Capacitors (AP Physics C: Electricity and Magnetism 10.3)

A camera flash needs a burst of energy far faster than its battery can deliver it. The fix is to spend a few seconds pulling electrons off one metal plate and piling them onto another, then let them rush back in a millisecond. Two conductors holding equal and opposite charge make a capacitor, and the energy is stored in the field between them.

Capacitance

A capacitor is any pair of conductors, separated by an insulator, that carry charges $+Q$ and $-Q$ . The simplest is a parallel-plate capacitor: two flat metal plates of area $A$ facing each other a distance $d$ apart. Connecting the plates to a battery moves electrons from one plate to the other through the battery, so the two plates always end up with equal amounts of opposite charge. When people say a capacitor "holds charge $Q$ ," they mean $+Q$ on one plate and $-Q$ on the other. The capacitor's net charge is zero.

Separated charge creates a potential difference $\Delta V$ between the conductors. Double the charge on each plate and the field everywhere doubles (superposition), so $\Delta V = \int E\,d\ell$ doubles too. Charge and potential difference are proportional, and the constant of proportionality is the capacitance.

$C = \frac{Q}{\Delta V}$ looks like it says "more charge means less capacitance," but it does not: adding charge raises $\Delta V$ in the same proportion, and the ratio stays fixed. Capacitance measures how much charge the device stores per volt. A farad is enormous: plates $1.0\ \text{mm}$ apart in air would need an area of about $10^{8}\ \text{m}^2$ to reach $1\ \text{F}$ . Lab capacitors are measured in microfarads ( $\mu\text{F}$ ) and picofarads ( $\text{pF}$ ).

The field between parallel plates

Treat each plate as a large sheet with surface charge density $\sigma = \frac{Q}{A}$ . Gauss's law (topic 8.6) gives each sheet's own field as $\frac{\sigma}{2\varepsilon_0}$ , pointing away from the positive plate and toward the negative plate. Between the plates both fields point from $+$ to $-$ , so they add; outside, they point opposite ways and cancel:

E_{\text{between}} = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}, \qquad E_{\text{outside}} = 0.

The same answer comes from the conductor result of topic 10.1: the plates' charges attract each other onto the inner faces, and the field just outside a conductor's surface is $\frac{\sigma}{\varepsilon_0}$ . The field between the plates is uniform, the same magnitude and direction everywhere, and it is proportional to the surface charge density. Near the edges it bulges outward and weakens. This fringing is negligible when $d$ is much smaller than the plates' width, which is the condition for all the formulas here.

Side view of a parallel-plate capacitor with plate separation d. The top plate carries plus Q on its inner face and the bottom plate minus Q on its inner face. Between the plates, evenly spaced straight field lines point down from the positive plate to the negative plate, labeled E equals sigma over epsilon zero. At each edge, a few field lines bow outward beyond the plates. The region outside the plates is labeled E approximately zero.

Now derive the capacitance. A uniform field gives $\Delta V = Ed$ , so

\Delta V = \frac{\sigma}{\varepsilon_0}d = \frac{Qd}{\varepsilon_0 A} \quad\Rightarrow\quad C = \frac{Q}{\Delta V} = \frac{\varepsilon_0 A}{d}.

Larger plates hold more charge at the same field, so $C \propto A$ . Wider spacing makes the same field produce a larger $\Delta V$ , so $C \propto \frac{1}{d}$ . For example, square plates $14\ \text{cm}$ on a side ( $A = (0.14\ \text{m})^2 = 0.0196\ \text{m}^2$ ) $1.0\ \text{mm}$ apart in air have $C = \frac{(8.85\times10^{-12})(0.0196)}{1.0\times10^{-3}} = 1.73\times10^{-10}\ \text{F} = 173\ \text{pF}$ .

Charges moving between the plates

A uniform field exerts a constant force $q\vec{E}$ on a charged particle between the plates, so the particle has constant acceleration $\vec{a} = \frac{q\vec{E}}{m}$ . That is the same situation as a ball thrown near the Earth's surface: a particle that enters moving parallel to the plates follows a parabola, with constant velocity along the plates and uniformly accelerated motion across them. Every projectile equation from mechanics applies, with $\frac{qE}{m}$ in place of $g$ . A positive charge accelerates toward the negative plate, a negative charge toward the positive plate. Gravity on an electron or proton is negligible next to the electric force, by many orders of magnitude in any realistic capacitor.

Energy offers a shortcut when you only need speeds. A charge $q$ moving from one plate to the other changes its potential energy by $q\,\Delta V$ (Unit 9), so a proton released from rest at the positive plate reaches the negative plate with kinetic energy $e\,\Delta V$ , whatever the plate spacing.

