- understand that an electric field is an example of a field of force and define electric field as force per unit positive charge
- recall and use $F = qE$ for the force on a charge in an electric field
- represent an electric field by means of field lines
Electric fields
A-Level Physics · Topic 18
18.1
Electric fields
Syllabus
Source: Cambridge International syllabus
Lightning is a giant spark driven by a huge electric field.
An electric field 电场 is a region where a charge feels a force 力 from other charges. The electric field strength 电场强度 $E$ at a point is the force per unit positive charge on a small positive test charge 检验电荷 placed there:
$$E = \frac{F}{q}.$$
Unit: $\text{N C}^{-1}$ (the same as $\text{V m}^{-1}$, as we will see). $E$ is a vector 矢量, pointing the way the force acts on a positive charge. The force on a charge $q$ is
$$F = qE,$$
opposite to the field if $q$ is negative.
Field lines
- field lines 场线 point the way the force acts on a positive test charge.
- lines start on positive charges and end on negative charges (or go to infinity).
- lines never cross; closer lines mean a stronger field.
Examples: a positive point charge 点电荷 has radial lines pointing out; a negative one has lines pointing in; two opposite charges (a dipole 偶极子) have lines curving from + to −; two parallel charged plates give a uniform field 匀强场 of equally spaced parallel lines.
Field-line patterns for parallel plates, a dipole, a point charge, and a charged sphere above an earthed plate
A Van de Graaff generator stores a large static charge on its metal dome, making a strong electric field around it
Explore
Electric fields
E ∝ Q / r²
A charge sets up a radial field — out for +, in for −, obeying the inverse-square law.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| electric field | 电场 | diàn chǎng |
| force | 力 | lì |
| electric field strength | 电场强度 | diàn chǎng qiáng dù |
| test charge | 检验电荷 | jiǎn yàn diàn hè |
| vector | 矢量 | shǐ liàng |
| field line | 场线 | chǎng xiàn |
| point charge | 点电荷 | diǎn diàn hè |
| dipole | 偶极子 | ǒu jí zi |
| uniform field | 匀强场 | yún qiáng chǎng |
18.2
Uniform electric fields
Syllabus
- recall and use $E = \Delta V / \Delta d$ to calculate the field strength of the uniform field between charged parallel plates
- describe the effect of a uniform electric field on the motion of charged particles
Source: Cambridge International syllabus
Between two parallel plates a distance $d$ apart with potential difference 电势差 $V$ between them, the field is uniform (apart from edge effects) with size
Between parallel plates the field is uniform, E = V/d
$$E = \frac{V}{d}.$$
It points from the higher-potential plate to the lower one. The unit $\text{V m}^{-1}$ comes straight from this and equals $\text{N C}^{-1}$.
Worked example. Two parallel plates $5.0\ \text{mm}$ apart have a p.d. of $200\ \text{V}$ between them. Find the field strength, and the force on an electron in the gap. ($e = 1.6 \times 10^{-19}\ \text{C}$.)
$$E = \frac{V}{d} = \frac{200}{5.0 \times 10^{-3}} = 4.0 \times 10^{4}\ \text{V m}^{-1}, \qquad F = qE = (1.6 \times 10^{-19})(4.0 \times 10^{4}) = 6.4 \times 10^{-15}\ \text{N}.$$
A charged particle in a uniform field
A charge $q$ in a uniform field feels a constant force $F = qE$, so a constant acceleration $a = qE/m$ — just like a mass in a uniform gravitational field.
- released at rest, it speeds up along the field (positive charge) or against it (negative charge), gaining kinetic energy 动能.
- entering at right angles to the field, it follows a parabolic 抛物线 path — like a projectile 抛体 in gravity. This is how a cathode-ray tube 阴极射线管 used to steer its beam.
