one in which charges would be able to flow with no change in potential difference. 11.2.A.3 A single circuit element may be part of multiple electrical loops. 11.2.A.4 Circuit schematics are representations used to describe and analyze electric circuits. 11.2.A.4.i The properties of an electric circuit are dependent on the physical arrangement of its constituent elements. 11.2.A.4.ii Circuit elements have common symbols that are used to create schematic diagrams. Variable elements are indicated by a diagonal strikethrough arrow across the standard symbol for that element. (Symbols: Battery, Bulb, Switch, Capacitor, Resistor, Ammeter, Voltmeter, Inductor.) Boundary statement: Unless otherwise specified, all circuit schematic diagrams will be drawn using conventional current.

Learning Objective

Essential Knowledge

11.2.A
Describe the behavior of a circuit.

11.2.A.1 A circuit is composed of electrical loops, which can include wires, batteries, resistors, lightbulbs, capacitors, inductors, switches, ammeters, and voltmeters.
11.2.A.2 A closed electrical loop is a closed path through which charges may flow.
- 11.2.A.2.i A closed circuit is one in which charges would be able to flow.
- 11.2.A.2.ii An open circuit is one in which charges would not be able to flow.
- 11.2.A.2.iii A short circuit is one in which charges would be able to flow with no change in potential difference.
11.2.A.3 A single circuit element may be part of multiple electrical loops.
11.2.A.4 Circuit schematics are representations used to describe and analyze electric circuits.
- 11.2.A.4.i The properties of an electric circuit are dependent on the physical arrangement of its constituent elements.
- 11.2.A.4.ii Circuit elements have common symbols that are used to create schematic diagrams. Variable elements are indicated by a diagonal strikethrough arrow across the standard symbol for that element. (Symbols: Battery, Bulb, Switch, Capacitor, Resistor, Ammeter, Voltmeter, Inductor.)

Boundary statement: Unless otherwise specified, all circuit schematic diagrams will be drawn using conventional current.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

A circuit is a set of closed loops built from wires, batteries, resistors, lightbulbs, capacitors, inductors, switches, and meters; charge can flow only around a closed path. One element can belong to several loops at once – that is what makes multi-loop problems interesting. Every analysis starts by reading the circuit diagram 电路图: trace each loop and identify which elements share the same current (series 串联) and which share the same potential difference (parallel 并联).

Bahasa Indonesia

Sebuah sirkuit adalah sekumpulan loop tertutup yang dibangun dari kawat, baterai, resistor, bohlam, kapasitor, induktor, sakelar, dan meter; muatan hanya dapat mengalir mengelilingi jalur tertutup. Satu elemen dapat menjadi bagian dari beberapa loop sekaligus – itulah yang membuat masalah multi-loop menarik. Setiap analisis dimulai dengan membaca diagram sirkuit: telusuri setiap loop dan identifikasi elemen mana yang memiliki arus yang sama (seri) dan mana yang memiliki beda potensial yang sama (paralel).

Explore · ⁨Jelajahi⁩

Build series and parallel circuits · ⁨Bangun rangkaian seri dan paralel⁩

In series the same current flows through every bulb and voltage divides; in parallel each branch gets the full voltage. Switch mode to see the bulbs' brightness change. · ⁨Dalam seri, arus yang sama mengalir melalui setiap bola lampu dan tegangan terbagi; dalam paralel, setiap cabang mendapat tegangan penuh. Ubah mode untuk melihat perubahan kecerahan bola lampu.⁩

11.3

Resistance, Resistivity, and Ohm's Law · ⁨Resistansi, Resistivitas, dan Hukum Ohm⁩

Syllabus · ⁨Silabus⁩

Learning Objective	Essential Knowledge
11.3.A Describe the resistance of an object using physical properties of that object.	11.3.A.1 Resistance is a measure of the degree to which an object opposes the movement of electric charge. 11.3.A.2 The resistance of a resistor with uniform geometry is proportional to its resistivity and length and is inversely proportional to its cross-sectional area. Equation: $R = \dfrac{\rho\ell}{A}$ 11.3.A.2.i Resistivity is a fundamental property of a material that depends on its atomic and molecular structure and quantifies how strongly the material opposes the motion of electric charge. 11.3.A.2.ii The resistivity of a conductor typically increases with temperature. 11.3.A.2.iii The total resistance of a resistor with uniform geometry, but that is made of a material whose resistivity varies along the length of the resistor, is given by $R = \int \dfrac{\rho(\ell)\,d\ell}{A}$.
11.3.B Describe the electrical characteristics of elements of a circuit.	11.3.B.1 Ohm's law relates current, resistance, and potential difference across a conductive element of a circuit. Equation: $I = \dfrac{\Delta V}{R}$ 11.3.B.1.i Materials that obey Ohm's law have constant resistance for all currents and are called ohmic materials. 11.3.B.1.ii The resistivity of an ohmic material is constant regardless of temperature. 11.3.B.1.iii Resistors can also convert electrical energy to thermal energy, which may change the temperature of both the resistor and the resistor's environment. 11.3.B.1.iv The resistance of an ohmic circuit element can be determined from the slope of a graph of the current in the element as a function of the potential difference across the element.

Learning Objective

Essential Knowledge

11.3.A
Describe the resistance of an object using physical properties of that object.

11.3.A.1 Resistance is a measure of the degree to which an object opposes the movement of electric charge.
11.3.A.2 The resistance of a resistor with uniform geometry is proportional to its resistivity and length and is inversely proportional to its cross-sectional area.
- Equation: $R = \dfrac{\rho\ell}{A}$
- 11.3.A.2.i Resistivity is a fundamental property of a material that depends on its atomic and molecular structure and quantifies how strongly the material opposes the motion of electric charge.
- 11.3.A.2.ii The resistivity of a conductor typically increases with temperature.
- 11.3.A.2.iii The total resistance of a resistor with uniform geometry, but that is made of a material whose resistivity varies along the length of the resistor, is given by $R = \int \dfrac{\rho(\ell)\,d\ell}{A}$.

11.3.B
Describe the electrical characteristics of elements of a circuit.

11.3.B.1 Ohm's law relates current, resistance, and potential difference across a conductive element of a circuit.
- Equation: $I = \dfrac{\Delta V}{R}$
- 11.3.B.1.i Materials that obey Ohm's law have constant resistance for all currents and are called ohmic materials.
- 11.3.B.1.ii The resistivity of an ohmic material is constant regardless of temperature.
- 11.3.B.1.iii Resistors can also convert electrical energy to thermal energy, which may change the temperature of both the resistor and the resistor's environment.
- 11.3.B.1.iv The resistance of an ohmic circuit element can be determined from the slope of a graph of the current in the element as a function of the potential difference across the element.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

Resistance 电阻 measures how strongly an object opposes charge flow. It grows with the material's resistivity 电阻率 and the conductor's length, and shrinks with its area:

$$R=\frac{\rho\,\ell}{A}.$$

Ohm's law 欧姆定律 relates the current through an element to the potential difference across it:

$$I=\frac{\Delta V}{R}.$$

Worked example. Stretch a wire to double its length: the volume is fixed, so the area halves, and $R=\rho\ell/A$ becomes $\rho(2\ell)/(A/2)=4R$ – four times the resistance. An element is ohmic 欧姆性 if $R$ stays constant (a straight line through the origin on an $I$–$\Delta V$ graph); a lightbulb filament, which heats up, is not.

Bahasa Indonesia

Resistansi mengukur seberapa kuat suatu objek melawan aliran muatan. Resistansi meningkat seiring dengan resistivitas material dan panjang konduktor, serta menurun seiring dengan luas permukaannya:

$$R=\frac{\rho\,\ell}{A}.$$

Hukum Ohm menghubungkan arus melalui suatu elemen dengan beda potensial di atasnya:

$$I=\frac{\Delta V}{R}.$$

Konduktor yang lebih panjang memiliki resistansi lebih besar; yang lebih lebar memiliki lebih sedikit

Contoh terpecahkan. Regangkan kawat hingga gandakan panjangnya: volumenya tetap, sehingga luasnya berkurang separuh, dan $R=\rho\ell/A$ menjadi $\rho(2\ell)/(A/2)=4R$ – empat kali resistansi. Suatu elemen disebut ohmik jika $R$ tetap konstan (garis lurus melalui titik asal pada grafik $I$–$\Delta V$); filamen bohlam, yang memanas, tidak.

Explore · ⁨Jelajahi⁩

Apply Ohm's law · ⁨Terapkan Hukum Ohm⁩

Ohm's law $V=IR$: for a fixed resistance, current is proportional to voltage. Raise the resistance and the same voltage pushes less current. · ⁨Hukum Ohm $V=IR$: untuk hambatan tetap, arus berbanding lurus dengan tegangan. Naikkan hambatan dan tegangan yang sama mendorong arus lebih sedikit.⁩

Vocabulary · ⁨Kosa kata⁩ Train · ⁨Latih⁩

English	Bahasa Indonesia
resistivity/ˌriːzɪˈstɪvəti/	resistivitas
Ohm's law/əʊmz lɔː/	Hukum Ohm
ohmic/ˈəʊmɪk/	Ohmik
equivalent resistance/ɪˈkwɪvələnt rɪˈzɪstəns/	resistansi ekuivalen
ideal battery/aɪˈdɪəl ˈbætəri/	Baterai ideal
internal resistance/ɪnˈtɜːnl rɪˈzɪstəns/	resistansi dalam
terminal voltage/ˈtɜːmɪnl ˈvəʊltɪdʒ/	Tegangan terminal

11.4

Electric Power · ⁨Daya Listrik⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 11.4.A: Describe the transfer of energy into, out of, or within an electric circuit, in terms of power.

11.4.A.1 The rate at which energy is transferred, converted, or dissipated by a circuit element depends on the current in the element and the electric potential difference across it.
- Equation: $P = I\Delta V$
- Equation: $P = I^2 R = \dfrac{\Delta V^2}{R}$
11.4.A.2 The brightness of a lightbulb increases with power, so power can be used to qualitatively predict the brightness of lightbulbs in a circuit.

Boundary statement: AP Physics C: Electricity & Magnetism only expects students to analyze the transfer of mechanical and electrical energy, although students should be aware that electrical energy can also be dissipated in the form of thermal energy.

Bahasa Indonesia

Tujuan Pembelajaran 11.4.A: Jelaskan transfer energi masuk, keluar, atau di dalam sebuah sirkuit listrik, berdasarkan daya.

11.4.A.1 Laju transfer, konversi, atau pelepasan energi oleh elemen sirkuit bergantung pada arus dalam elemen tersebut dan beda potensial listrik melintasinya.
- Persamaan: $P = I\Delta V$
- Persamaan: $P = I^2 R = \dfrac{\Delta V^2}{R}$
11.4.A.2 Kehebatan mentol meningkat dengan kuasa, maka kuasa boleh digunakan untuk meramal kehebatan mentol dalam litar secara kualitatif.

Pernyataan sempadan: AP Physics C: Elektrik & Magnetisme hanya mengharapkan pelajar menganalisis pemindahan tenaga mekanikal dan elektrik, walaupun pelajar harus menyedari bahawa tenaga elektrik juga boleh terlarap dalam bentuk tenaga haba.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

A charge $q$ falling through a potential difference $\Delta V$ gives up energy $q\Delta V$, so the rate of energy transfer in an element is

$$P=I\,\Delta V=I^2R=\frac{(\Delta V)^2}{R}.$$

In a resistor all of it becomes heat. Use the form whose quantities you actually know – and use power to rank lightbulb brightness: brighter = more power, not necessarily more resistance. In series, the larger resistance is brighter ($P=I^2R$, same $I$); in parallel, the smaller one is ($P=\Delta V^2/R$, same $\Delta V$).

Bahasa Indonesia

Muatan $q$ yang jatuh melewati beda potensial $\Delta V$ melepaskan energi $q\Delta V$, sehingga laju transfer energi dalam suatu elemen adalah

$$P=I\,\Delta V=I^2R=\frac{(\Delta V)^2}{R}.$$

Dalam resistor semuanya berubah menjadi panas. Gunakan bentuk yang besaran-besarannya Anda ketahui sebenarnya – dan gunakan daya untuk peringkat kecerahan bohlam: lebih terang = lebih banyak daya, tidak selalu resistansi lebih besar. Dalam seri, resistansi yang lebih besar lebih terang ($P=I^2R$, arus $I$ sama); dalam paralel, yang lebih kecil yang lebih terang ($P=\Delta V^2/R$, beda potensial $\Delta V$ sama).

Explore · ⁨Jelajahi⁩

Read an I-V characteristic · ⁨Baca karakteristik I-V⁩

Power is $P=IV$. A resistor's I-V line is straight, but a lamp curves as it heats and its resistance rises. The area under I-V relates to the energy delivered. · ⁨Daya adalah $P=IV$. Garis I-V resistor lurus, tetapi lampu melengkung saat memanas dan hambatannya naik. Area di bawah I-V berkaitan dengan energi yang disalurkan.⁩

11.5

Compound Direct Current Circuits · ⁨Sirkuit Arus Langsung Majemuk⁩

Syllabus · ⁨Silabus⁩

Learning Objective	Essential Knowledge
11.5.A Describe the equivalent resistance of multiple resistors connected in a circuit.	11.5.A.1 Circuit elements may be connected in series and/or in parallel. 11.5.A.1.i A series connection is one in which any charge passing through one circuit element must proceed through all elements in that connection and has no other path available. The current in each element in series must be the same. 11.5.A.1.ii A parallel connection is one in which charges may pass through one of two or more paths. Across each path, the potential difference is the same. 11.5.A.2 A collection of resistors in a circuit may be analyzed as though it were a single resistor with an equivalent resistance $R_{\text{eq}}$. 11.5.A.2.i The equivalent resistance of a set of resistors in series is the sum of the individual resistances. Equation: $R_{\text{eq},s} = \sum_{i} R_i$ 11.5.A.2.ii The inverse of the equivalent resistance of a set of resistors connected in parallel is equal to the sum of the inverses of the individual resistances. Equation: $\dfrac{1}{R_{\text{eq},p}} = \sum_{i} \dfrac{1}{R_i}$ 11.5.A.2.iii When resistors are connected in parallel, the number of paths available to charges increases, and the equivalent resistance of the group of resistors decreases.
11.5.B Describe a circuit with resistive wires and a battery with internal resistance.	11.5.B.1 Ideal batteries have negligible internal resistance. Ideal wires have negligible resistance. 11.5.B.1.i The resistance of wires that are good conductors may normally be neglected, because their resistance is much smaller than that of other elements of a circuit. 11.5.B.1.ii The resistance of wires may only be neglected if the circuit contains other elements that do have resistance. 11.5.B.1.iii The potential difference a battery would supply if it were ideal is the potential difference measured across the terminals when there is no current in the battery and is sometimes referred to as its $\mathrm{emf}$ ($\mathcal{E}$). 11.5.B.2 The internal resistance of a nonideal battery may be treated as the resistance of a resistor in series with an ideal battery and the remainder of the circuit. 11.5.B.3 When there is current in a nonideal battery with internal resistance $r$, the potential difference across the terminals of the battery is reduced relative to the potential difference when there is no current in the battery. Equation: $\Delta V_{\text{terminal}} = \mathcal{E} - Ir$
11.5.C Describe the measurement of current and potential difference in a circuit.	11.5.C.1 Ammeters are used to measure current at a specific point in a circuit. 11.5.C.1.i Ammeters must be connected in series with the element in which current is being measured. 11.5.C.1.ii Ideal ammeters have zero resistance so that they do not affect the current in the element that they are in series with. 11.5.C.2 Voltmeters are used to measure electric potential difference between two points in a circuit. 11.5.C.2.i Voltmeters must be connected in parallel with the element across which potential difference is being measured. 11.5.C.2.ii Ideal voltmeters have infinite resistance so that no charge flows through them. 11.5.C.3 Nonideal ammeters and voltmeters will change the properties of the circuit being measured. Boundary statement: Unless otherwise stated, all batteries, wires, and meters are assumed to be ideal. Circuits with batteries of different potential differences connected in parallel will not be assessed.

