Conductors and Capacitors

When a person's heart stops beating in a regular rhythm, a defibrillator can restart it. It charges a capacitor for a few seconds, then dumps about 200 joules through the chest in a few milliseconds. A battery could never deliver energy that fast; a capacitor can, because it stores energy in the electric field between two conductors. This unit starts with how charge behaves on conductors and builds up to capacitors. 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. This is why the metal body of a car protects the people inside from lightning, and why your phone loses signal inside a lift.

The graphs show all four properties at once for a sphere of radius $R$ carrying $+Q$. Outside, Gauss's law gives the point-charge field $E=kQ/r^2$, so $V=kQ/r$. Inside, $E=0$, so $V$ cannot change: it keeps its surface value $kQ/R$ all the way to the centre. A common exam error is to draw $V$ falling to zero inside.

Worked example (a sphere inside a shell). A solid metal sphere of radius $a$ carries $+Q$. Around it, with the same centre, is a neutral metal shell with inner radius $b$ and outer radius $c$. Use Gauss's law in each region:

• $r<a$ (inside the sphere's metal): $E=0$.
• $a<r<b$ (the gap): the Gaussian sphere encloses $+Q$, so $E=kQ/r^2$.
• $b<r<c$ (inside the shell's metal): $E=0$, so the Gaussian sphere must enclose zero net charge. The shell's inner surface therefore carries $-Q$. The shell is neutral, so its outer surface carries $+Q$.
• $r>c$: the total enclosed is $+Q$, so $E=kQ/r^2$ again.

If the shell is grounded, the $+Q$ on its outer surface flows away to Earth, and the field outside becomes zero. The $-Q$ on the inner surface stays, held by the attraction of the sphere.

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.

Worked example (charging by induction). A neutral metal sphere stands on an insulating base.

1. Bring a negatively charged rod near, without touching. The sphere's electrons are pushed to the far side; the near side is left positive. The sphere is still neutral overall.
2. Touch the far side to ground. The repelled electrons flow away to the ground.
3. Remove the ground connection first, while the rod is still near.
4. Now take the rod away. The sphere is left with a net positive charge, which spreads evenly over its surface.

The induced charge always has the opposite sign to the nearby charge. If you removed the rod before breaking the ground connection, electrons would flow back and the sphere would end up neutral.

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:

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

This $E\to\Delta V\to C$ chain is a standard FRQ derivation – learn it as three steps, and quote each one.

Worked derivation (concentric spheres). An inner sphere of radius $a$ carries $+Q$ and an outer shell of radius $b$ carries $-Q$. Between them, a spherical Gaussian surface encloses only $+Q$, so $E=kQ/r^2$. Then

When the gap $b-a$ is small compared with $a$, the area is about $4\pi a^2$ and this becomes $\varepsilon_0A/d$ again, which is a good check. 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,

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: $6\ \mu\text{F}$.

Exam skill (measuring a capacitance). For a parallel-plate capacitor $C=\kappa\varepsilon_0A/d$, so a graph of $C$ against $1/d$ is a straight line through the origin with slope $\kappa\varepsilon_0A$. To find the dielectric constant of paper, put sheets of paper between two metal plates, measure $C$ for different numbers of sheets, and plot $C$ against $1/d$: then $\kappa=\dfrac{\text{slope}}{\varepsilon_0A}$. Larger plates make a steeper line.

Worked example (the defibrillator). A defibrillator must store 200 J at 2000 V. The capacitance needed is

holding $Q=C\Delta V=0.20$ C. If the pulse lasts 5 ms, the average power is $200/0.005=40{,}000$ W, far more than a battery of the same size could supply.

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:

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:

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

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.

Worked example (a half-filled gap). A slab of dielectric $\kappa$ fills the lower half of the gap $d$ of a parallel-plate capacitor of area $A$. The field passes through the slab and then the air, one after the other, so treat them as two capacitors in series, each with gap $d/2$:

For $\kappa=3$, $C=1.5\,C_0$, less than the $3C_0$ of a full slab. If instead the slab filled the left half of the plate area across the whole gap, the two parts would share the same $\Delta V$: they are in parallel, and $C=\tfrac{\kappa+1}{2}C_0=2C_0$.

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 $U=\tfrac12 C(\Delta V)^2$.

Worked example (a dielectric that changes across the gap). A parallel-plate capacitor with area $A$ and gap $d$ is filled with a material whose dielectric constant rises steadily from 1 at one plate to 2 at the other: $\kappa(x)=1+x/d$. Cut the gap into thin slices of thickness $dx$. Each slice is a capacitor $\dfrac{\kappa\varepsilon_0A}{dx}$, and the slices are in series, so their reciprocals add:

That lies between the empty value ($\kappa=1$) and the value for $\kappa=2$ throughout, as it must.

• 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.

Common mistakes

• Not stating what stays fixed when a dielectric goes in. Isolated: $Q$ is fixed and $V$ falls. Connected to a battery: $V$ is fixed and $Q$ rises.
• Using the resistor rules for capacitors. Capacitors in parallel add; in series the reciprocals add.
• Saying the field inside a conductor in equilibrium is small. It is exactly zero, and all excess charge sits on the surface.
• Using $U = QV$ for a charged capacitor. The stored energy is $\tfrac12 QV = \tfrac12 CV^2$.
• Drawing the potential inside a charged conductor as zero. It is constant and equal to the surface value.
• Saying charge spreads so that touching spheres hold equal charges. They reach equal potentials; the larger sphere holds more charge.