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Electric Charges, Fields, and Gauss's Law

AP Physics C: Electricity and Magnetism · Topic 8

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8.1

Electric Charge and Electric Force

Syllabus
Learning ObjectiveEssential Knowledge

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.

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.

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.

Source: College Board AP Course and Exam Description

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.

Explore

Explore the field of a source charge

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.

Vocabulary Train
English Chinese Pinyin
Electric charge 电荷 diàn hè
quantized 量子化 liàng zǐ huà
elementary charge 基本电荷 jī běn diàn hè
conservation of charge 电荷守恒 diàn hè shǒu héng
point charges 点电荷 diǎn diàn hè
Coulomb's law 库仑定律 kù lún dìng lǜ
inverse-square 平方反比 píng fāng fǎn bǐ
superposition 叠加 dié jiā
fundamental property 基本属性 jī běn shǔ xìng
permittivity of free space 真空介电常数 zhēn kōng jiè diàn cháng shù
8.2

Electric Charge and the Process of Charging

Syllabus
Learning ObjectiveEssential 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

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.

Explore

Explore charging by rubbing

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.

Vocabulary Train
English Chinese Pinyin
conductor 导体 dǎo tǐ
insulator 绝缘体 jué yuán tǐ
Charging by friction 摩擦起电 mó cā qǐ diàn
Charging by conduction 传导起电 chuán dǎo qǐ diàn
Charging by induction 感应起电 gǎn yìng qǐ diàn
ground 接地 jiē dì
polarization 极化 jí huà
electroscope 验电器 yàn diàn qì
8.3

Electric Fields

Syllabus
Learning ObjectiveEssential Knowledge

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.

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.

Source: College Board AP Course and Exam Description

The electric field of a dipole

A Tesla coil topped by a metal ring, throwing long purple sparks out into the air A Tesla coil throws sparks: it raises the voltage so high that the electric field ionizes the surrounding air, and charge leaps across as a visible discharge

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.

Electric field-line patterns for parallel plates, a dipole, and a point charge Electric field-line patterns for parallel plates, a dipole, and a point charge

Electric field lines of a dipole point from the positive to the negative charge Electric field lines of a dipole point from the positive to the negative charge

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.

A charge crossing a uniform field follows a parabolic path A charge crossing a uniform field follows a parabolic path

A person touching a Van de Graaff generator with their hair standing up Charge spreads over the dome and onto the hair; like charges repel, so the strands push apart

Explore

Explore field lines of a point charge

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.

Vocabulary Train
English Chinese Pinyin
electric field 电场 diàn chǎng
test charge 检验电荷 jiǎn yàn diàn hè
Field lines 电场线 diàn chǎng xiàn
uniform 均匀 jūn yún
Exercise sheet
8.4

Electric Fields of Charge Distributions

Syllabus
Learning ObjectiveEssential Knowledge

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.

Source: College Board AP Course and Exam Description

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.

Vocabulary Train
English Chinese Pinyin
continuous charge distribution 连续电荷分布 lián xù diàn hè fēn bù
linear charge density 线电荷密度 xiàn diàn hè mì dù
surface charge density 面电荷密度 miàn diàn hè mì dù
volume charge density 体电荷密度 tǐ diàn hè mì dù
symmetry 对称性 duì chèn xìng
Exercise sheet
8.5

Electric Flux

Syllabus
Learning ObjectiveEssential Knowledge

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

Source: College Board AP Course and Exam Description

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.

Electric flux through a surface depends on the angle between the field and the surface normal Electric flux through a surface depends on the angle between the field and the surface normal

Vocabulary Train
English Chinese Pinyin
Electric flux 电通量 diàn tōng liàng
area vector 面积矢量 miàn jī shǐ liàng
8.6

Gauss's Law

Syllabus
Learning ObjectiveEssential Knowledge

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.

Source: College Board AP Course and Exam Description

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").

A Gaussian surface is chosen to match the symmetry of the charge A Gaussian surface is chosen to match the symmetry of the charge

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.

Vocabulary Train
English Chinese Pinyin
Gauss's law 高斯定律 gāo sī dìng lǜ
Gaussian surface 高斯面 gāo sī miàn
spherical symmetry 球对称 qiú duì chèn
cylindrical symmetry 柱对称 zhù duì chèn
planar symmetry 平面对称 píng miàn duì chèn
8.6

Exam tips

  • 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.

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