Learning Objective

Essential Knowledge

11.5.A
Describe the equivalent resistance of multiple resistors connected in a circuit.

11.5.A.1 Circuit elements may be connected in series and/or in parallel.
- 11.5.A.1.i A series connection is one in which any charge passing through one circuit element must proceed through all elements in that connection and has no other path available. The current in each element in series must be the same.
- 11.5.A.1.ii A parallel connection is one in which charges may pass through one of two or more paths. Across each path, the potential difference is the same.
11.5.A.2 A collection of resistors in a circuit may be analyzed as though it were a single resistor with an equivalent resistance $R_{\text{eq}}$.
- 11.5.A.2.i The equivalent resistance of a set of resistors in series is the sum of the individual resistances.
  - Equation: $R_{\text{eq},s} = \sum_{i} R_i$
- 11.5.A.2.ii The inverse of the equivalent resistance of a set of resistors connected in parallel is equal to the sum of the inverses of the individual resistances.
  - Equation: $\dfrac{1}{R_{\text{eq},p}} = \sum_{i} \dfrac{1}{R_i}$
- 11.5.A.2.iii When resistors are connected in parallel, the number of paths available to charges increases, and the equivalent resistance of the group of resistors decreases.

11.5.B
Describe a circuit with resistive wires and a battery with internal resistance.

11.5.B.1 Ideal batteries have negligible internal resistance. Ideal wires have negligible resistance.
- 11.5.B.1.i The resistance of wires that are good conductors may normally be neglected, because their resistance is much smaller than that of other elements of a circuit.
- 11.5.B.1.ii The resistance of wires may only be neglected if the circuit contains other elements that do have resistance.
- 11.5.B.1.iii The potential difference a battery would supply if it were ideal is the potential difference measured across the terminals when there is no current in the battery and is sometimes referred to as its $\mathrm{emf}$ ($\mathcal{E}$).
11.5.B.2 The internal resistance of a nonideal battery may be treated as the resistance of a resistor in series with an ideal battery and the remainder of the circuit.
11.5.B.3 When there is current in a nonideal battery with internal resistance $r$, the potential difference across the terminals of the battery is reduced relative to the potential difference when there is no current in the battery.
- Equation: $\Delta V_{\text{terminal}} = \mathcal{E} - Ir$

11.5.C
Describe the measurement of current and potential difference in a circuit.

11.5.C.1 Ammeters are used to measure current at a specific point in a circuit.
- 11.5.C.1.i Ammeters must be connected in series with the element in which current is being measured.
- 11.5.C.1.ii Ideal ammeters have zero resistance so that they do not affect the current in the element that they are in series with.
11.5.C.2 Voltmeters are used to measure electric potential difference between two points in a circuit.
- 11.5.C.2.i Voltmeters must be connected in parallel with the element across which potential difference is being measured.
- 11.5.C.2.ii Ideal voltmeters have infinite resistance so that no charge flows through them.
11.5.C.3 Nonideal ammeters and voltmeters will change the properties of the circuit being measured.

Boundary statement: Unless otherwise stated, all batteries, wires, and meters are assumed to be ideal. Circuits with batteries of different potential differences connected in parallel will not be assessed.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

Reduce resistor networks to an equivalent resistance 等效电阻: series resistances add ($R_{\text{eq}}=R_1+R_2+\cdots$), while parallel resistances add as reciprocals ($\tfrac{1}{R_{\text{eq}}}=\tfrac{1}{R_1}+\tfrac{1}{R_2}+\cdots$ – always less than the smallest branch). Collapse the network step by step to find the battery current, then expand back out to find each element's current and voltage.

Worked example. A $12\ \text{V}$ battery drives a $4.0\ \Omega$ and a $2.0\ \Omega$ resistor in series: $R_{\text{eq}}=6.0\ \Omega$, $I=2.0\ \text{A}$, the voltages split $8.0\ \text{V}$ and $4.0\ \text{V}$, and the $4.0\ \Omega$ resistor dissipates $P=I^2R=16\ \text{W}$.

Real batteries are not ideal. Model a battery as an ideal battery 理想电池 of emf $\varepsilon$ in series with its own internal resistance 内阻 $r$. When current flows, some emf is used up inside, so the terminal voltage 端电压 – what a voltmeter across the battery actually reads – drops:

$$\Delta V_{\text{terminal}}=\varepsilon-Ir.$$

Worked example. A battery with $\varepsilon=12\ \text{V}$ and $r=0.50\ \Omega$ supplies $2.0\ \text{A}$: the terminals sit at $\Delta V=12-2.0(0.50)=11\ \text{V}$. With no current, a voltmeter reads the full $12\ \text{V}$.

Meters: an ammeter 电流表 goes in series at the point whose current you want (ideal ammeter: zero resistance); a voltmeter 电压表 goes in parallel across the element (ideal voltmeter: infinite resistance). Non-ideal meters disturb the circuit they measure – a real ammeter adds series resistance, a real voltmeter steals current.

Bahasa Indonesia

Sederhanakan jaringan resistor menjadi resistansi ekuivalen: resistansi seri ditambahkan ($R_{\text{eq}}=R_1+R_2+\cdots$), sementara resistansi paralel ditambahkan sebagai resiprokal ($\tfrac{1}{R_{\text{eq}}}=\tfrac{1}{R_1}+\tfrac{1}{R_2}+\cdots$ – selalu kurang dari cabang terkecil). Runtuhkan jaringan langkah demi langkah untuk menemukan arus baterai, lalu kembangkan kembali untuk menemukan arus dan tegangan setiap elemen.

Contoh terpecahkan. Sebuah baterai $12\ \text{V}$ mendorong arus melalui resistor $4.0\ \Omega$ dan $2.0\ \Omega$ secara seri: $R_{\text{eq}}=6.0\ \Omega$, $I=2.0\ \text{A}$, tegangan terbagi menjadi $8.0\ \text{V}$ dan $4.0\ \text{V}$, dan resistor $4.0\ \Omega$ mendissipasi $P=I^2R=16\ \text{W}$.

Resistor dalam paralel bergabung menjadi resistansi ekuivalen yang lebih kecil

Baterai nyata tidak ideal. Modelkan baterai sebagai baterai ideal dengan GGL $\varepsilon$ yang dirangkai seri dengan resistansi internal sendiri $r$. Ketika arus mengalir, sebagian GGL habis di dalam, sehingga tegangan terminal – apa yang dibaca voltmeter melintasi baterai sebenarnya – turun:

$$\Delta V_{\text{terminal}}=\varepsilon-Ir.$$

Contoh terpecahkan. Sebuah baterai dengan $\varepsilon=12\ \text{V}$ dan $r=0.50\ \Omega$ menyediakan $2.0\ \text{A}$: terminal-nya berada pada $\Delta V=12-2.0(0.50)=11\ \text{V}$. Tanpa arus, voltmeter menunjukkan $12\ \text{V}$ penuh.

Meteran: ammeter dipasang seri pada titik di mana arusnya ingin Anda ukur (ammeter ideal: nol resistansi); voltmeter dipasang paralel melintasi elemen (voltmeter ideal: resistansi tak terhingga). Meteran non-ideal mengganggu sirkuit yang diukurnya – ammeter nyata menambah resistansi seri, voltmeter nyata mengambil arus.

Vocabulary · ⁨Kosa kata⁩ Train · ⁨Latih⁩

English	Bahasa Indonesia
parallel/ˈpærəlel/	paralel
Resistance/rɪˈzɪstəns/	Penolakan
ammeter/ˈæmiːtə/	amperemeter
voltmeter/ˈvəʊltmiːtə/	voltmeter
Kirchhoff's loop rule/ˈkɜːkhɒfs luːp ruːl/	Aturan loop Kirchhoff
Kirchhoff's junction rule/ˈkɜːkhɒfs ˈdʒʌŋkʃn ruːl/	Aturan titik simpul Kirchhoff
junction/ˈdʒʌŋkʃn/	sambungan
equivalent capacitance/ɪˈkwɪvələnt kəˈpæsɪtəns/	Kapasitansi ekuivalen
RC circuit/ˌɑː ˈsiː ˈsɜːkɪt/	sirkuit RC
time constant/taɪm ˈkɒnstənt/	konstanta waktu
steady state/ˈstedi steɪt/	keadaan tunak

11.6

Kirchhoff's Loop Rule · ⁨Aturan Loop Kirchhoff⁩

Syllabus · ⁨Silabus⁩

Learning Objective	Essential Knowledge
11.6.A Describe a circuit or elements of a circuit by applying Kirchhoff's loop rule.	11.6.A.1 Energy changes in simple electrical circuits may be represented in terms of charges moving through electric potential differences within circuit elements. Equation: $\Delta U_E = q\Delta V$ 11.6.A.2 Kirchhoff's loop rule is a consequence of the conservation of energy. 11.6.A.2.i Kirchhoff's loop rule states that the sum of potential differences across all circuit elements in a single closed loop must equal zero. Equation: $\sum \Delta V = 0$ 11.6.A.2.ii The values of electric potential at points in a circuit can be represented by a graph of electric potential as a function of position within a loop.

Learning Objective

Essential Knowledge

11.6.A
Describe a circuit or elements of a circuit by applying Kirchhoff's loop rule.

11.6.A.1 Energy changes in simple electrical circuits may be represented in terms of charges moving through electric potential differences within circuit elements.
- Equation: $\Delta U_E = q\Delta V$
11.6.A.2 Kirchhoff's loop rule is a consequence of the conservation of energy.
- 11.6.A.2.i Kirchhoff's loop rule states that the sum of potential differences across all circuit elements in a single closed loop must equal zero.
  - Equation: $\sum \Delta V = 0$
- 11.6.A.2.ii The values of electric potential at points in a circuit can be represented by a graph of electric potential as a function of position within a loop.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

Charges moving through potential differences exchange energy ($\Delta U_E=q\Delta V$), and energy must balance around any closed path. That is Kirchhoff's loop rule 基尔霍夫回路定则:

$$\sum\Delta V=0\ \text{around any closed loop}.$$

Sign discipline wins these problems: crossing a battery from $-$ to $+$ is $+\varepsilon$; crossing a resistor with the assumed current is $-IR$ (against it, $+IR$). Write one equation per independent loop.

Bahasa Indonesia

Muatan yang bergerak melalui beda potensial bertukar energi ($\Delta U_E=q\Delta V$), dan energi harus seimbang di sekitar jalur tertutup apa pun. Itu adalah aturan loop Kirchhoff:

$$\sum\Delta V=0\ \text{around any closed loop}.$$

Disiplin tanda memenangkan masalah ini: menyeberangi baterai dari $-$ ke $+$ adalah $+\varepsilon$; menyeberangi resistor searah dengan arus yang diasumsikan adalah $-IR$ (melawannya, $+IR$). Tulis satu persamaan per loop independen.

11.7

Kirchhoff's Junction Rule · ⁨Aturan Simpul Kirchhoff⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 11.7.A: Describe a circuit or elements of a circuit by applying Kirchhoff's junction rule.

11.7.A.1 Kirchhoff's junction rule is a consequence of the conservation of electric charge.
11.7.A.2 Kirchhoff's junction rule states that the total amount of charge entering a junction per unit time must equal the total amount of charge exiting that junction per unit time.
- Equation: $\sum I_{\text{in}} = \sum I_{\text{out}}$

Bahasa Indonesia

Tujuan Pembelajaran 11.7.A: Deskripsikan sebuah sirkuit atau elemen-elemen sirkuit dengan menerapkan aturan titik simpul Kirchhoff.

11.7.A.1 Aturan titik simpul Kirchhoff merupakan konsekuensi dari kekekalan muatan listrik.
11.7.A.2 Aturan titik simpul Kirchhoff menyatakan bahwa total jumlah muatan yang masuk ke sebuah titik simpul per satuan waktu harus sama dengan total jumlah muatan yang keluar dari titik simpul tersebut per satuan waktu.
- Persamaan: $\sum I_{\text{in}} = \sum I_{\text{out}}$

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

Kirchhoff's junction rule 基尔霍夫节点定则 is conservation of charge at a junction 节点:

$$\sum I_{\text{in}}=\sum I_{\text{out}}.$$

Together the two rules solve any multi-loop circuit: assign a current to each branch, write junction equations, then loop equations, and solve. A negative answer just means that current flows opposite to your assumed direction.

Worked example. In the circuit above, $\varepsilon_1=12\ \text{V}$ with $R_1=1.0\ \Omega$ on the left, $\varepsilon_2=9.0\ \text{V}$ with $R_2=1.0\ \Omega$ on the right, and a shared middle resistor $R_3=2.0\ \Omega$ carrying $I_3=I_1+I_2$ (junction rule). The two loop equations are

$$12=I_1+2(I_1+I_2)=3I_1+2I_2,\qquad 9=I_2+2(I_1+I_2)=2I_1+3I_2.$$

Solving: $I_1=3.6\ \text{A}$, $I_2=0.60\ \text{A}$, so $I_3=4.2\ \text{A}$ through the middle. Check with the second loop: $2(3.6)+3(0.60)=9.0$ ✓.

Bahasa Indonesia

Aturan simpul Kirchhoff adalah konservasi muatan di simpul:

$$\sum I_{\text{in}}=\sum I_{\text{out}}.$$

Arus terbagi di simpul: apa yang masuk sama dengan apa yang keluar

Bersama-sama kedua aturan tersebut menyelesaikan sirkuit multi-loop apa pun: tentukan arus untuk setiap cabang, tulis persamaan simpul, lalu persamaan loop, dan selesaikan. Jawaban negatif hanya berarti arus mengalir berlawanan dengan arah asumsi Anda.

Dua persamaan loop dan satu persamaan simpul menyelesaikan sirkuit dua-baterai ini

Contoh terpecahkan. Pada sirkuit di atas, $\varepsilon_1=12\ \text{V}$ dengan $R_1=1.0\ \Omega$ di sebelah kiri, $\varepsilon_2=9.0\ \text{V}$ dengan $R_2=1.0\ \Omega$ di sebelah kanan, dan resistor tengah bersama $R_3=2.0\ \Omega$ yang membawa $I_3=I_1+I_2$ (aturan simpul). Dua persamaan loop adalah

$$12=I_1+2(I_1+I_2)=3I_1+2I_2,\qquad 9=I_2+2(I_1+I_2)=2I_1+3I_2.$$

Menyelesaikan: $I_1=3.6\ \text{A}$, $I_2=0.60\ \text{A}$, sehingga $I_3=4.2\ \text{A}$ melalui bagian tengah. Periksa dengan loop kedua: $2(3.6)+3(0.60)=9.0$ ✓.