A charge entering a uniform field at right angles follows a parabolic path, like a projectile
Explore
Uniform electric field lab
Follow how a charge behaves between parallel plates.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| potential difference | 电势差 | diàn shì chà |
| kinetic energy | 动能 | dòng néng |
| parabolic | 抛物线 | pāo wù xiàn |
| projectile | 抛体 | pāo tǐ |
| cathode-ray tube | 阴极射线管 | yīn jí shè xiàn guǎn |
18.3
Coulomb's law
Syllabus
- understand that, for a point outside a spherical conductor, the charge on the sphere may be considered to be a point charge at its centre
- recall and use Coulomb’s law $F = Q_1Q_2 / (4\pi\varepsilon_0 r^2)$ for the force between two point charges in free space
Source: Cambridge International syllabus
For two point charges $Q_{1}$ and $Q_{2}$ a distance $r$ apart in free space, each feels a force of size
$$F = \frac{Q_{1} Q_{2}}{4\pi\varepsilon_{0} r^{2}}.$$
This is Coulomb's law 库仑定律. Here $\varepsilon_{0} = 8.85 \times 10^{-12}\ \text{F m}^{-1}$ is the permittivity of free space 真空电容率, and $1/(4\pi\varepsilon_{0}) \approx 9.0 \times 10^{9}\ \text{N m}^{2}\ \text{C}^{-2}$. The force is along the line joining the charges: repulsive for like charges, attractive for opposite charges.
Like charges repel; opposite charges attract — the force is along the line joining them
Worked example. Find the electrostatic force between point charges of $+2.0\ \text{nC}$ and $+3.0\ \text{nC}$ placed $4.0\ \text{cm}$ apart. ($1/(4\pi\varepsilon_{0}) = 9.0 \times 10^{9}\ \text{N m}^{2}\ \text{C}^{-2}$.)
$$F = \frac{Q_{1}Q_{2}}{4\pi\varepsilon_{0} r^{2}} = \frac{(9.0 \times 10^{9})(2.0 \times 10^{-9})(3.0 \times 10^{-9})}{(0.040)^{2}} \approx 3.4 \times 10^{-5}\ \text{N (repulsive).}$$
Spheres treated as point charges
A spherical conductor 导体 with total charge $Q$ gives, at any point outside, the same field as a point charge $Q$ at its centre (measure $r$ from the centre). Inside a hollow charged conductor the field is zero, so the conductor is an equipotential 等势面.
Explore
Coulomb's law
F ∝ Qq / r²
Like charges repel, unlike attract — and the force follows 1/r².
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| Coulomb's law | 库仑定律 | kù lún dìng lǜ |
| permittivity of free space | 真空电容率 | zhēn kōng diàn róng lǜ |
| conductor | 导体 | dǎo tǐ |
| equipotential | 等势面 | děng shì miàn |
18.4
Electric field due to a point charge
Syllabus
- recall and use $E = Q / (4\pi\varepsilon_0 r^2)$ for the electric field strength due to a point charge in free space
Source: Cambridge International syllabus
Coulomb's law gives the force between two charges. Divide it by the test charge ($E = F/q$) and you are left with the field of the source charge $Q$ alone. The field at distance $r$ from a point charge $Q$ is
$$E = \frac{Q}{4\pi\varepsilon_{0} r^{2}}.$$
It points out from a positive $Q$, in towards a negative $Q$, and falls as $1/r^{2}$ — just like gravitational field 重力场 strength, except gravity is always attractive. For several charges, add the fields as a vector sum 矢量和.
Explore
Field around a point charge
Change the charge. Field lines point away from positive and toward negative, and crowd together where the field is strongest.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| gravitational field | 重力场 | zhòng lì chǎng |
| vector sum | 矢量和 | shǐ liàng hé |
18.5
Electric potential
Syllabus
- define electric potential at a point as the work done per unit positive charge in bringing a small test charge from infinity to the point
- recall and use the fact that the electric field at a point is equal to the negative of potential gradient at that point
- use $V = Q / (4\pi\varepsilon_0 r)$ for the electric potential in the field due to a point charge
- understand how the concept of electric potential leads to the electric potential energy of two point charges and use $E_P = Qq / (4\pi\varepsilon_0 r)$
Source: Cambridge International syllabus
Electric potential 电势 $V$ at a point is the work done per unit positive charge in bringing a small positive test charge from infinity 无穷远 to that point:
$$V = \frac{W}{q}.$$
Unit: $\text{V}$. The potential is zero at infinity. For a positive source charge $V > 0$ everywhere outside; for a negative source charge $V < 0$.
Potential due to a point charge
$$V = \frac{Q}{4\pi\varepsilon_{0} r}.$$
Note the $1/r$ here (compared with $1/r^{2}$ for the field). $V$ is a scalar 标量; for several charges, add the potentials (with sign).
