电力、电场与电势

Electric charge 电荷 is a fundamental property 基本属性 of matter, and it comes in two kinds, positive and negative; like charges 同种电荷 repel 排斥 and opposite charges 异种电荷 attract 吸引. Charge is conserved 守恒 and quantized 量子化 (a multiple of the elementary charge 元电荷 $e$). The force between two point charges 点电荷 is Coulomb's law 库仑定律:

directed along the line joining them – an inverse-square law 平方反比定律 like gravity, but it can push or pull.

The constant $k$ hides a property of the medium: the permittivity 介电常数. Free space has a fixed permittivity of free space 真空介电常数 $\varepsilon_0$ (with $k = 1/4\pi\varepsilon_0$), while the permittivity of matter differs from $\varepsilon_0$, depending on the material's composition and arrangement - which is why a material between two charges changes the force. Note also that although the electric force is vastly stronger than gravity, gravity dominates at large scales, because large objects are usually electrically neutral 中性 (equal + and −), leaving only gravity to act.

Worked example. Two point charges, $+3.0\ \mu\text{C}$ and $-2.0\ \mu\text{C}$, sit $0.10\ \text{m}$ apart ($k=9.0\times10^{9}$). The force between them is

attractive, because the charges have opposite signs. (Use the magnitudes for the size and decide the direction from the signs.)

Objects charge by moving electrons. A conductor 导体 lets charge move freely; an insulator 绝缘体 holds it in place. Three methods:

• Friction 摩擦起电: rubbing transfers electrons.
• Conduction 接触起电: touching shares charge.
• Induction 感应起电: a nearby charge rearranges charge in a neutral object, which can then be grounded 接地 to leave it charged.

An electric field 电场 $\vec{E}$ is the force per unit charge that a small positive test charge 试探电荷 would feel:

Electric field lines 电场线 point away from positive charges and toward negative ones; where they are denser, the field is stronger. A charge in a field feels $\vec{F}=q\vec{E}$.

Worked example. Find the electric field $0.20\ \text{m}$ from a $+5.0\ \mu\text{C}$ point charge: $E=\dfrac{kQ}{r^2}=\dfrac{9.0\times10^{9}\times5.0\times10^{-6}}{(0.20)^2}=1.1\times10^{6}\ \text{N/C}$, pointing away from the charge. A $+2\ \text{nC}$ charge placed there would feel $F=q\,E=2\times10^{-9}\times1.1\times10^{6}=2.2\times10^{-3}\ \text{N}$.

Worked example (adding fields). A $+4.0\ \mu\text{C}$ charge sits at $x = 0$ and a $-1.0\ \mu\text{C}$ charge at $x = 0.30\ \text{m}$. Find (a) the field at the midpoint M and (b) the point on the line where the field is zero.

(a) Each charge is $0.15\ \text{m}$ from M.

The field of the positive charge points away from it ($+x$); the field of the negative charge points toward it, also $+x$. They add: $E = 2.0\times10^6\ \text{N/C}$ in the $+x$ direction.

(b) Between the charges both fields point the same way, so they cannot cancel there. To the left of $x = 0$ the larger charge is always closer, so its field always wins. The zero must be to the right of the smaller charge, at some $x > 0.30\ \text{m}$, where the two fields point in opposite directions:

The zero-field point is always nearer the smaller charge, outside the pair when the charges have opposite signs.

Worked example (uniform field between plates). Two parallel plates 平行板 $0.020\ \text{m}$ apart are held at a potential difference 电势差 of $500\ \text{V}$. The field between them is uniform 均匀的:

An electron between the plates feels $F = qE = (1.6\times10^{-19})(2.5\times10^4) = 4.0\times10^{-15}\ \text{N}$, toward the positive plate, the same anywhere in the gap. (One $\text{V/m}$ is the same as one $\text{N/C}$.)

Two charges have electric potential energy 电势能 stored in their arrangement:

Like charges pushed together store positive energy; opposite charges have negative energy (bound). Moving a charge changes $U$, and the electric force does work equal to $-\Delta U$.

Electric potential 电势 $V$ is the potential energy per unit charge – a scalar field, measured in volts:

The potential difference (voltage 电压) between two points is the work per unit charge to move between them: $\Delta V=\dfrac{\Delta U}{q}$, and $U=qV$. Positive charges move from high to low potential on their own. Because potential is a scalar 标量, adding the potentials from several charges is much easier than adding field vectors 矢量.

Worked example. For the two charges in the field example above, the potential at the midpoint M is

Keep the sign of each charge and add; there are no directions to worry about. The field at M is not zero, but that tells you nothing about $V$: field and potential are different quantities.