11.8

Resistor-Capacitor (RC) Circuits · ⁨Sirkuit Resistor-Kapasitor (RC)⁩

Syllabus · ⁨Silabus⁩

Learning Objective	Essential Knowledge
11.8.A Describe the equivalent capacitance of multiple capacitors.	11.8.A.1 A collection of capacitors in a circuit may be analyzed as though it was a single capacitor with an equivalent capacitance $C_{\text{eq}}$. 11.8.A.1.i The inverse of the equivalent capacitance of a set of capacitors connected in series is equal to the sum of the inverses of the individual capacitances. Equation: $\dfrac{1}{C_{\text{eq},s}} = \sum_{i} \dfrac{1}{C_i}$ 11.8.A.1.ii The equivalent capacitance of a set of capacitors in series is less than the capacitance of the smallest capacitor. 11.8.A.1.iii The equivalent capacitance of a set of capacitors in parallel is the sum of the individual capacitances. Equation: $C_{\text{eq},p} = \sum_{i} C_i$ 11.8.A.2 As a result of conservation of charge, each of the capacitors in series must have the same magnitude of charge on each plate.
11.8.B Describe the behavior of a circuit containing combinations of resistors and capacitors.	11.8.B.1 The charge on a capacitor or the current in a resistor in an RC circuit can be described by a fundamental differential equation derived from Kirchhoff's loop rule. Equation: $\mathcal{E} = \dfrac{dq}{dt}R + \dfrac{q}{C}$ 11.8.B.2 The time constant ($\tau$) is a significant feature of an RC circuit. 11.8.B.2.i The time constant of an RC circuit is a measure of how quickly the capacitor will charge or discharge and is defined as $\tau = R_{\text{eq}}C_{\text{eq}}$. 11.8.B.2.ii For a charging capacitor, the time constant represents the time required for the capacitor's charge to increase from zero to approximately 63 percent of its final asymptotic value. 11.8.B.2.iii For a discharging capacitor, the time constant represents the time required for the capacitor's charge to decrease from fully charged to approximately 37 percent of its initial value. 11.8.B.3 The potential difference across a capacitor and the current in the branch of the circuit containing the capacitor each change over time as the capacitor charges and discharges, but both will reach a steady state after a long time interval. 11.8.B.3.i Immediately after being placed in a circuit, an uncharged capacitor acts like a wire, and charge can easily flow to or from the plates of the capacitor. 11.8.B.3.ii As a capacitor charges, changes to the potential difference across the capacitor affect the charge on the plates of the capacitor, the current in the circuit branch in which the capacitor is located, and the electric potential energy stored in the capacitor. 11.8.B.3.iii The potential difference across a capacitor, the current in the circuit branch in which the capacitor is located, and the electric potential energy stored in the capacitor all change with respect to time and asymptotically approach steady state conditions. 11.8.B.3.iv After a long time, a charging capacitor approaches a state of being fully charged, reaching a maximum potential difference at which there is zero current in the circuit branch in which the capacitor is located. 11.8.B.3.v Immediately after a charged capacitor begins discharging, the amount of charge on the capacitor and the energy stored in the capacitor begin to decrease. 11.8.B.3.vi As a capacitor discharges, the amount of charge on the capacitor, the potential difference across the capacitor, and the current in the circuit branch in which the capacitor is located all decrease until a steady state is reached. 11.8.B.3.vii After either charging or discharging for times much greater than the time constant, the capacitor and the relevant circuit branch may be modeled using steady-state conditions.

Learning Objective

Essential Knowledge

11.8.A
Describe the equivalent capacitance of multiple capacitors.

11.8.A.1 A collection of capacitors in a circuit may be analyzed as though it was a single capacitor with an equivalent capacitance $C_{\text{eq}}$.
- 11.8.A.1.i The inverse of the equivalent capacitance of a set of capacitors connected in series is equal to the sum of the inverses of the individual capacitances.
  - Equation: $\dfrac{1}{C_{\text{eq},s}} = \sum_{i} \dfrac{1}{C_i}$
- 11.8.A.1.ii The equivalent capacitance of a set of capacitors in series is less than the capacitance of the smallest capacitor.
- 11.8.A.1.iii The equivalent capacitance of a set of capacitors in parallel is the sum of the individual capacitances.
  - Equation: $C_{\text{eq},p} = \sum_{i} C_i$
11.8.A.2 As a result of conservation of charge, each of the capacitors in series must have the same magnitude of charge on each plate.

11.8.B
Describe the behavior of a circuit containing combinations of resistors and capacitors.

11.8.B.1 The charge on a capacitor or the current in a resistor in an RC circuit can be described by a fundamental differential equation derived from Kirchhoff's loop rule.
- Equation: $\mathcal{E} = \dfrac{dq}{dt}R + \dfrac{q}{C}$
11.8.B.2 The time constant ($\tau$) is a significant feature of an RC circuit.
- 11.8.B.2.i The time constant of an RC circuit is a measure of how quickly the capacitor will charge or discharge and is defined as $\tau = R_{\text{eq}}C_{\text{eq}}$.
- 11.8.B.2.ii For a charging capacitor, the time constant represents the time required for the capacitor's charge to increase from zero to approximately 63 percent of its final asymptotic value.
- 11.8.B.2.iii For a discharging capacitor, the time constant represents the time required for the capacitor's charge to decrease from fully charged to approximately 37 percent of its initial value.
11.8.B.3 The potential difference across a capacitor and the current in the branch of the circuit containing the capacitor each change over time as the capacitor charges and discharges, but both will reach a steady state after a long time interval.
- 11.8.B.3.i Immediately after being placed in a circuit, an uncharged capacitor acts like a wire, and charge can easily flow to or from the plates of the capacitor.
- 11.8.B.3.ii As a capacitor charges, changes to the potential difference across the capacitor affect the charge on the plates of the capacitor, the current in the circuit branch in which the capacitor is located, and the electric potential energy stored in the capacitor.
- 11.8.B.3.iii The potential difference across a capacitor, the current in the circuit branch in which the capacitor is located, and the electric potential energy stored in the capacitor all change with respect to time and asymptotically approach steady state conditions.
- 11.8.B.3.iv After a long time, a charging capacitor approaches a state of being fully charged, reaching a maximum potential difference at which there is zero current in the circuit branch in which the capacitor is located.
- 11.8.B.3.v Immediately after a charged capacitor begins discharging, the amount of charge on the capacitor and the energy stored in the capacitor begin to decrease.
- 11.8.B.3.vi As a capacitor discharges, the amount of charge on the capacitor, the potential difference across the capacitor, and the current in the circuit branch in which the capacitor is located all decrease until a steady state is reached.
- 11.8.B.3.vii After either charging or discharging for times much greater than the time constant, the capacitor and the relevant circuit branch may be modeled using steady-state conditions.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

Discharging a capacitor: τ = RC

Charging a capacitor (RC)

Capacitor networks reduce like resistors but with the rules swapped: parallel capacitances add ($C_{\text{eq}}=C_1+C_2$), series add as reciprocals – and capacitors in series must carry the same charge on each plate, by conservation of charge. Use the equivalent capacitance 等效电容 to analyse the network, then expand back.

In an RC circuit RC电路, Kirchhoff's loop rule gives the differential equation

$$\varepsilon=R\frac{dq}{dt}+\frac{q}{C},$$

whose solutions are exponentials with time constant 时间常数 $\tau=RC$:

$$q(t)=Q\big(1-e^{-t/RC}\big)\ \text{(charging)},\qquad q(t)=Q\,e^{-t/RC}\ \text{(discharging)},\qquad i(t)=\frac{\varepsilon}{R}e^{-t/RC}.$$

The current is largest at the first instant and decays – it never "waits" for the capacitor. Learn the two limits: at $t=0$ an uncharged capacitor acts like a plain wire (maximum current); after a long time it is fully charged, no current flows in its branch, and it acts like a break. In any steady state 稳态, cover the capacitor branch with your finger, solve the resistor circuit, then read the capacitor's voltage from the element it sits across.

Worked example. With $R=5.0\ \text{k}\Omega$, $C=200\ \mu\text{F}$, and a $10\ \text{V}$ battery: $\tau=RC=1.0\ \text{s}$; initial current $\dfrac{\varepsilon}{R}=2.0\ \text{mA}$; after one time constant the charge is $q=CV(1-e^{-1})\approx1.3\times10^{-3}\ \text{C}$, about $63\%$ of full charge, and the current has fallen to $37\%$ of its initial value.

Bahasa Indonesia

Mengosongkan kapasitor: τ = RC

Mengisi kapasitor (RC)

Jaringan kapasitor disederhanakan seperti resistor tetapi dengan aturan yang ditukar: kapasitansi paralel ditambahkan ($C_{\text{eq}}=C_1+C_2$), seri ditambahkan sebagai resiprokal – dan kapasitor dalam seri harus membawa muatan yang sama di setiap pelat, oleh karena konservasi muatan. Gunakan kapasitansi ekuivalen untuk menganalisis jaringan, kemudian kembangkan kembali.

Dalam sirkuit RC RC, aturan loop Kirchhoff memberikan persamaan diferensial

$$\varepsilon=R\frac{dq}{dt}+\frac{q}{C},$$

yang solusinya adalah eksponensial dengan konstanta waktu $\tau=RC$:

$$q(t)=Q\big(1-e^{-t/RC}\big)\ \text{(charging)},\qquad q(t)=Q\,e^{-t/RC}\ \text{(discharging)},\qquad i(t)=\frac{\varepsilon}{R}e^{-t/RC}.$$

Arus terbesar pada saat pertama dan meluruh – ia tidak pernah "menunggu" kapasitor. Pelajari dua batas: pada $t=0$ kapasitor tanpa muatan bertindak seperti kawat biasa (arus maksimum); setelah waktu lama ia terisi penuh, tidak ada arus mengalir di cabangnya, dan ia bertindak seperti pemutusan. Dalam keadaan stabil apa pun, tutup cabang kapasitor dengan jari Anda, selesaikan sirkuit resistor, lalu baca tegangan kapasitor dari elemen yang dilintasi olehnya.

Muatan pada kapasitor meluruh secara eksponensial saat melepaskan muatan

Contoh terpecahkan. Dengan $R=5.0\ \text{k}\Omega$, $C=200\ \mu\text{F}$, dan baterai $10\ \text{V}$: $\tau=RC=1.0\ \text{s}$; arus awal $\dfrac{\varepsilon}{R}=2.0\ \text{mA}$; setelah satu konstanta waktu muatannya adalah $q=CV(1-e^{-1})\approx1.3\times10^{-3}\ \text{C}$, sekitar $63\%$ dari muatan penuh, dan arus telah turun ke $37\%$ dari nilainya awal.

Sirkuit elektronik yang dibangun di atas papan prototipe — Sirkuit nyata di papan breadboard: resistor dan kapasitor bersama-sama menentukan seberapa cepat tegangan naik dan turun

11.8

Exam tips · ⁨Tips ujian⁩

English

Relate current to charge flow $I=\tfrac{dQ}{dt}$ and use $J=\sigma E$, $R=\tfrac{\rho L}{A}$ for resistance.
Apply Kirchhoff's laws (junction: charge; loop: energy) with consistent sign conventions.
Analyse RC circuits with calculus: charging/discharging give exponentials with time constant $\tau=RC$.
Combine resistors (series add, parallel reciprocal) and track power $P=IV=I^2R$.
At $t=0$ a capacitor acts like a wire; after a long time ($t\to\infty$) it acts like an open branch.

Bahasa Indonesia

Hubungkan arus dengan aliran muatan $I=\tfrac{dQ}{dt}$ dan gunakan $J=\sigma E$, $R=\tfrac{\rho L}{A}$ untuk hambatan.
Terapkan hukum Kirchhoff (simpul: muatan; loop: energi) dengan konvensi tanda yang konsisten.
Analisis sirkuit RC dengan kalkulus: pengisian/pengosongan menghasilkan eksponensial dengan konstanta waktu $\tau=RC$.
Gabungkan resistor (seri tambahkan, paralel resiprokal) dan lacak daya $P=IV=I^2R$.
Pada $t=0$ kapasitor bertindak seperti kawat; setelah waktu lama ($t\to\infty$) ia bertindak seperti cabang terbuka.

Magnetic Fields and Electromagnetism · ⁨Medan Magnet dan Elektromagnetisme⁩

Watch lesson · ⁨Tonton pelajaran⁩

Open this topic on its own page · ⁨Buka topik ini di halaman sendiri⁩

12.1

Magnetic Fields · ⁨Medan Magnet⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 12.1.A: Describe the properties of a magnetic field.

12.1.A.1 A magnetic field is a vector field that can be used to determine the magnetic force exerted on moving electric charges, electric currents, or magnetic materials.
- 12.1.A.1.i Magnetic fields can be produced by magnetic dipoles or combinations of dipoles, but never by monopoles.
- 12.1.A.1.ii Magnetic dipoles have north and south polarity.
12.1.A.2 A magnetic field is a vector quantity and can be represented using vector field maps.
12.1.A.3 Magnetic field lines must form closed loops, as described by Gauss's law for magnetism.
- 12.1.A.3.i Maxwell's equations are the collection of equations that fully describe electromagnetism. Gauss's law for magnetism is Maxwell's second equation.
  - Equation: $\oint \vec{B} \cdot d\vec{A} = 0$
- 12.1.A.3.ii Magnetic fields in a bar magnet form closed loops, with the external magnetic field pointing away from one end (defined as the north pole) and returning to the other end (defined as the south pole).

Learning Objective 12.1.B: Describe the magnetic behavior of a material as a result of the configuration of magnetic dipoles in the material.

12.1.B.1 Magnetic dipoles result from the circular or rotational motion of electric charges. In magnetic materials, this can be the motion of electrons.
- 12.1.B.1.i Permanent magnetism and induced magnetism are system properties that both result from the alignment of magnetic dipoles within a system.
- 12.1.B.1.ii No magnetic north pole is ever found in isolation from a south pole. For example, if a bar magnet is broken in half, both halves are magnetic dipoles.
- 12.1.B.1.iii Magnetic poles of the same polarity will repel; magnetic poles of opposite polarity will attract.
- 12.1.B.1.iv The magnitude of the magnetic field from a magnetic dipole decreases with increasing distance from the dipole.
12.1.B.2 A magnetic dipole, such as a magnetic compass, placed in a magnetic field will tend to align with the magnetic field.
12.1.B.3 A material's composition influences its magnetic behavior in the presence of an external magnetic field.
- 12.1.B.3.i Ferromagnetic materials such as iron, nickel, and cobalt can be permanently magnetized by an external field that causes the alignment of magnetic domains or atomic magnetic dipoles.
- 12.1.B.3.ii Paramagnetic materials such as aluminum, titanium, and magnesium interact weakly with an external magnetic field, in that the magnetic dipoles of the material do not remain aligned after the external field is removed.
- 12.1.B.3.iii All materials have the property of diamagnetism, in that their electronic structure creates a usually weak alignment of the dipole moments of the material opposite the external magnetic field.
12.1.B.4 Earth's magnetic field may be approximated as a magnetic dipole.

Learning Objective 12.1.C: Describe the magnetic permeability of a material.

12.1.C.1 Magnetic permeability is a measurement of the amount of magnetization in a material in response to an external magnetic field.
12.1.C.2 Free space has a constant value of magnetic permeability, known as the vacuum permeability $\mu_0$, that appears in equations representing physical relationships.
12.1.C.3 The permeability of matter has values different from that of free space and arises from the matter's composition and arrangement. It is not a constant for a material and varies based on many factors, including temperature, orientation, and strength of the external field.

Bahasa Indonesia

Tujuan Pembelajaran 12.1.A: Jelaskan sifat-sifat medan magnet.