Spherical and cylindrical capacitors

Beyond parallel plates, the AP Physics C course covers two other shapes quantitatively (and no others): a spherical capacitor (a metal sphere inside a concentric spherical shell) and a coaxial cylindrical capacitor (a wire or rod inside a coaxial tube, the geometry of a coaxial cable). The method is the same for every shape:

Put $+Q$ on one conductor and $-Q$ on the other.
Use Gauss's law to find $E$ in the gap.
Integrate across the gap to get $\Delta V = \int E\,dr$ .
Divide: $C = \frac{Q}{\Delta V}$ . The charge always cancels.

For a sphere of radius $a$ inside a shell of inner radius $b$ , a Gaussian sphere of radius $r$ in the gap encloses $Q$ , so $E = \frac{kQ}{r^2}$ and

\Delta V = \int_a^b \frac{kQ}{r^2}\,dr = kQ\left(\frac{1}{a} - \frac{1}{b}\right) = \frac{kQ(b - a)}{ab} \quad\Rightarrow\quad C = \frac{ab}{k(b - a)} = \frac{4\pi\varepsilon_0 ab}{b - a}.

Check the limit: when the gap $b - a$ is tiny, $C \approx \frac{4\pi\varepsilon_0 a^2}{b - a}$ , which is $\frac{\varepsilon_0 A}{d}$ with $A = 4\pi a^2$ . A thin spherical gap is a parallel-plate capacitor wrapped into a ball. The cylindrical case is worked in Example 4.

Energy stored in a capacitor

Charging a capacitor means moving charge from the negative plate to the positive plate against the field that the charge already there creates. The energy stored equals the work done to separate the charge. When the plates hold $q$ , the potential difference is $\frac{q}{C}$ , and moving the next small charge $dq$ across takes work $dW = \frac{q}{C}\,dq$ . Charging from zero to $Q$ :

U_C = \int_0^Q \frac{q}{C}\,dq = \frac{Q^2}{2C}.

Graph of potential difference against charge q while a capacitor charges. The line rises straight from the origin with slope 1 over C to the point Q, delta V. A thin vertical strip of width dq under the line represents the work to move one small charge. The shaded triangle under the line has area one half Q delta V, which is the stored energy.

The graph shows where the $\frac{1}{2}$ comes from. The first bits of charge cross almost no potential difference; only the last bits cross the full $\Delta V$ . On average the charge crosses half the final potential difference, so the work is half of $Q\,\Delta V$ . Using $Q = C\,\Delta V$ , the three equivalent forms are

Worked examples

Worked exampleExample 1: a parallel-plate capacitor on a battery

A parallel-plate capacitor has plates of area $0.050\ \text{m}^2$ separated by $1.5\ \text{mm}$ of air and is connected to a $12\ \text{V}$ battery. Use $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/(\text{N}\cdot\text{m}^2)$ . Find (a) the capacitance, (b) the charge on each plate, (c) the field between the plates, and (d) the energy stored.

Show solution

(a) $C = \dfrac{\varepsilon_0 A}{d} = \dfrac{(8.85\times10^{-12})(0.050)}{1.5\times10^{-3}} = 2.95\times10^{-10}\ \text{F}$ (295 pF).

(b) $Q = C\,\Delta V = (2.95\times10^{-10}\ \text{F})(12\ \text{V}) = 3.54\times10^{-9}\ \text{C}$ : $+3.54\ \text{nC}$ on the plate connected to the positive terminal, $-3.54\ \text{nC}$ on the other.

(c) The field is uniform, so $E = \dfrac{\Delta V}{d} = \dfrac{12\ \text{V}}{1.5\times10^{-3}\ \text{m}} = 8.0\times10^{3}\ \text{V/m}$ , from the positive plate toward the negative plate. As a check, $\sigma = \frac{Q}{A} = 7.08\times10^{-8}\ \text{C/m}^2$ , and $\frac{\sigma}{\varepsilon_0} = 8.0\times10^{3}\ \text{V/m}$ .

(d) $U_C = \frac{1}{2}C(\Delta V)^2 = \frac{1}{2}(2.95\times10^{-10})(12)^2 = 2.12\times10^{-8}\ \text{J}$ .

Worked exampleExample 2: pulling the plates apart

A parallel-plate capacitor with capacitance $C_0$ is charged to potential difference $\Delta V_0$ . The plate separation is then doubled. Find how $C$ , $Q$ , $\Delta V$ , $E$ and $U$ change (a) if the capacitor was disconnected from the battery first and (b) if it stays connected.