A capacitor 电容器 stores charge and energy on two conductors separated by a gap. Its capacitance 电容 relates charge to voltage:

and the stored energy is $U=\tfrac{1}{2}CV^2$. Capacitance depends on the plates' geometry 几何形状 and the material between them, not on the charge placed on it.

Worked example. A $100\ \mu\text{F}$ capacitor is charged to $12\ \text{V}$. It holds $Q=CV=100\times10^{-6}\times12=1.2\times10^{-3}\ \text{C}$ of charge and stores $U=\tfrac12 CV^2=\tfrac12\times100\times10^{-6}\times12^2=7.2\times10^{-3}\ \text{J}$ of energy.

Energy is conserved for charges just as for masses. A charge released in a field converts electric potential energy into kinetic energy 动能:

Worked example. An electron ($q=1.6\times10^{-19}\ \text{C}$, $m=9.1\times10^{-31}\ \text{kg}$) is accelerated from rest through a potential difference of $100\ \text{V}$. Its final speed is

This is exactly how the electron gun 电子枪 in an old television or an electron microscope works.

Answer each before reading the answers below. Take $k=9.0\times10^{9}\ \text{N m}^2/\text{C}^2$ and $e=1.6\times10^{-19}\ \text{C}$.

1. Two $+2.0\ \mu\text{C}$ charges are $0.30\ \text{m}$ apart. Find the force between them, and say whether it attracts or repels.
2. The distance in question 1 is tripled. Find the new force without starting again.
3. Find the electric field $0.50\ \text{m}$ from a $-8.0\ \text{nC}$ point charge, with its direction.
4. Two parallel plates $0.010\ \text{m}$ apart have a potential difference of $200\ \text{V}$. Find the field between them, the force on a proton there, and the kinetic energy a proton gains crossing from the positive plate to the negative plate.
5. A student says: "At a point where the electric potential is zero, the electric field must be zero too." Correct the student.

1. $F=\dfrac{kq_1q_2}{r^2}=\dfrac{(9.0\times10^{9})(2.0\times10^{-6})^2}{0.30^2}=0.40\ \text{N}$, repulsive (like charges).
2. $F\propto1/r^2$, so tripling $r$ divides $F$ by 9: $0.40/9=0.044\ \text{N}$.
3. $E=\dfrac{kQ}{r^2}=\dfrac{(9.0\times10^{9})(8.0\times10^{-9})}{0.50^2}=2.9\times10^{2}\ \text{N/C}$, pointing toward the charge, because it is negative.
4. $E=\dfrac{\Delta V}{d}=\dfrac{200}{0.010}=2.0\times10^{4}\ \text{V/m}$. $F=eE=(1.6\times10^{-19})(2.0\times10^{4})=3.2\times10^{-15}\ \text{N}$. Energy gained $=e\,\Delta V=(1.6\times10^{-19})(200)=3.2\times10^{-17}\ \text{J}$.
5. Potential and field are different quantities. Midway between equal charges $+q$ and $-q$, the two potentials cancel, so $V=0$. But both fields there point from $+q$ toward $-q$ and add, so $E$ is not zero. The field depends on how fast $V$ changes with position, not on the value of $V$.

• Coulomb's law and the point-charge field are inverse-square — doubling the separation makes the force (or field) four times smaller.
• Use magnitudes for the size of a force and decide direction from the signs; the field points the way a positive test charge would move.
• Potential ($V$) is a scalar so potentials from several charges simply add; the field ($E$) is a vector and must be added by direction.
• For a charge accelerated through a voltage use energy conservation $qV=\tfrac12 mv^2$.
• Capacitor relations: $C=Q/V$ (capacitance is fixed by geometry) and energy $U=\tfrac12 CV^2$.

Common mistakes

• Adding field magnitudes without thinking about direction. Draw each field at the point first; fields from a $+$ and a $-$ charge point the same way between them.
• Putting the sign of a charge into $E = kQ/r^2$ and then also adding a direction. Use magnitudes for size, then choose the direction from the sign.
• Using $1/r$ for the field or $1/r^2$ for the potential. Field and force go as $1/r^2$; potential and potential energy as $1/r$.
• Saying that where $E = 0$ the potential must be zero (or the reverse). They are independent: at M above, $E \ne 0$ and $V \ne 0$; at the midpoint of two equal opposite charges $V = 0$ but $E \ne 0$.
• Forgetting the micro- prefix: $1\ \mu\text{C} = 10^{-6}\ \text{C}$, $1\ \text{nC} = 10^{-9}\ \text{C}$.
• Thinking capacitance changes when the charge or voltage changes. $C$ is set by the plates and the material between them; $Q$ and $V$ change together.