12.1.A.1 Medan magnet adalah medan vektor yang dapat digunakan untuk menentukan gaya magnet yang bekerja pada muatan listrik bergerak, arus listrik, atau material magnet.
- 12.1.A.1.i Medan magnet dapat dihasilkan oleh dipol magnet atau kombinasi dipol, tetapi tidak pernah oleh monopole.
- 12.1.A.1.ii Dipol magnet memiliki kutub utara dan selatan.
12.1.A.2 Medan magnet adalah besaran vektor dan dapat direpresentasikan menggunakan peta medan vektor.
12.1.A.3 Garis-garis medan magnet harus membentuk loop tertutup, sebagaimana dijelaskan oleh hukum Gauss untuk magnetisme.
- 12.1.A.3.i Persamaan Maxwell adalah kumpulan persamaan yang secara penuh mendeskripsikan elektromagnetisme. Hukum Gauss untuk magnetisme adalah persamaan kedua Maxwell.
  - Persamaan: $\oint \vec{B} \cdot d\vec{A} = 0$
- 12.1.A.3.ii Medan magnet pada magnet batang membentuk loop tertutup, dengan medan magnet eksternal mengarah menjauhi salah satu ujung (didefinisikan sebagai kutub utara) dan kembali ke ujung lainnya (didefinisikan sebagai kutub selatan).

Tujuan Pembelajaran 12.1.B: Jelaskan perilaku magnetik suatu material akibat konfigurasi dipol magnet di dalamnya.

12.1.B.1 Dipol magnet berasal dari gerakan melingkar atau rotasi muatan listrik. Dalam material magnetik, hal ini dapat berupa gerakan elektron.
- 12.1.B.1.i Kemagnetan permanen dan kemagnetan induksi adalah sifat sistem yang keduanya berasal dari penyelarasan dipol magnet di dalam sistem.
- 12.1.B.1.ii Tidak pernah ditemukan kutub utara magnet secara terpisah dari kutub selatan. Sebagai contoh, jika sebuah magnet batang dipatahkan menjadi dua bagian, kedua bagian tersebut tetap merupakan dipol magnet.
- 12.1.B.1.iii Kutub magnet dengan polaritas yang sama akan tolak-menolak; kutub magnet dengan polaritas yang berlawanan akan tarik-menarik.
- 12.1.B.1.iv Besar medan magnet dari sebuah dipol magnet berkurang seiring bertambahnya jarak dari dipol tersebut.
12.1.B.2 Dipol magnet, seperti kompas magnet, yang diletakkan dalam medan magnet cenderung sejajar dengan medan magnet tersebut.
12.1.B.3 Komposisi suatu material memengaruhi perilaku magnetiknya di hadapan medan magnet eksternal.
- 12.1.B.3.i Material feromagnetik seperti besi, nikel, dan kobalt dapat dimagnetkan secara permanen oleh medan eksternal yang menyebabkan penyelarasan domain magnet atau dipol magnet atom.
- 12.1.B.3.ii Material paramagnetik seperti aluminium, titanium, dan magnesium berinteraksi lemah dengan medan magnet eksternal, di mana dipol magnet material tersebut tidak tetap menyelaraskan diri setelah medan eksternal dihilangkan.
- 12.1.B.3.iii Semua material memiliki sifat diamagnetisme, di mana struktur elektron mereka menciptakan penyelarasan momen dipol material yang biasanya lemah dan berlawanan arah dengan medan magnet eksternal.
12.1.B.4 Medan magnet Bumi dapat didekati sebagai dipol magnet.

Tujuan Pembelajaran 12.1.C: Jelaskan permeabilitas magnetik suatu material.

12.1.C.1 Permeabilitas magnetik adalah pengukuran jumlah magnetisasi dalam suatu material sebagai respons terhadap medan magnet eksternal.
12.1.C.2 Ruang hampa memiliki nilai konstan permeabilitas magnetik, yang dikenal sebagai permeabilitas vakum $\mu_0$, yang muncul dalam persamaan yang merepresentasikan hubungan fisik.
12.1.C.3 Permeabilitas materi memiliki nilai berbeda dari ruang bebas dan timbul dari komposisi serta susunan materinya. Nilai ini bukan konstanta untuk suatu material dan bervariasi berdasarkan banyak faktor, termasuk suhu, orientasi, dan kekuatan medan eksternal.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

A magnetic field 磁场 $\vec{B}$ is a vector field 矢量场 that surrounds magnets, moving charges, and currents; it determines the magnetic force on any moving charge placed in it. Magnetic field lines 磁感线 must form closed loops: they leave the north pole, return to the south pole, and continue through the magnet. Denser lines mean a stronger field.

Closed loops are the content of Gauss's law for magnetism, the second of Maxwell's equations 麦克斯韦方程组:

$$\oint \vec{B}\cdot d\vec{A}=0.$$

The net magnetic flux through any closed surface is zero – as many field lines leave as enter. That is exactly the statement that isolated magnetic monopoles 磁单极子 do not exist. Every source of magnetism is a magnetic dipole 磁偶极子 (a north–south pair), made by circulating charge – in materials, the motion of electrons. Break a bar magnet in half and you get two smaller dipoles, never a lone pole. Like poles repel, opposite poles attract, and a free dipole – a compass 指南针 – rotates to line up with the local field. Earth's own field is approximately a dipole, which is why a compass works. Dipole fields weaken with distance.

How a material responds to an external field depends on how its internal dipoles behave:

Ferromagnetic 铁磁性 (iron, nickel, cobalt): an external field aligns whole magnetic domains 磁畴, and the alignment survives after the field is removed – a permanent magnet.
Paramagnetic 顺磁性 (aluminum, titanium): dipoles align weakly with the field but relax when it is removed.
Diamagnetic 抗磁性 (all materials): the electronic structure produces a usually weak alignment opposite to the field.

The strength of a material's response is its magnetic permeability 磁导率. Free space has the constant value $\mu_0$ (the vacuum permeability); the permeability of matter differs from $\mu_0$ and is not even constant – it varies with temperature, orientation, and field strength.

Bahasa Indonesia

Medan magnetik $\vec{B}$ adalah medan vektor yang mengelilingi magnet, muatan bergerak, dan arus; medan ini menentukan gaya magnetik pada setiap muatan bergerak yang diletakkan di dalamnya. Garis-garis medan magnetik harus membentuk loop tertutup: keluar dari kutub utara, kembali ke kutub selatan, dan berlanjut melalui magnet. Garis yang lebih rapat berarti medan yang lebih kuat.

Garis medan berjalan dari N ke S di luar magnet batang

Loop tertutup adalah isi hukum Gauss untuk kemagnetan, yang kedua dari persamaan Maxwell:

$$\oint \vec{B}\cdot d\vec{A}=0.$$

Fluks magnetik bersih melalui permukaan tertutup apa pun adalah nol – sebanyak garis medan keluar sama dengan yang masuk. Itu tepat merupakan pernyataan bahwa monopole magnetik terisolasi tidak ada. Setiap sumber kemagnetan adalah dipol magnetik (pasangan utara-selatan), dibuat oleh muatan yang berputar – dalam material, gerakan elektron. Pecah magnet batang menjadi dua dan Anda akan mendapatkan dua dipol yang lebih kecil, bukan satu kutub tunggal. Kutup serupa tolak-menolak, kutup berlawanan tarik-menarik, dan dipol bebas – sebuah kompas – berotasi untuk sejajar dengan medan lokal. Medan Bumi sendiri kira-kira berupa dipol, itulah sebabnya kompas berfungsi. Medan dipol melemah seiring jarak.

Bagaimana suatu respons terhadap medan eksternal tergantung pada bagaimana dipol internalnya berperilaku:

Feromagnetik (besi, nikel, kobalt): medan eksternal menyelaraskan seluruh domin magnetik, dan penyelarasan bertahan setelah medan dihilangkan – magnet permanen.
Paramagnetik (aluminium, titanium): dipol menyelaraskan lemah dengan medan tetapi rileks saat medan dihilangkan.
Diamagnetik (semua material): struktur elektronik menghasilkan penyelarasan biasanya lemah berlawanan dengan medan.

Kekuatan respons material adalah permeabilitas magnetiknya. Ruang hampa memiliki nilai konstan $\mu_0$ (permeabilitas vakum); permeabilitas materi berbeda dari $\mu_0$ dan bahkan tidak konstan – itu bervariasi dengan suhu, orientasi, dan kekuatan medan.

Aurora: partikel bermuatan dari Matahari diarahkan oleh medan magnetik Bumi menuju kutub

Explore · ⁨Jelajahi⁩

See a magnet's field lines · ⁨Lihat garis medan magnet⁩

Magnetic field lines run from north to south outside a magnet; where they crowd together the field is strongest. · ⁨Garis medan magnet berjalan dari utara ke selatan di luar magnet; di mana mereka mengkerat, medannya paling kuat.⁩

Watch lesson · ⁨Tonton pelajaran⁩

12.2

Magnetism and Moving Charges · ⁨Kemagnetan dan Muatan Bergerak⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 12.2.A: Describe the magnetic field produced by moving charged objects.

12.2.A.1 A single moving charged object produces a magnetic field.
- 12.2.A.1.i The magnetic field at a particular point produced by a moving charged object depends on the object's velocity and the distance between the point and the object.
- 12.2.A.1.ii At a point in space, the direction of the magnetic field produced by a moving charged object is perpendicular to both the velocity of the object and the position vector from the object to that point in space and can be determined using the right-hand rule.
- 12.2.A.1.iii The magnitude of the magnetic field is a maximum when the velocity vector and the position vector from the object to that point in space are perpendicular.

Learning Objective 12.2.B: Describe the force exerted on moving charged objects by a magnetic field.

12.2.B.1 A magnetic field will exert a force on a charged object moving within that field, with magnitude and direction that depend on the cross-product of the charge's velocity and the magnetic field.
- Equation: $\vec{F}_B = q\left(\vec{v} \times \vec{B}\right)$
12.2.B.2 In a region containing both a magnetic field and an electric field, a moving charged object will experience independent forces from each field.
12.2.B.3 The Hall effect describes the potential difference created in a conductor by an external magnetic field that has a component perpendicular to the direction of charges moving in the conductor.

Bahasa Indonesia

Tujuan Pembelajaran 12.2.A: Deskripsikan medan magnet yang dihasilkan oleh objek bermuatan bergerak.

12.2.A.1 Sebuah objek bermuatan tunggal yang bergerak menghasilkan medan magnet.
- 12.2.A.1.i Medan magnet pada titik tertentu yang dihasilkan oleh objek bermuatan bergerak bergantung pada kecepatan objek dan jarak antara titik tersebut dengan objek.
- 12.2.A.1.ii Pada suatu titik di ruang, arah medan magnet yang dihasilkan oleh objek bermuatan bergerak tegak lurus terhadap baik kecepatan objek maupun vektor posisi dari objek ke titik tersebut di ruang, dan dapat ditentukan menggunakan aturan tangan kanan.
- 12.2.A.1.iii Besar medan magnet maksimum ketika vektor kecepatan dan vektor posisi dari objek ke titik tersebut di ruang saling tegak lurus.

Tujuan Pembelajaran 12.2.B: Deskripsikan gaya yang diberikan oleh medan magnet pada objek bermuatan bergerak.

12.2.B.1 Medan magnet akan memberikan gaya pada benda bermuatan yang bergerak di dalam medan tersebut, dengan besaran dan arah yang bergantung pada hasil kali silang kecepatan muatan dan medan magnet.
- Persamaan: $\vec{F}_B = q\left(\vec{v} \times \vec{B}\right)$
12.2.B.2 Di daerah yang berisi medan magnet dan medan listrik, benda bermuatan yang bergerak akan mengalami gaya independen dari masing-masing medan.
12.2.B.3 Efek Hall menggambarkan beda potensial yang tercipta dalam konduktor akibat medan magnet eksternal yang memiliki komponen tegak lurus terhadap arah pergerakan muatan dalam konduktor tersebut.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

A moving charge in a magnetic field

A moving charge does two things: it creates a magnetic field, and it feels a force in an external one. The field it creates at a point is perpendicular 垂直 to both its velocity and the position vector from charge to point (right-hand rule again), and is largest where those two are perpendicular – zero directly ahead of or behind the motion.

The force on a charge moving in a field $\vec{B}$ is the cross product

$$\vec{F}_B=q\,\vec{v}\times\vec{B},\qquad F=qvB\sin\theta,$$

perpendicular to both $\vec{v}$ and $\vec{B}$ – point the fingers of your right hand along $\vec{v}$, curl toward $\vec{B}$, and the thumb gives the force on a positive charge (reverse it for negative). Because $\vec{F}_B\perp\vec{v}$, the magnetic force does no work on the charge: it changes direction, never speed.

A charge moving perpendicular to a uniform field therefore travels in a circle, the magnetic force supplying the centripetal force 向心力:

$$qvB=\frac{mv^2}{r}\quad\Rightarrow\quad r=\frac{mv}{qB},\qquad T=\frac{2\pi m}{qB}.$$

Notice the period $T$ does not depend on the speed – faster particles ride larger circles in the same time.

Worked example. A proton ($q=1.6\times10^{-19}\ \text{C}$, $m=1.67\times10^{-27}\ \text{kg}$) enters a $0.50\ \text{T}$ field at $2.0\times10^{5}\ \text{m/s}$, perpendicular to it. The magnetic force $F=qvB=1.6\times10^{-14}\ \text{N}$ bends it into a circle of radius $r=\dfrac{mv}{qB}=\dfrac{1.67\times10^{-27}(2.0\times10^{5})}{1.6\times10^{-19}(0.50)}=4.2\times10^{-3}\ \text{m}$.

In a region with both an electric and a magnetic field, the two forces act independently and add as vectors. Balancing them makes a velocity selector 速度选择器: with $\vec{E}$ and $\vec{B}$ crossed, only charges with $qE=qvB$, i.e. $v=E/B$, pass straight through. The Hall effect 霍尔效应 is the same physics inside a conductor: a field with a component perpendicular to the current pushes the moving carriers sideways, charge builds up on one face, and a measurable potential difference appears across the conductor – its sign reveals whether the carriers are positive or negative.

Bahasa Indonesia

Muatan yang bergerak dalam medan magnet

Muatan bergerak melakukan dua hal: ia menciptakan medan magnetik, dan ia merasakan gaya dalam medan eksternal. Medan yang diciptakannya pada suatu titik adalah tegak lurus terhadap kecepatan dan vektor posisi dari muatan ke titik tersebut (aturan tangan kanan lagi), dan terbesar di mana keduanya tegak lurus – nol tepat di depan atau di belakang gerakan.

Gaya pada muatan yang bergerak dalam medan $\vec{B}$ adalah hasil kali silang

$$\vec{F}_B=q\,\vec{v}\times\vec{B},\qquad F=qvB\sin\theta,$$

tegak lurus terhadap baik $\vec{v}$ maupun $\vec{B}$ – arahkan jari tangan kanan Anda sepanjang $\vec{v}$, melengkung menuju $\vec{B}$, dan ibu jari memberikan gaya pada muatan positif (balikkan untuk negatif). Karena $\vec{F}_B\perp\vec{v}$, gaya magnetik tidak melakukan kerja pada muatan: mengubah arah, tidak pernah kecepatan.

Muatan yang bergerak tegak lurus terhadap medan seragam sehingga bergerak dalam lingkaran, gaya magnetik menyediakan gaya sentripetal:

$$qvB=\frac{mv^2}{r}\quad\Rightarrow\quad r=\frac{mv}{qB},\qquad T=\frac{2\pi m}{qB}.$$

Perhatikan periode $T$ tidak bergantung pada kecepatan – partikel lebih cepat menaiki lingkaran lebih besar dalam waktu yang sama.

Partikel bermuatan yang bergerak melintasi medan magnet mengikuti lintasan lingkaran

Contoh dikerjakan. Proton ($q=1.6\times10^{-19}\ \text{C}$, $m=1.67\times10^{-27}\ \text{kg}$) memasuki medan $0.50\ \text{T}$ pada $2.0\times10^{5}\ \text{m/s}$, tegak lurus dengannya. Gaya magnetik $F=qvB=1.6\times10^{-14}\ \text{N}$ membungkukkannya ke dalam lingkaran berjari-jari $r=\dfrac{mv}{qB}=\dfrac{1.67\times10^{-27}(2.0\times10^{5})}{1.6\times10^{-19}(0.50)}=4.2\times10^{-3}\ \text{m}$.