Worked example. Find the electric potential $4.0\ \text{cm}$ from a point charge of $+3.0\ \text{nC}$. ($1/(4\pi\varepsilon_{0}) = 9.0 \times 10^{9}\ \text{N m}^{2}\ \text{C}^{-2}$.)
$$V = \frac{Q}{4\pi\varepsilon_{0} r} = \frac{(9.0 \times 10^{9})(3.0 \times 10^{-9})}{0.040} \approx 6.8 \times 10^{2}\ \text{V}.$$
Because potential is a scalar, the potential from several charges is simply their sum (each with its own sign) — no directions to resolve.
The potential near a point charge varies as 1/r — positive for a positive charge, negative for a negative one
Link between field and potential
The field equals the negative potential gradient 电势梯度:
$$E = -\frac{dV}{dx}.$$
Between parallel plates $V$ changes evenly with position, giving $E = V/d$ as before. The minus sign means the field points towards lower potential. For a point charge, $-\dfrac{dV}{dr} = \dfrac{Q}{4\pi\varepsilon_{0} r^{2}} = E$.
In a uniform field the potential falls steadily with distance, so the field strength $V/d$ is constant
Electric potential energy
A charge $q$ at a point of potential $V$ has electric potential energy 电势能 $E_{\text{P}} = qV$. For two point charges $Q$ and $q$ a distance $r$ apart:
$$E_{\text{P}} = \frac{Qq}{4\pi\varepsilon_{0} r}.$$
- like charges: $E_{\text{P}} > 0$ — stored energy that would be released if they flew apart.
- opposite charges: $E_{\text{P}} < 0$ — a bound 束缚 system; energy must be supplied to separate them.
In both cases $E_{\text{P}} \to 0$ as $r \to \infty$.
Electric PE of two charges: positive for like charges, a negative well for opposite charges
Worked-example pattern
An electron 电子 orbits a nucleus 原子核 of charge $+Ze$ at distance $r$. The Coulomb attraction provides the centripetal force 向心力:
$$\frac{Z e^{2}}{4\pi\varepsilon_{0} r^{2}} = \frac{m_{e} v^{2}}{r}, \qquad v = \sqrt{\frac{Z e^{2}}{4\pi\varepsilon_{0} m_{e} r}}.$$
The total energy is kinetic plus potential:
$$E_{\text{total}} = \tfrac{1}{2} m_{e} v^{2} - \frac{Z e^{2}}{4\pi\varepsilon_{0} r} = -\frac{Z e^{2}}{8\pi\varepsilon_{0} r},$$
which is negative (a bound state).
Gravitational versus electric
The two field theories look alike:
| Quantity | Gravitational | Electric |
|---|---|---|
| Source | mass $M$ (always positive) | charge $Q$ (can be ±) |
| Field strength | $g = GM/r^{2}$ | $E = Q/(4\pi\varepsilon_{0} r^{2})$ |
| Force on test object | $F = mg$ | $F = qE$ |
| Potential | $\phi = -GM/r$ | $V = Q/(4\pi\varepsilon_{0} r)$ |
| PE of two | $-GMm/r$ | $Qq/(4\pi\varepsilon_{0} r)$ |
| Nature | always attractive | attractive or repulsive |
The minus sign in the gravitational potential 引力势 reflects that gravity is always attractive; the electric potential takes the sign of the source charge.
Explore
Electric potential
V = kQ / r
Potential ∝ 1/r around a charge — steep near it, flattening out.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| electric potential | 电势 | diàn shì |
| infinity | 无穷远 | wú qióng yuǎn |
| scalar | 标量 | biāo liàng |
| potential gradient | 电势梯度 | diàn shì tī dù |
| electric potential energy | 电势能 | diàn shì néng |
| bound | 束缚 | shù fù |
| electron | 电子 | diàn zi |
| nucleus | 原子核 | yuán zǐ hé |
| centripetal force | 向心力 | xiàng xīn lì |
| gravitational potential | 引力势 | yǐn lì shì |
18.5
Exam tips
- Coulomb's law $F = Q_1 Q_2 / 4\pi\varepsilon_0 r^2$ (inverse-square, but can attract or repel).
- Distinguish a uniform field ($E = V/d$, between plates) from a radial field ($E = Q/4\pi\varepsilon_0 r^2$).
- Electric potential $V = Q/4\pi\varepsilon_0 r$ (sign follows the charge); the field points from high to low potential.