Di daerah dengan medan listrik dan magnetik, kedua gaya bekerja secara independen dan ditambahkan sebagai vektor. Menyeimbangkan mereka membuat selektor kecepatan: dengan $\vec{E}$ dan $\vec{B}$ bersilangan, hanya muatan dengan $qE=qvB$, yaitu $v=E/B$, yang melewati lurus. Efek Hall adalah fisika yang sama di dalam konduktor: medan dengan komponen tegak lurus terhadap arus mendorong pembawa yang bergerak menyamping, muatan menumpuk di satu sisi, dan perbedaan potensial yang dapat diukur muncul di seluruh konduktor – tandanya mengungkapkan apakah pembawanya positif atau negatif.

Explore · ⁨Jelajahi⁩

Force on a moving charge · ⁨Gaya pada muatan bergerak⁩

A charge moving through a magnetic field feels a force $F=qvB$ at right angles to both its velocity and the field — the basis of the motor effect. Reverse either and the force flips. · ⁨Muatan yang bergerak melalui medan magnet mengalami gaya $F=qvB$ tegak lurus terhadap kecepatan dan medannya — dasar efek motor. Balik salah satunya dan gayanya berbalik arah.⁩

Watch lesson · ⁨Tonton pelajaran⁩

12.3

Magnetic Fields of Current-Carrying Wires · ⁨Medan Magnetik Kawat Berarus⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 12.3.A: Describe the magnetic field produced by a current-carrying wire.

12.3.A.1 The Biot-Savart law defines the magnitude and direction of a magnetic field created by an electrical current.
- Equation: $d\vec{B} = \dfrac{\mu_0}{4\pi} \dfrac{I(d\vec{\ell} \times \hat{r})}{r^2}$
12.3.A.2 The magnetic field vectors around a small segment of a current-carrying wire are tangent to concentric circles centered on that wire. The field has no component toward, away from, or parallel to the segment of the current-carrying wire.
12.3.A.3 The Biot-Savart law can be used to derive the magnitudes and directions of magnetic fields around segments of current-carrying wires, for example at the center of a circular loop of wire.
- Equation: $B_{\text{center of loop}} = \dfrac{\mu_0 I}{2R}$

Learning Objective 12.3.B: Describe the force exerted on current-carrying wires by a magnetic field.

12.3.B.1 A magnetic field will exert a force on a current-carrying wire.
- Equation: $\vec{F}_B = \int I\left(d\vec{\ell} \times \vec{B}\right)$

Boundary statement: AP Physics C: Electricity & Magnetism only expects students to perform quantitative analysis of certain cases of current-carrying conductors using the Biot-Savart law, such as at a location along the perpendicular bisector of a straight conductor, at a location along the central axis of a circular loop, or at the center of a segment of a circular loop.

Bahasa Indonesia

Tujuan Pembelajaran 12.3.A: Deskripsikan medan magnet yang dihasilkan oleh kawat pembawa arus.

12.3.A.1 Hukum Biot-Savart mendefinisikan besaran dan arah medan magnet yang dihasilkan oleh arus listrik.
- Persamaan: $d\vec{B} = \dfrac{\mu_0}{4\pi} \dfrac{I(d\vec{\ell} \times \hat{r})}{r^2}$
12.3.A.2 Vektor-vektor medan magnet di sekitar segmen kecil kawat berarus menyinggung lingkaran konsentris yang berpusat pada kawat tersebut. Medan ini tidak memiliki komponen menuju, menjauhi, atau sejajar dengan segmen kawat berarus.
12.3.A.3 Hukum Biot-Savart dapat digunakan untuk menurunkan besaran dan arah medan magnet di sekitar segmen kawat berarus, misalnya di pusat sebuah loop kawat melingkar.
- Persamaan: $B_{\text{center of loop}} = \dfrac{\mu_0 I}{2R}$

Tujuan Pembelajaran 12.3.B: Deskripsikan gaya yang diberikan pada kawat berarus oleh medan magnet.

12.3.B.1 Medan magnet akan memberikan gaya pada kawat berarus.
- Persamaan: $\vec{F}_B = \int I\left(d\vec{\ell} \times \vec{B}\right)$

Pernyataan Batas: AP Physics C: Listrik & Magnetisme hanya mengharapkan siswa melakukan analisis kuantitatif atas kasus-kasus tertentu konduktor berarus menggunakan hukum Biot-Savart, seperti pada lokasi sepanjang garis bagi tegak lurus konduktor lurus, pada lokasi sepanjang sumbu pusat loop melingkar, atau di tengah segmen loop melingkar.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

The magnetic field around a current

A current is a stream of moving charges, so it creates a magnetic field. The Biot–Savart law 毕奥-萨伐尔定律 adds up the field of each current element:

$$d\vec{B}=\frac{\mu_0}{4\pi}\frac{I\,d\vec{l}\times\hat{r}}{r^2}.$$

Around any straight segment the field vectors are tangent 相切 to concentric circles 同心圆 centred on the wire – no component toward, away from, or along the wire. Curl your right hand around the wire with the thumb along the current: your fingers give the field direction.

The Biot–Savart integrals AP expects: a long straight wire ($B=\dfrac{\mu_0 I}{2\pi r}$, or a point on the perpendicular bisector of a finite wire), and the centre of a circular loop,

$$B_{\text{centre of loop}}=\frac{\mu_0 I}{2R},$$

where every element $d\vec{l}$ is perpendicular to $\hat{r}$ and equally distant $R$ – say that in an FRQ derivation. An arc that is a fraction of a full circle contributes that same fraction of $\mu_0 I/2R$ at its centre.

A field also pushes on a current-carrying wire, element by element:

$$\vec{F}_B=\int I\,d\vec{l}\times\vec{B}\qquad(\vec{F}=I\vec{L}\times\vec{B}\ \text{for a straight wire in a uniform field}).$$

Worked example. Two long parallel wires a distance $d=0.10\ \text{m}$ apart each carry $5.0\ \text{A}$ in the same direction. Wire 1's field at wire 2 is $B=\dfrac{\mu_0 I}{2\pi d}=1.0\times10^{-5}\ \text{T}$, so wire 2 feels $\dfrac{F}{L}=I B=5.0\times10^{-5}\ \text{N/m}$, pulled toward wire 1. Same-direction currents attract; opposite currents repel.

Bahasa Indonesia

Medan magnet di sekitar arus listrik

Arus adalah aliran muatan bergerak, sehingga menciptakan medan magnetik. Hukum Biot–Savart menjumlahkan medan dari setiap elemen arus:

$$d\vec{B}=\frac{\mu_0}{4\pi}\frac{I\,d\vec{l}\times\hat{r}}{r^2}.$$

Di sekitar segmen lurus apa pun vektor medan adalah tangen terhadap lingkaran konsentris yang berpusat pada kawat – tidak ada komponen menuju, menjauhi, atau sepanjang kawat. Melingkari tangan kanan Anda di sekitar kawat dengan ibu jari sepanjang arus: jari Anda menunjukkan arah medan.

Garis-garis medan melingkar konsentris mengelilingi kawat pembawa arus lurus

Integral Biot–Savart yang diharapkan AP: kawat lurus panjang ($B=\dfrac{\mu_0 I}{2\pi r}$, atau titik pada garis bagi tegak lurus kawat terbatas), dan pusat lingkaran,

$$B_{\text{centre of loop}}=\frac{\mu_0 I}{2R},$$

di mana setiap elemen $d\vec{l}$ tegak lurus terhadap $\hat{r}$ dan berjarak sama $R$ – katakanlah dalam derivasi FRQ. Busur yang merupakan pecahan dari lingkaran penuh berkontribusi pecahan yang sama dari $\mu_0 I/2R$ di pusatnya.

Medan juga mendorong pada kawat berarus, elemen demi elemen:

$$\vec{F}_B=\int I\,d\vec{l}\times\vec{B}\qquad(\vec{F}=I\vec{L}\times\vec{B}\ \text{for a straight wire in a uniform field}).$$

Contoh dikerjakan. Dua kawat sejajar panjang berjarak $d=0.10\ \text{m}$ masing-masing membawa $5.0\ \text{A}$ dalam arah yang sama. Medan Wire 1 pada wire 2 adalah $B=\dfrac{\mu_0 I}{2\pi d}=1.0\times10^{-5}\ \text{T}$, jadi wire 2 merasakan $\dfrac{F}{L}=I B=5.0\times10^{-5}\ \text{N/m}$, ditarik menuju wire 1. Arus searah menarik; arus berlawanan menolak.

Serbuk besi di sekitar kumpalan berarus — Kumparan pembawa arus (solenoid) menghasilkan medan magnet berbentuk persis seperti batang magnet

Vocabulary · ⁨Kosa kata⁩ Train · ⁨Latih⁩

English	Bahasa Indonesia
magnetic field/mæɡˈnetɪk fiːld/	medan magnet
vector field/ˈvektə fiːld/	Medan vektor
magnetic field lines/mæɡˈnetɪk fiːld laɪnz/	Garis medan magnetik
Maxwell's equations/ˈmækswelz ɪˈkweɪʒnz/	Persamaan Maxwell
magnetic monopoles/mæɡˈnetɪk ˈmɒnəʊpəʊlz/	Monopole magnetik
magnetic dipole/mæɡˈnetɪk ˈdaɪpəʊl/	Dipole magnetik
compass/ˈkʌmpəs/	kompas
Ferromagnetic/ˌferəməɡˈnetɪk/	Feromagnetik
magnetic domains/mæɡˈnetɪk dəˈmeɪnz/	Domain magnetik
Paramagnetic/ˌpærəməɡˈnetɪk/	Paramagnetik
Diamagnetic/ˌdaɪəməɡˈnetɪk/	Diamagnetik
magnetic permeability/mæɡˈnetɪk ˌpɜːməˈbɪlɪti/	Permeabilitas magnetik
perpendicular/ˌpɜːpənˈdɪkjʊlə/	tegak lurus
centripetal force/senˈtrɪpɪtl fɔːs/	kekuatan sentripetal
velocity selector/vəˈlɒsɪti sɪˈlektə/	selektor kecepatan
Hall effect/hɔːl ɪˈfekt/	efek Hall
Biot–Savart law/ˈbɪɒt ˈsævɑːt lɔː/	hukum Biot–Savart
tangent/ˈtændʒənt/	tangens
concentric circles/kənˈsentrɪk ˈsɜːklz/	Lingkaran konsentris
Ampère's law/ˈæmpeəz lɔː/	Hukum Ampère
Amperian loop/æmˈpɪərɪən luːp/	loop Amperian
current density/ˈkʌrənt ˈdensɪti/	Kepadatan arus
superposition/ˌsuːpəpəˈzɪʃn/	superposisi
solenoid/ˈsəʊlənɔɪd/	solenaoid

Watch lesson · ⁨Tonton pelajaran⁩

12.4

Ampère's Law · ⁨Hukum Ampère⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 12.4.A: Use Ampère's law to describe the magnetic field created by a moving charge carrier.

12.4.A.1 Ampère's law relates the magnitude of the magnetic field to the current enclosed by a closed imaginary path called an Amperian loop.
- Equation: $\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\text{enc}}$
- 12.4.A.1.i Ampère's law can be used to determine the magnetic field near a long, straight current-carrying wire.
  - Equation: $B_{\text{wire}} = \dfrac{\mu_0}{2\pi} \dfrac{I}{r}$
- 12.4.A.1.ii Unless otherwise stated, all solenoids are assumed to be very long, with uniform magnetic fields inside the solenoids and negligible magnetic fields outside the solenoids.
- 12.4.A.1.iii Ampère's law can be used to determine the magnetic field inside of a long solenoid.
  - Equation: $B_{\text{sol}} = \mu_0 n I$
12.4.A.2 An Amperian loop is a closed path around a current-carrying conductor.
12.4.A.3 The principle of superposition can be used to determine the net magnetic field at a point in space created by various combinations of current-carrying conductors, or conducting loops, segments, or cylinders.
12.4.A.4 Maxwell's equations are the collection of equations that fully describe electromagnetism. Maxwell's fourth equation is Ampère's law with Maxwell's addition; it states that magnetic fields can be generated by electric current (Ampère's law) and that a changing electric field creates a magnetic field, similar to the way a moving charge creates a magnetic field (Maxwell's addition).
- Equation: $\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I + \mu_0 \varepsilon_0 \dfrac{d\Phi_E}{dt}$

Boundary statement: AP Physics C: Electricity & Magnetism only expects quantitative application of Ampère's law limited to situations involving symmetrical magnetic fields. Long straight wires, long solenoids carrying currents, as well as conductive slabs or cylindrical conductors carrying a current density, are the types of shapes to which Ampère's law will be applied on the AP Physics C: Electricity & Magnetism Exam.

Boundary statement: AP Physics C: Electricity & Magnetism does not expect students to use Maxwell's fourth equation with a changing electric field. However, students should understand that a changing electric field generates a magnetic field.

Bahasa Indonesia

Tujuan Pembelajaran 12.4.A: Gunakan hukum Ampère untuk mendeskripsikan medan magnet yang dihasilkan oleh pembawa muatan yang bergerak.

12.4.A.1 Hukum Ampère menghubungkan besaran medan magnet dengan arus yang dibatasi oleh lintasan imajiner tertutup yang disebut loop Amperian.
- Persamaan: $\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\text{enc}}$
- 12.4.A.1.i Hukum Ampère dapat digunakan untuk menentukan medan magnet di dekat kawat berarus lurus panjang.
  - Persamaan: $B_{\text{wire}} = \dfrac{\mu_0}{2\pi} \dfrac{I}{r}$
- 12.4.A.1.ii Kecuali dinyatakan sebaliknya, semua solenoid diasumsikan sangat panjang, dengan medan magnet seragam di dalam solenoid dan medan magnet yang dapat diabaikan di luar solenoid.
- 12.4.A.1.iii Hukum Ampère dapat digunakan untuk menentukan medan magnet di dalam solenoid panjang.
  - Persamaan: $B_{\text{sol}} = \mu_0 n I$
12.4.A.2 Loop Amperian adalah lintasan tertutup di sekitar konduktor berarus.
12.4.A.3 Prinsip superposisi dapat digunakan untuk menentukan medan magnet total pada suatu titik dalam ruang yang dihasilkan oleh berbagai kombinasi konduktor berarus, atau loop, segmen, atau silinder penghantar.
12.4.A.4 Persamaan Maxwell adalah kumpulan persamaan yang secara penuh mendeskripsikan elektromagnetisme. Persamaan keempat Maxwell adalah hukum Ampère dengan penambahan Maxwell; persamaan ini menyatakan bahwa medan magnet dapat dihasilkan oleh arus listrik (hukum Ampère) dan bahwa medan listrik yang berubah menghasilkan medan magnet, serupa dengan cara muatan yang bergerak menghasilkan medan magnet (penambahan Maxwell).
- Persamaan: $\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I + \mu_0 \varepsilon_0 \dfrac{d\Phi_E}{dt}$

Pernyataan Batas: AP Physics C: Listrik & Magnetisme hanya mengharapkan aplikasi kuantitatif hukum Ampère yang terbatas pada situasi yang melibatkan medan magnet simetris. Kawat lurus panjang, solenoid panjang yang mengalirkan arus, serta pelat konduktif atau konduktor silinder yang mengalirkan kerapatan arus, adalah jenis bentuk-bentuk yang akan diterapkan hukum Ampère pada Ujian AP Physics C: Listrik & Magnetisme.

Pernyataan Batas: AP Physics C: Listrik & Magnetisme tidak mengharapkan siswa menggunakan persamaan keempat Maxwell dengan medan listrik yang berubah. Namun, siswa harus memahami bahwa medan listrik yang berubah menghasilkan medan magnet.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

For symmetric current distributions, Ampère's law 安培定律 finds the field far faster than Biot–Savart:

$$\oint \vec{B}\cdot d\vec{l}=\mu_0 I_{\text{enc}}.$$

Choose an Amperian loop 安培环路 that matches the symmetry, so that $B$ is constant along the loop (and parallel to it) and comes out of the integral. AP applies it to long straight wires, long solenoids, and conductors carrying a current density 电流密度 (slabs and cylinders); combinations are handled by superposition 叠加.

Worked example. Find the field $0.10\ \text{m}$ from a long wire carrying $5.0\ \text{A}$. A circular loop of radius $r$ shares the field's symmetry, so $B(2\pi r)=\mu_0 I$ and

$$B=\frac{\mu_0 I}{2\pi r}=\frac{(4\pi\times10^{-7})(5.0)}{2\pi(0.10)}=1.0\times10^{-5}\ \text{T}.$$

A long solenoid 螺线管 (assumed: uniform field inside, negligible field outside) is the other classic. Take a rectangular loop with one side of length $L$ inside, parallel to the axis: only that side contributes to the integral, so $BL=\mu_0 (nL) I$ and

$$B_{\text{sol}}=\mu_0 n I,$$

with $n$ the turns per metre.

Worked example (cylinder). A solid cylindrical conductor of radius $R$ carries current $I$ spread uniformly over its cross-section. Inside ($r), a circular loop encloses $I_{\text{enc}}=I\dfrac{r^2}{R^2}$, so $B=\dfrac{\mu_0 I r}{2\pi R^2}$ – the field grows linearly to the surface, then falls off as $1/r$ outside.

Exam skill. Every Ampère's-law answer earns its marks in the setup: name the loop, state why $B$ is constant and parallel to it (symmetry), and count $I_{\text{enc}}$ carefully before you solve. Maxwell's fourth equation adds one more idea you should know qualitatively: a changing electric field also creates a magnetic field – but AP will not ask you to compute with that term.

Bahasa Indonesia

Untuk distribusi arus simetris, hukum Ampère menemukan medan jauh lebih cepat daripada Biot–Savart:

$$\oint \vec{B}\cdot d\vec{l}=\mu_0 I_{\text{enc}}.$$

Pilih lingkaran Amperian yang sesuai dengan simetri, sehingga $B$ konstan sepanjang lingkaran (dan sejajar dengannya) serta dapat dikeluarkan dari integral. AP menerapkannya pada kawat lurus panjang, solenoida panjang, dan konduktor yang membawa kerapatan arus (pelat dan silinder); kombinasi ditangani melalui superposisi.

Contoh terpecahkan. Temukan medan $0.10\ \text{m}$ dari kawat panjang yang membawa $5.0\ \text{A}$. Lingkaran berdiameter $r$ memiliki simetri yang sama dengan medannya, sehingga $B(2\pi r)=\mu_0 I$ dan

$$B=\frac{\mu_0 I}{2\pi r}=\frac{(4\pi\times10^{-7})(5.0)}{2\pi(0.10)}=1.0\times10^{-5}\ \text{T}.$$

Solenoida panjang (diasumsikan: medan seragam di dalam, medan dapat diabaikan di luar) adalah contoh klasik lainnya. Ambil lingkaran persegi panjang dengan salah satu sisinya sepanjang $L$ di bagian dalam, sejajar dengan sumbu: hanya sisi tersebut yang berkontribusi terhadap integral, sehingga $BL=\mu_0 (nL) I$ dan

$$B_{\text{sol}}=\mu_0 n I,$$

dengan $n$ jumlah lilitan per meter.

Lingkaran Amperian persegi panjang menurunkan medan seragam di dalam solenoida

Contoh terpecahkan (silinder). Konduktor silinder pejal berjari-jari $R$ membawa arus $I$ yang tersebar merata di seluruh penampang lintangnya. Di dalam ($r), sebuah loop melingkupi $I_{\text{enc}}=I\dfrac{r^2}{R^2}$, sehingga $B=\dfrac{\mu_0 I r}{2\pi R^2}$ – medan tumbuh secara linear hingga permukaan, lalu menurun sebanding $1/r$ di luar.

Keterampilan ujian. Setiap jawaban hukum Ampère mendapatkan nilainya pada tahap penyusunan: sebutkan nama lingkarannya, jelaskan mengapa $B$ konstan dan sejajar dengannya (simetri), dan hitung $I_{\text{enc}}$ dengan teliti sebelum Anda menyelesaikan. Persamaan keempat Maxwell menambahkan satu ide lagi yang harus Anda ketahui secara kualitatif: medan listrik yang berubah juga menciptakan medan magnet – tetapi AP tidak akan meminta Anda menghitung dengan suku tersebut.

Watch lesson · ⁨Tonton pelajaran⁩

12.4

Exam tips · ⁨Tips ujian⁩

English

The magnetic force $\vec F=q\vec v\times\vec B$ is perpendicular to velocity — use the right-hand rule and note it does no work.
A charge in a uniform field moves in a circle with $r=\tfrac{mv}{qB}$.
Find the field of currents with Biot–Savart $d\vec B=\tfrac{\mu_0}{4\pi}\tfrac{I\,d\vec l\times\hat r}{r^2}$ or Ampère's law $\oint \vec B\cdot d\vec l=\mu_0 I_{enc}$ when there is symmetry.
Force on a wire is $\vec F=I\vec L\times\vec B$; keep the cross-product direction straight.
Distinguish the force on a charge from the field the current creates.

Bahasa Indonesia

Gaya magnetik $\vec F=q\vec v\times\vec B$ tegak lurus terhadap kecepatan – gunakan aturan tangan kanan dan perhatikan bahwa gaya ini tidak melakukan usaha.
Muatan dalam medan seragam bergerak dalam lingkaran dengan $r=\tfrac{mv}{qB}$.
Temukan medan arus dengan Biot–Savart $d\vec B=\tfrac{\mu_0}{4\pi}\tfrac{I\,d\vec l\times\hat r}{r^2}$ atau hukum Ampère $\oint \vec B\cdot d\vec l=\mu_0 I_{enc}$ ketika ada simetri.
Gaya pada kawat adalah $\vec F=I\vec L\times\vec B$; jaga arah perkalian silang tetap benar.
Bedakan gaya pada muatan dari medan yang dihasilkan arus ke muatan.

Electromagnetic Induction · ⁨Induksi Elektromagnetik⁩

Watch lesson · ⁨Tonton pelajaran⁩

Open this topic on its own page · ⁨Buka topik ini di halaman sendiri⁩

13.1

Magnetic Flux · ⁨Fluks Magnetik⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 13.1.A: Describe the magnetic flux through an arbitrary area or geometric shape.

13.1.A.1 For a magnetic field $\vec{B}$ that is constant across an area $\vec{A}$, the magnetic flux through the area is defined as $\Phi_B = \vec{B} \cdot \vec{A}$ .
- 13.1.A.1.i The area vector is defined as perpendicular to the plane of the surface and outward from a closed surface.
- 13.1.A.1.ii The sign of flux is given by the dot product of the magnetic field vector and the area vector.
13.1.A.2 The total magnetic flux passing through a surface is defined by the surface integral of the magnetic field over the surface area.
- Equation: $\Phi_B = \displaystyle\int \vec{B} \cdot d\vec{A}$

Bahasa Indonesia

Tujuan Pembelajaran 13.1.A: Deskripsikan fluks magnetik melalui area sembarang atau bentuk geometris.

13.1.A.1 Untuk medan magnet $\vec{B}$ yang konstan di seluruh area $\vec{A}$, fluks magnetik melalui area tersebut didefinisikan sebagai $\Phi_B = \vec{B} \cdot \vec{A}$ .
- 13.1.A.1.i Vektor luas didefinisikan tegak lurus terhadap bidang permukaan dan mengarah keluar dari permukaan tertutup.
- 13.1.A.1.ii Tanda fluks ditentukan oleh hasil kali titik antara vektor medan magnet dan vektor luas.
13.1.A.2 Total fluks magnetik yang melewati suatu permukaan didefinisikan oleh integral permukaan dari medan magnet terhadap luas permukaan.
- Persamaan: $\Phi_B = \displaystyle\int \vec{B} \cdot d\vec{A}$

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

Magnetic flux 磁通量 measures how much magnetic field passes through a surface. For a uniform field through a flat loop it is the dot product $\Phi_B=\vec{B}\cdot\vec{A}=BA\cos\theta$; in general it is the surface integral 面积分

$$\Phi_B=\int \vec{B}\cdot d\vec{A}.$$

The area vector 面积矢量 is perpendicular to the surface (outward from a closed one), and the sign of the flux comes from the dot product. Flux changes if $B$ changes, the area changes, or the loop turns – keep all three routes in mind, because each one is an exam question.

Bahasa Indonesia

Fluks magnetik mengukur seberapa banyak medan magnet yang menembus suatu permukaan. Untuk medan seragam melalui loop datar, fluksnya adalah hasil titik $\Phi_B=\vec{B}\cdot\vec{A}=BA\cos\theta$; secara umum fluksnya adalah integral permukaan

$$\Phi_B=\int \vec{B}\cdot d\vec{A}.$$

Vektor luas tegak lurus terhadap permukaan (mengarah keluar dari permukaan tertutup), dan tanda fluks berasal dari hasil titik. Fluks berubah jika $B$ berubah, luas berubah, atau loop berputar – ingat ketiga kemungkinan ini, karena masing-masing bisa menjadi soal ujian.

13.2

Electromagnetic Induction · ⁨Induksi Elektromagnetik⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 13.2.A: Describe the induced electric potential difference resulting from a change in magnetic flux.

13.2.A.1 Faraday's law describes the relationship between changing magnetic flux and the resulting induced emf in a system.
- Equation: $\mathcal{E} = -\dfrac{d\Phi_B}{dt} = -\dfrac{d\left(\vec{B} \cdot \vec{A}\right)}{dt}$
- 13.2.A.1.i When the area of the surface being considered is constant, the induced emf is equal to the area multiplied by the rate of change in the component of the magnetic field perpendicular to the surface.
- 13.2.A.1.ii When the magnetic field is constant, the induced emf is equal to the magnetic field multiplied by the rate of change in area perpendicular to the magnetic field.
- 13.2.A.1.iii When an emf is induced in a long solenoid, the total induced emf is equal to the induced emf in a single loop multiplied by the number of loops in the solenoid.
  - Equation: $\left|\mathcal{E}_{\text{sol}}\right| = N\left|\dfrac{d\Phi_B}{dt}\right|$
13.2.A.2 Lenz's law is used to determine the direction of an induced emf resulting from a changing magnetic flux.
- 13.2.A.2.i An induced emf generates a current that creates a magnetic field that opposes the change in magnetic flux.
- 13.2.A.2.ii The right-hand rule is used to determine the relationships between current, emf, and magnetic flux.
13.2.A.3 Maxwell's equations are the collection of equations that fully describe electromagnetism. Maxwell's third equation is Faraday's law of induction, which describes the relationship between a changing magnetic flux and an induced electric field.
- Equation: $\mathcal{E} = \oint \vec{E} \cdot d\vec{\ell} = -\dfrac{d\Phi_B}{dt}$
13.2.A.4 Maxwell's equations can be used to show that electric and magnetic fields obey wave equations and that electromagnetic waves travel at a constant speed in free space.
- Equation (derived): $c = \dfrac{1}{\sqrt{\varepsilon_0 \mu_0}}$

Boundary statement: AP Physics C: Electricity & Magnetism does not expect students to mathematically derive the speed of light in free space from Maxwell's equations. This relationship is included above solely as an indication of the further applications, implications, and connections to physical phenomena that students may study in more advanced physics courses.

Bahasa Indonesia

Tujuan Pembelajaran 13.2.A: Deskripsikan beda potensial listrik induksi yang timbul akibat perubahan fluks magnetik.

13.2.A.1 Hukum Faraday mendeskripsikan hubungan antara perubahan fluks magnetik dan emf induksi yang dihasilkan dalam suatu sistem.
- Persamaan: $\mathcal{E} = -\dfrac{d\Phi_B}{dt} = -\dfrac{d\left(\vec{B} \cdot \vec{A}\right)}{dt}$
- 13.2.A.1.i Ketika luas permukaan yang dipertimbangkan tetap konstan, emf induksi sama dengan luas dikalikan dengan laju perubahan komponen medan magnet yang tegak lurus terhadap permukaan.
- 13.2.A.1.ii Ketika medan magnet konstan, GGL induksi sama dengan medan magnet dikalikan dengan laju perubahan luas yang tegak lurus terhadap medan magnet.
- 13.2.A.1.iii Ketika GGL diinduksi dalam solenoida panjang, total GGL induksi sama dengan GGL induksi pada satu lingkaran dikalikan dengan jumlah lingkaran dalam solenoida.
  - Persamaan: $\left|\mathcal{E}_{\text{sol}}\right| = N\left|\dfrac{d\Phi_B}{dt}\right|$
13.2.A.2 Hukum Lenz digunakan untuk menentukan arah GGL induksi yang dihasilkan dari perubahan fluks magnet.
- 13.2.A.2.i GGL induksi menghasilkan arus yang menciptakan medan magnet yang berlawanan dengan perubahan fluks magnet.
- 13.2.A.2.ii Aturan tangan kanan digunakan untuk menentukan hubungan antara arus, GGL, dan fluks magnet.
13.2.A.3 Persamaan Maxwell adalah kumpulan persamaan yang secara penuh mendeskripsikan elektromagnetisme. Persamaan ketiga Maxwell adalah hukum induksi Faraday, yang menggambarkan hubungan antara fluks magnet yang berubah dan medan listrik terinduksi.
- Persamaan: $\mathcal{E} = \oint \vec{E} \cdot d\vec{\ell} = -\dfrac{d\Phi_B}{dt}$
13.2.A.4 Persamaan Maxwell dapat digunakan untuk menunjukkan bahwa medan listrik dan magnet mematuhi persamaan gelombang dan bahwa gelombang elektromagnetik merambat dengan kecepatan konstan di ruang hampa.
- Persamaan (turunan): $c = \dfrac{1}{\sqrt{\varepsilon_0 \mu_0}}$

Pernyataan Batas: AP Physics C: Listrik & Magnetisme tidak mengharapkan siswa menurunkan secara matematis kecepatan cahaya di ruang hampa dari persamaan Maxwell. Hubungan ini disertakan di atas semata-mata sebagai indikasi penerapan lebih lanjut, implikasi, dan koneksi ke fenomena fisika yang mungkin dipelajari siswa dalam mata pelajaran fisika yang lebih maju.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

Electromagnetic induction

A changing flux induces an emf 电动势 – Faraday's law 法拉第定律:

$$\varepsilon=-\frac{d\Phi_B}{dt}.$$

With constant area, $\varepsilon=-A\,\dfrac{dB_\perp}{dt}$; with constant field, $\varepsilon=-B\,\dfrac{dA_\perp}{dt}$. A coil of $N$ turns multiplies the single-loop emf: $|\varepsilon_{\text{sol}}|=N\left|\dfrac{d\Phi_B}{dt}\right|$.

Lenz's law 楞次定律 is the minus sign: the induced current 感应电流 flows so that its own magnetic field opposes the change in flux that created it. Push a magnet toward a loop and the loop pushes back; pull it away and the loop pulls it in. Use the right-hand rule 右手定则 to turn "oppose the change" into a current direction.

A rod of length $L$ sliding at speed $v$ across a field is the special case worth memorising – the motional emf 动生电动势 $\varepsilon=BLv$.

Worked example. A $0.20\ \text{m}$ rod slides at $3.0\ \text{m/s}$ across a $0.50\ \text{T}$ field: $\varepsilon=BLv=0.50(0.20)(3.0)=0.30\ \text{V}$. Equivalently, if a single loop's flux drops from $0.020\ \text{Wb}$ to $0.008\ \text{Wb}$ in $0.030\ \text{s}$, the average emf is $\varepsilon=\dfrac{0.012}{0.030}=0.40\ \text{V}$.

Faraday's law is also the third of Maxwell's equations 麦克斯韦方程组, in a deeper form: a changing magnetic flux creates a circulating electric field, $\oint\vec{E}\cdot d\vec{l}=-\dfrac{d\Phi_B}{dt}$ – that field is what pushes the charges around the loop. Together, Maxwell's equations predict electromagnetic waves 电磁波 travelling at $c=1/\sqrt{\varepsilon_0\mu_0}$ (you should know this connection, but AP will not ask you to derive it).

Bahasa Indonesia

Induksi elektromagnetik

Perubahan fluks menginduksi ggl – hukum Faraday:

$$\varepsilon=-\frac{d\Phi_B}{dt}.$$

Dengan luas konstan, $\varepsilon=-A\,\dfrac{dB_\perp}{dt}$; dengan medan konstan, $\varepsilon=-B\,\dfrac{dA_\perp}{dt}$. Kumparan dengan $N$ lilitan mengalikan ggl satu loop tunggal: $|\varepsilon_{\text{sol}}|=N\left|\dfrac{d\Phi_B}{dt}\right|$.

Hukum Lenz adalah tanda minus: arus terinduksi mengalir sedemikian rupa sehingga medan magnetnya sendiri melawan perubahan fluks yang menyebabkannya. Dorong magnet ke arah loop dan loop mendorong kembali; tarik menjauh dan loop menarik masuk. Gunakan aturan tangan kanan untuk mengubah "melawan perubahan" menjadi arah arus.

Memindahkan magnet masuk ke dalam kumparan menginduksi g.g.b yang menggerakkan arus

Sebuah batang sepanjang $L$ yang bergerak meluncur dengan kecepatan $v$ melintasi medan adalah kasus khusus yang perlu dihafal – yaitu gagal listrik gerak $\varepsilon=BLv$.

Contoh terpecahkan. Batang $0.20\ \text{m}$ meluncur dengan $3.0\ \text{m/s}$ melintasi medan $0.50\ \text{T}$: $\varepsilon=BLv=0.50(0.20)(3.0)=0.30\ \text{V}$. Setara, jika fluks satu loop turun dari $0.020\ \text{Wb}$ menjadi $0.008\ \text{Wb}$ dalam waktu $0.030\ \text{s}$, maka ggl rata-ratanya adalah $\varepsilon=\dfrac{0.012}{0.030}=0.40\ \text{V}$.

Hukum Faraday juga merupakan persamaan ketiga Maxwell, dalam bentuk yang lebih mendalam: perubahan fluks magnetik menciptakan medan listrik siklik, $\oint\vec{E}\cdot d\vec{l}=-\dfrac{d\Phi_B}{dt}$ – medan inilah yang mendorong muatan mengelilingi loop. Bersama-sama, persamaan Maxwell memprediksi gelombang elektromagnetik yang merambat dengan kecepatan $c=1/\sqrt{\varepsilon_0\mu_0}$ (Anda harus tahu koneksi ini, tetapi AP tidak akan meminta Anda menurunkannya).

Rotor berpilin pada motor listrik — Kumparan kawat berputar dalam medan magnet: menggerakkannya melewati medan menginduksi arus, dasar sebuah generator

Transformator gardu listrik: perubahan fluks magnet dalam kumparan menginduksi tegangan yang menyalurkan listrik ke jaringan

Explore · ⁨Jelajahi⁩

Induce a current by moving a magnet · ⁨Induksikan arus dengan menggerakkan magnet⁩

A changing magnetic flux through a coil induces an EMF (Faraday's law); its direction opposes the change (Lenz's law). Move the magnet faster for a bigger EMF. · ⁨Fluks magnet yang berubah melalui kumpalan menginduksi GGL (hukum Faraday); arahnya berlawanan dengan perubahan (hukum Lenz). Gerakkan magnet lebih cepat untuk GGL yang lebih besar.⁩

Vocabulary · ⁨Kosa kata⁩ Train · ⁨Latih⁩

English	Bahasa Indonesia
generator/ˈdʒenəreɪtə/	generator
inductance/ɪnˈdʌktəns/	induktansi
inductor/ɪnˈdʌktə/	induktor

13.3

Induced Currents and Magnetic Forces · ⁨Arus Terinduksi dan Gaya Magnetik⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 13.3.A: Describe the force exerted on a conductor due to the interaction between an external magnetic field and an induced current within that conductor.

13.3.A.1 When an induced current is created in a conductive loop, the already-present magnetic field will exert a magnetic force on the moving charge carriers within the loop.
- Equation: $\vec{F}_B = \displaystyle\int I\left(d\vec{\ell} \times \vec{B}\right)$
13.3.A.2 When current is induced in a conducting loop, magnetic forces are only exerted on the segments of the loop that are within the external magnetic field. These magnetic forces may cause translational or rotational acceleration.
13.3.A.3 The force on a conducting loop is proportional to the induced current in the loop, which depends on the rate of change of magnetic flux, the resistance of the loop, and the velocity of the loop.
13.3.A.4 Newton's second law can be applied to a conducting loop moving in a magnetic field as it experiences an induced emf.

Bahasa Indonesia

Tujuan Pembelajaran 13.3.A: Deskripsikan gaya yang bekerja pada konduktor akibat interaksi antara medan magnet eksternal dan arus terinduksi di dalam konduktor tersebut.

13.3.A.1 Ketika arus terinduksi tercipta dalam lingkar konduktif, medan magnet yang sudah ada akan memberikan gaya magnet pada pembawa muatan yang bergerak di dalam lingkar tersebut.
- Persamaan: $\vec{F}_B = \displaystyle\int I\left(d\vec{\ell} \times \vec{B}\right)$
13.3.A.2 Ketika arus diinduksi dalam lingkar konduktif, gaya magnet hanya diberikan pada segmen-segmen lingkar yang berada di dalam medan magnet eksternal. Gaya-gaya magnet ini dapat menyebabkan akselerasi translasi atau rotasi.
13.3.A.3 Gaya pada lingkar konduktif sebanding dengan arus terinduksi dalam lingkar, yang bergantung pada laju perubahan fluks magnet, resistansi lingkar, dan kecepatan lingkar.
13.3.A.4 Hukum kedua Newton dapat diterapkan pada lingkar konduktif yang bergerak dalam medan magnet saat mengalami GGL induksi.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

Once an induced current flows, the external field exerts forces on it ($\vec{F}=\int I\,d\vec{l}\times\vec{B}$) – and by Lenz's law those forces always resist the motion that causes the induction. Only the segments of the loop actually inside the field feel a force, which can make a loop accelerate, rotate, or brake. This is the origin of eddy currents 涡电流 braking, and you can apply Newton's second law to a moving loop or rod like any other mechanics problem.

The classic setup is a rod sliding on conducting rails connected by a resistor:

Worked example (the full chain). Rails $L=0.20\ \text{m}$ apart with resistance $R=0.60\ \Omega$ sit in a $0.50\ \text{T}$ field into the page. The rod is pushed at a constant $3.0\ \text{m/s}$. Then: $\varepsilon=BLv=0.30\ \text{V}$; $I=\varepsilon/R=0.50\ \text{A}$; the field pushes back on the rod with $F=BIL=0.50(0.50)(0.20)=0.050\ \text{N}$. The pushing force does work at $P=Fv=0.15\ \text{W}$ – exactly the $P=I^2R=0.15\ \text{W}$ dissipated in the resistor. Mechanical work becomes electrical energy: that is a generator 发电机, and energy is conserved. Released with no push, the rod slows exponentially: $ma=-\dfrac{B^2L^2}{R}v$.

Exam skill. FRQs walk this exact chain: flux $\to$ emf $\to$ current $\to$ force $\to$ Newton's second law. Write each link separately and check the direction with Lenz's law at the end.

Bahasa Indonesia

Setelah arus terinduksi mengalir, medan eksternal memberikan gaya padanya ($\vec{F}=\int I\,d\vec{l}\times\vec{B}$) – dan menurut hukum Lenz, gaya-gaya tersebut selalu menentang gerakan yang menyebabkan induksi. Hanya segmen loop yang benar-benar berada di dalam medan yang merasakan gaya, yang dapat membuat loop berakselerasi, berputar, atau mengerem. Ini adalah asal mula pengereman arus pusar, dan Anda dapat menerapkan hukum kedua Newton pada loop atau batang yang bergerak seperti masalah mekanika lainnya.

Arus pusar terinduksi menentang gerakan, cepat meredam pelat logam yang berayun dalam medan

Pengaturan klasik adalah batang yang meluncur di atas rel konduktif yang dihubungkan oleh resistor:

Batang meluncur di atas rel: arus terinduksi merasakan gaya yang menentang gerakan

Contoh terpecahkan (rangkaian lengkap). Rel $L=0.20\ \text{m}$ terpisah dengan hambatan $R=0.60\ \Omega$ berada dalam medan $0.50\ \text{T}$ yang mengarah ke dalam bidang. Batang didorong dengan kecepatan konstan $3.0\ \text{m/s}$. Maka: $\varepsilon=BLv=0.30\ \text{V}$; $I=\varepsilon/R=0.50\ \text{A}$; medan mendorong balik batang dengan gaya $F=BIL=0.50(0.50)(0.20)=0.050\ \text{N}$. Gaya dorong melakukan kerja pada laju $P=Fv=0.15\ \text{W}$ – tepat sebesar $P=I^2R=0.15\ \text{W}$ yang dihamburkan pada resistor. Kerja mekanis berubah menjadi energi listrik: itu adalah generator, dan energi kekal. Dilepaskan tanpa dorongan, batang melambat secara eksponensial: $ma=-\dfrac{B^2L^2}{R}v$.

Keterampilan ujian. FRQ mengikuti rantai persis ini: fluks $\to$ ggl $\to$ arus $\to$ gaya $\to$ Hukum Kedua Newton. Tulis setiap hubungan secara terpisah dan periksa arahnya dengan hukum Lenz di akhir.

13.4

Inductance · ⁨Induktansi⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 13.4.A: Describe the physical and electrical properties of an inductor.

13.4.A.1 Inductance is the tendency of a conductor to oppose a change in electrical current.
- 13.4.A.1.i Inductance of a conductor depends on the physical properties of the conductor. Straight wires are typically modeled as having zero inductance.
- 13.4.A.1.ii An inductor, such as a solenoid, is a circuit element that has significant inductance.
- 13.4.A.1.iii The inductance of a solenoid is dependent on the total number of turns, the length of the solenoid, the cross-sectional area of the solenoid, and magnetic permeability of the solenoid's core.
  - Equation: $L_{\text{sol}} = \dfrac{\mu_{\text{core}} N^2 A}{\ell}$
13.4.A.2 Inductors store energy in the magnetic field that is generated by current in the inductor.
- Equation: $U_L = \dfrac{1}{2} L I^2$
- 13.4.A.2.i The energy stored in the magnetic field generated by an inductor in which current is flowing can be dissipated through a resistor or used to charge a capacitor.
- 13.4.A.2.ii The transfer of energy generated in an inductor to other forms of energy obeys conservation laws.
13.4.A.3 By applying Faraday's law to an inductor and using the definition of inductance, induced emf can be related to inductance and the rate of change of current.
- Equation: $\mathcal{E}_i = -L\dfrac{dI}{dt}$

Bahasa Indonesia

Tujuan Pembelajaran 13.4.A: Deskripsikan sifat fisik dan elektrik dari sebuah induktor.

13.4.A.1 Induktansi adalah kecenderungan konduktor untuk melawan perubahan arus listrik.
- 13.4.A.1.i Induktansi suatu konduktor bergantung pada sifat fisik konduktor tersebut. Kawat lurus biasanya dimodelkan memiliki induktansi nol.
- 13.4.A.1.ii Sebuah induktor, seperti solenoida, adalah elemen sirkuit yang memiliki induktansi signifikan.
- 13.4.A.1.iii Induktansi solenoida bergantung pada total jumlah lilitan, panjang solenoida, luas penampang solenoida, dan permeabilitas magnetik inti solenoida.
  - Persamaan: $L_{\text{sol}} = \dfrac{\mu_{\text{core}} N^2 A}{\ell}$
13.4.A.2 Induktor menyimpan energi dalam medan magnet yang dihasilkan oleh arus dalam induktor.
- Persamaan: $U_L = \dfrac{1}{2} L I^2$
- 13.4.A.2.i Energi yang tersimpan dalam medan magnet yang dihasilkan oleh induktor di mana arus mengalir dapat didispersikan melalui resistor atau digunakan untuk mengisi kapasitor.
- 13.4.A.2.ii Transfer energi yang dihasilkan dalam induktor ke bentuk energi lain mematuhi hukum kekekalan.
13.4.A.3 Dengan menerapkan hukum Faraday pada induktor dan menggunakan definisi induktansi, GGL induksi dapat dikaitkan dengan induktansi dan laju perubahan arus.
- Persamaan: $\mathcal{E}_i = -L\dfrac{dI}{dt}$

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

Inductance 电感 is a conductor's tendency to oppose a change in its own current: changing current changes its own flux, which self-induces an emf. From Faraday's law,

$$\varepsilon=-L\frac{dI}{dt}.$$

An inductor 电感器 is a circuit element built to have large inductance – usually a solenoid 螺线管, where geometry gives

$$L_{\text{sol}}=\frac{\mu_{\text{core}}N^2A}{\ell},$$

with $N$ total turns, area $A$, length $\ell$, and the magnetic permeability 磁导率 of the core. (Straight wires are modelled as having zero inductance.) An inductor carrying current $I$ stores energy in its magnetic field:

$$U_L=\tfrac{1}{2}LI^2,$$

which can later be dissipated in a resistor or moved into a capacitor – conservation of energy applies as usual.

Bahasa Indonesia

Induktansi adalah kecenderungan konduktor untuk melawan perubahan arus miliknya: arus yang berubah mengubah fluksnya sendiri, yang menginduksi ggl sendiri. Dari hukum Faraday,

$$\varepsilon=-L\frac{dI}{dt}.$$

Sebuah induktor adalah elemen rangkaian yang dirancang untuk memiliki induktansi besar – biasanya berupa solenoida, di mana geometri memberikan

$$L_{\text{sol}}=\frac{\mu_{\text{core}}N^2A}{\ell},$$

dengan $N$ jumlah lilitan total, luas $A$, panjang $\ell$, dan permeabilitas magnetik inti. (Kawat lurus dimodelkan memiliki induktansi nol.) Induktor yang mengalirkan arus $I$ menyimpan energi dalam medan magnetnya:

$$U_L=\tfrac{1}{2}LI^2,$$

yang kemudian dapat didisipasikan pada resistor atau dipindahkan ke kapasitor – hukum kekekalan energi berlaku seperti biasa.

Vocabulary · ⁨Kosa kata⁩ Train · ⁨Latih⁩

English	Bahasa Indonesia
solenoid/ˈsəʊlənɔɪd/	solenaoid
magnetic permeability/mæɡˈnetɪk ˌpɜːməˈbɪlɪti/	Permeabilitas magnetik
RL circuit/ˌɑː ˈel ˈsɜːkɪt/	Rangkaian RL
differential equation/ˌdɪfəˈrenʃl ɪˈkweɪʒn/	persamaan diferensial
time constant/taɪm ˈkɒnstənt/	konstanta waktu
LC circuit/ˌel ˈsiː ˈsɜːkɪt/	Rangkaian LC
oscillates/ˈɒsɪleɪts/	berosilasi
simple harmonic motion/ˈsɪmpl hɑːˈmɒnɪk ˈməʊʃn/	gerak harmonik sederhana
conservation of energy/ˌkɒnsəˈveɪʃn ɒv ˈenədʒi/	hukum kekekalan energi

13.5

Circuits with Resistors and Inductors · ⁨Rangkaian dengan Resistor dan Induktor⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 13.5.A: Describe the physical and electrical properties of a circuit containing a combination of resistors and a single inductor.

13.5.A.1 A resistor will dissipate energy that was stored in an inductor as the current changes.
13.5.A.2 Kirchhoff's loop rule can be applied to a series LR circuit with a battery of emf $\mathcal{E}$, resulting in a differential equation that describes the current in the loop.
- Equation (derived): $\mathcal{E} = IR + L\dfrac{dI}{dt}$
13.5.A.3 The time constant is a significant feature of the behavior of an LR circuit.
- 13.5.A.3.i The time constant of a circuit is a measure of how quickly an LR circuit will reach a steady state and is described with the equation $\tau = \dfrac{L}{R_{\text{eq}}}$ .
- 13.5.A.3.ii The time constant represents the time an LR circuit would take to reach a steady state if the system continued to change at the initial rate of change.
- 13.5.A.3.iii For an inductor that has zero initial current, the time constant represents the time required for the current in the inductor to reach approximately 63 percent of its final asymptotic value.
- 13.5.A.3.iv For an inductor with an initial current, the time constant represents the time required for the current in the inductor to reach approximately 37 percent of its initial value.
13.5.A.4 The electric properties of inductors change during the time interval in which the current in the inductor changes, but will exhibit steady state behavior after a long time interval.
- 13.5.A.4.i When a switch is initially closed or opened in a circuit containing an inductor, the induced emf will be equal in magnitude and opposite in direction to the applied potential difference across the branch containing the inductor.
- 13.5.A.4.ii The potential difference across an inductor, the current in the inductor, and the energy stored in the inductor are exponential with respect to time and have asymptotes that are determined by the initial conditions of the circuit.
- 13.5.A.4.iii After a time much greater than the time constant of the circuit, an inductor will behave as a conducting wire with zero resistance.

Bahasa Indonesia

Tujuan Pembelajaran 13.5.A: Deskripsikan sifat fisik dan elektrik dari sirkuit yang mengandung kombinasi resistor dan satu induktor.

13.5.A.1 Resistor akan mendispersikan energi yang tersimpan dalam induktor seiring dengan perubahan arus.
13.5.A.2 Aturan loop Kirchhoff dapat diterapkan pada sirkuit LR seri dengan baterai GGL $\mathcal{E}$, menghasilkan persamaan diferensial yang mendeskripsikan arus dalam loop.
- Persamaan (turunan): $\mathcal{E} = IR + L\dfrac{dI}{dt}$
13.5.A.3 Konstanta waktu merupakan fitur penting dari perilaku sirkuit LR.
- 13.5.A.3.i Konstanta waktu dari sebuah sirkuit adalah ukuran seberapa cepat sirkuit LR akan mencapai keadaan tunak dan dijelaskan dengan persamaan $\tau = \dfrac{L}{R_{\text{eq}}}$.
- 13.5.A.3.ii Konstanta waktu mewakili waktu yang dibutuhkan sirkuit LR untuk mencapai keadaan tunak jika sistem terus berubah pada laju awal perubahan.
- 13.5.A.3.iii Untuk induktor yang memiliki arus awal nol, konstanta waktu mewakili waktu yang diperlukan agar arus dalam induktor mencapai sekitar 63 persen dari nilai asimtotik akhirnya.
- 13.5.A.3.iv Untuk induktor dengan arus awal, konstanta waktu mewakili waktu yang diperlukan agar arus dalam induktor mencapai sekitar 37 persen dari nilainya awal.
13.5.A.4 Sifat elektrik induktor berubah selama interval waktu di mana arus dalam induktor berubah, namun akan menampilkan perilaku keadaan tunak setelah interval waktu yang lama.
- 13.5.A.4.i Ketika sakelar awalnya ditutup atau dibuka dalam sirkuit yang mengandung induktor, GGL induksi akan sama besar dan berlawanan arah dengan beda potensial terapan pada cabang yang mengandung induktor.
- 13.5.A.4.ii Beda potensial pada induktor, arus dalam induktor, dan energi yang tersimpan dalam induktor bersifat eksponensial terhadap waktu dan memiliki asimtot yang ditentukan oleh kondisi awal rangkaian.
- 13.5.A.4.iii Setelah waktu yang jauh lebih besar dari konstanta waktu rangkaian, sebuah induktor akan berperilaku sebagai kawat penghantar dengan hambatan nol.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

In an RL circuit RL电路, Kirchhoff's loop rule for a battery $\varepsilon$, resistor $R$, and inductor $L$ in series gives a differential equation 微分方程:

$$\varepsilon=IR+L\frac{dI}{dt}\quad\Rightarrow\quad I(t)=\frac{\varepsilon}{R}\big(1-e^{-t/\tau}\big),\qquad \tau=\frac{L}{R}.$$

The time constant 时间常数 $\tau$ sets the pace: after one $\tau$ a rising current reaches about $63\%$ of its final value (a decaying one falls to $37\%$); it is also how long the change would take at the initial rate. Learn the two limits – they answer most conceptual questions:

Just after the switch closes ($t=0$): current cannot jump, so the inductor momentarily blocks it, self-inducing an emf equal and opposite to the applied potential difference.
Long after ($t\gg\tau$): the current is steady, $dI/dt=0$, and the inductor behaves as a plain wire – the exact opposite of a capacitor.

Worked example. $\varepsilon=12\ \text{V}$, $R=6.0\ \Omega$, $L=3.0\ \text{H}$: final current $\varepsilon/R=2.0\ \text{A}$, $\tau=L/R=0.50\ \text{s}$. At $t=0.50\ \text{s}$ the current is $2.0(1-e^{-1})\approx1.3\ \text{A}$. At $t=0$ the inductor's potential difference is the full $12\ \text{V}$; as $t\to\infty$ it falls to zero. Current, inductor voltage, and stored energy are all exponential in time.

Bahasa Indonesia

Dalam rangkaian RL RL, aturan loop Kirchhoff untuk baterai $\varepsilon$, resistor $R$, dan induktor $L$ yang dirangkai seri menghasilkan persamaan diferensial:

$$\varepsilon=IR+L\frac{dI}{dt}\quad\Rightarrow\quad I(t)=\frac{\varepsilon}{R}\big(1-e^{-t/\tau}\big),\qquad \tau=\frac{L}{R}.$$

Konstanta waktu $\tau$ menetapkan kecepatan: setelah satu $\tau$, arus yang naik mencapai sekitar $63\%$ dari nilai akhirnya (arus yang turun turun menjadi $37\%$); ini juga merupakan durasi perubahan jika terjadi pada laju awal. Pelajari dua batas limit tersebut – mereka menjawab sebagian besar pertanyaan konseptual:

Baru saja setelah sakelar ditutup ($t=0$): arus tidak bisa melompat mendadak, sehingga induktor sementara memblokir arus, menginduksi emf sendiri yang sama besar dan berlawanan arah dengan beda potensial yang diterapkan.
Lama setelah ($t\gg\tau$): arus sudah stabil, $dI/dt=0$, dan induktor berperilaku seperti kawat biasa – kebalikan sempurna dari kapasitor.

Arus dalam rangkaian RL meningkat secara eksponensial, mencapai 63% pada satu konstanta waktu

Contoh terpecahkan. $\varepsilon=12\ \text{V}$, $R=6.0\ \Omega$, $L=3.0\ \text{H}$: arus akhir $\varepsilon/R=2.0\ \text{A}$, $\tau=L/R=0.50\ \text{s}$. Pada $t=0.50\ \text{s}$ arusnya adalah $2.0(1-e^{-1})\approx1.3\ \text{A}$. Pada $t=0$ beda potensial induktor adalah penuh $12\ \text{V}$; saat $t\to\infty$ ia turun menjadi nol. Arus, tegangan induktor, dan energi tersimpan semuanya berubah secara eksponensial terhadap waktu.

Vocabulary · ⁨Kosa kata⁩ Train · ⁨Latih⁩

English	Bahasa Indonesia
Magnetic flux/mæɡˈnetɪk flʌks/	Fluks magnetik
surface integral/ˈsɜːfɪs ˈɪntɪɡrəl/	Integral permukaan
area vector/ˈeərɪə ˈvektə/	vektor luas
emf/emf/	Emf
Faraday's law/ˈfærədeɪz lɔː/	Hukum Faraday
Lenz's law/ˈlentsɪz lɔː/	Hukum Lenz
induced current/ɪnˈdjuːst ˈkʌrənt/	arus induksi
right-hand rule/raɪt hænd ruːl/	aturan tangan kanan
motional emf/ˈməʊʃənl emf/	Emf gerak
Maxwell's equations/ˈmækswelz ɪˈkweɪʒnz/	Persamaan Maxwell
electromagnetic waves/ɪˌlektrəʊməɡˈnetɪk weɪvz/	gelombang elektromagnetik
eddy currents/ˈedi ˈkʌrənts/	arus eddy

13.6

Circuits with Capacitors and Inductors · ⁨Rangkaian dengan Kapasitor dan Induktor⁩

Syllabus · ⁨Silabus⁩

English

Learning Objective 13.6.A: Describe the physical and electrical properties of a circuit containing a combination of capacitors and a single inductor.

13.6.A.1 In circuits containing only a charged capacitor and an inductor (LC circuits), the maximum current in the inductor can be determined using conservation of energy within the circuit.
13.6.A.2 In LC circuits, the time dependence of the charge stored in the capacitor can be modeled as simple harmonic motion.
- Equation (derived): $\dfrac{d^2 q}{dt^2} = -\dfrac{1}{LC} q$
13.6.A.3 The angular frequency of an oscillating LC circuit can be derived from the differential equation that describes an LC circuit.
- Equation (derived): $\omega = \dfrac{1}{\sqrt{LC}}$

Bahasa Indonesia

Tujuan Pembelajaran 13.6.A: Jelaskan sifat fisik dan listrik dari suatu rangkaian yang mengandung kombinasi kapasitor dan satu induktor tunggal.

13.6.A.1 Dalam rangkaian yang hanya berisi kapasitor bermuatan dan induktor (rangkaian LC), arus maksimum dalam induktor dapat ditentukan menggunakan hukum kekekalan energi di dalam rangkaian.
13.6.A.2 Dalam rangkaian LC, ketergantungan waktu muatan yang tersimpan dalam kapasitor dapat dimodelkan sebagai gerak harmonik sederhana.
- Persamaan (turunan): $\dfrac{d^2 q}{dt^2} = -\dfrac{1}{LC} q$
13.6.A.3 Frekuensi sudut dari rangkaian LC bergetar dapat diturunkan dari persamaan diferensial yang menggambarkan rangkaian LC.
- Persamaan (turunan): $\omega = \dfrac{1}{\sqrt{LC}}$

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

LC circuit: energy sloshes like a spring

An LC circuit LC电路 has no resistance, so nothing dissipates energy: it oscillates 振荡, sloshing energy between the capacitor's electric field and the inductor's magnetic field. The loop rule gives

$$\frac{d^2q}{dt^2}=-\frac{1}{LC}\,q,$$

the same equation as a mass on a spring – simple harmonic motion 简谐运动 with charge playing the role of displacement and

$$\omega=\frac{1}{\sqrt{LC}}.$$

The total energy $\dfrac{q^2}{2C}+\tfrac12LI^2$ stays constant: all in the capacitor at maximum charge, all in the inductor at maximum current.

Worked example. $L=2.0\ \text{H}$, $C=8.0\ \mu\text{F}$: $\omega=\dfrac{1}{\sqrt{2.0(8.0\times10^{-6})}}=250\ \text{rad/s}$, period $T=2\pi/\omega\approx0.025\ \text{s}$. If the capacitor starts charged to $12\ \text{V}$, then $U=\tfrac12CV^2=5.8\times10^{-4}\ \text{J}$, and the maximum current follows from $\tfrac12LI_{\max}^2=U$: $I_{\max}=\sqrt{2U/L}=0.024\ \text{A}$ – conservation of energy 能量守恒, no calculus needed.

Bahasa Indonesia

Rangkaian LC: energi bergelombang seperti pegas

Sebuah rangkaian LC LC tidak memiliki hambatan, sehingga tidak ada energi yang terdisipasi: rangkaian ini berosilasi, memindahkan energi antara medan listrik kapasitor dan medan magnet induktor. Aturan loop memberikan

$$\frac{d^2q}{dt^2}=-\frac{1}{LC}\,q,$$

persamaan yang sama dengan massa pada pegas – gerak harmonik sederhana dengan muatan berperan sebagai perpindahan dan

$$\omega=\frac{1}{\sqrt{LC}}.$$

Energi total $\dfrac{q^2}{2C}+\tfrac12LI^2$ tetap konstan: seluruhnya pada kapasitor saat muatan maksimum, seluruhnya pada induktor saat arus maksimum.

Contoh terpecahkan. $L=2.0\ \text{H}$, $C=8.0\ \mu\text{F}$: $\omega=\dfrac{1}{\sqrt{2.0(8.0\times10^{-6})}}=250\ \text{rad/s}$, periode $T=2\pi/\omega\approx0.025\ \text{s}$. Jika kapasitor dimulai dengan bermuatan $12\ \text{V}$, maka $U=\tfrac12CV^2=5.8\times10^{-4}\ \text{J}$, dan arus maksimum diturunkan dari $\tfrac12LI_{\max}^2=U$: $I_{\max}=\sqrt{2U/L}=0.024\ \text{A}$ – kekekalan energi, tanpa perlu kalkulus.

13.6

Exam tips · ⁨Tips ujian⁩

English

Compute flux $\Phi_B=\int \vec B\cdot d\vec A$ and get the induced EMF from Faraday's law $\varepsilon=-\tfrac{d\Phi_B}{dt}$.
Use Lenz's law (the minus sign) to fix the direction: the induced current opposes the change in flux.
Flux changes three ways — changing $B$, changing area, or changing angle — identify which and differentiate.
For a rod of length $L$ moving at speed $v$, the motional EMF is $BLv$.
An inductor stores energy $\tfrac12 LI^2$ and resists changes in current (RL time constant $\tau=L/R$).

Bahasa Indonesia

Hitung fluks magnet $\Phi_B=\int \vec B\cdot d\vec A$ dan dapatkan GGL induksi dari hukum Faraday $\varepsilon=-\tfrac{d\Phi_B}{dt}$.
Gunakan hukum Lenz (tanda minus) untuk menentukan arah: arus induksi berlawanan dengan perubahan fluks.
Fluks berubah dengan tiga cara – mengubah $B$, mengubah luas, atau mengubah sudut – identifikasi mana yang terjadi dan lakukan diferensiasi.
Untuk batang sepanjang $L$ bergerak dengan kecepatan $v$, GGL gerak adalah $BLv$.
Induktor menyimpan energi $\tfrac12 LI^2$ dan menentang perubahan arus (konstanta waktu RL $\tau=L/R$).