Show solution

In both cases $C = \frac{\varepsilon_0 A}{d}$ , so doubling $d$ halves the capacitance: $C = \frac{C_0}{2}$ . What else changes depends on which quantity is held fixed.

(a) Disconnected, the charge has nowhere to go, so $Q = Q_0$ . Then $\Delta V = \frac{Q}{C} = 2\Delta V_0$ . The field is $\frac{\sigma}{\varepsilon_0}$ and $\sigma$ has not changed, so $E$ is unchanged (twice the distance at the same field gives twice the potential difference, consistent). The energy is $U = \frac{Q^2}{2C} = 2U_0$ . The extra energy comes from you: the plates attract each other, and pulling them apart takes positive work.

(b) Connected, the battery holds $\Delta V = \Delta V_0$ . Then $Q = C\,\Delta V = \frac{Q_0}{2}$ : half the charge flows back through the battery. The field is $\frac{\Delta V}{d}$ , so it halves. The energy is $U = \frac{1}{2}C(\Delta V)^2 = \frac{U_0}{2}$ .

	$C$	$Q$	$\Delta V$	$E$	$U$
Disconnected	$\times\frac{1}{2}$	same	$\times 2$	same	$\times 2$
Connected	$\times\frac{1}{2}$	$\times\frac{1}{2}$	same	$\times\frac{1}{2}$	$\times\frac{1}{2}$

The first move in any "what changes" question is to decide whether $Q$ or $\Delta V$ is fixed.

Worked exampleExample 3: an electron fired between the plates

Two horizontal plates $5.0\ \text{cm}$ long and $1.0\ \text{cm}$ apart have a potential difference of $50\ \text{V}$ , with the upper plate positive. An electron enters midway between the plates moving horizontally at $2.0\times10^{7}\ \text{m/s}$ . Use $e = 1.60\times10^{-19}\ \text{C}$ and $m_e = 9.11\times10^{-31}\ \text{kg}$ ; ignore gravity and fringing.

(a) Find the electron's acceleration.

(b) Does the electron leave the plates without hitting one? Justify your answer.

Show solution

(a) The field is $E = \frac{\Delta V}{d} = \frac{50\ \text{V}}{0.010\ \text{m}} = 5.0\times10^{3}\ \text{V/m}$ , pointing down (from the positive upper plate). The electron's charge is negative, so the force is up, toward the positive plate:

a = \frac{eE}{m_e} = \frac{(1.60\times10^{-19})(5.0\times10^{3})}{9.11\times10^{-31}} = 8.78\times10^{14}\ \text{m/s}^2, \text{ upward.}

(That is about $10^{14}$ times $g$ , so ignoring gravity is safe.)

(b) Treat it as projectile motion. Horizontally there is no force, so the time between the plates is $t = \frac{0.050\ \text{m}}{2.0\times10^{7}\ \text{m/s}} = 2.5\times10^{-9}\ \text{s}$ . Vertically it starts at rest, so in that time it rises

y = \tfrac{1}{2}at^2 = \tfrac{1}{2}(8.78\times10^{14})(2.5\times10^{-9})^2 = 2.7\times10^{-3}\ \text{m} = 2.7\ \text{mm}.

It started $5.0\ \text{mm}$ below the upper plate, and $2.7\ \text{mm} < 5.0\ \text{mm}$ , so it leaves without hitting the plate.

(c) The vertical velocity at exit is $v_y = at = (8.78\times10^{14})(2.5\times10^{-9}) = 2.2\times10^{6}\ \text{m/s}$ . The angle above the horizontal is $\tan^{-1}\frac{2.2\times10^{6}}{2.0\times10^{7}} = 6.3^\circ$ .

Worked exampleExample 4: a coaxial cable (free-response style)

A coaxial cable of length $L$ has a solid inner conductor of radius $a$ and a thin outer conducting tube of radius $b$ , with air between them. The inner conductor carries charge $+Q$ and the outer tube $-Q$ . Ignore end effects.

(a) Using Gauss's law, derive an expression for the electric field between the conductors at distance $r$ from the axis.

(b) Derive an expression for the capacitance of the cable.

(c) For $a = 0.50\ \text{mm}$ , $b = 2.0\ \text{mm}$ and $L = 10\ \text{m}$ , calculate the capacitance and the energy stored when the potential difference is $100\ \text{V}$ . Use $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/(\text{N}\cdot\text{m}^2)$ .

(d) A second cable is identical except that $b = 4.0\ \text{mm}$ . Does it have more or less capacitance than the first? By what factor?

Show solution

(a) By cylindrical symmetry, the field in the gap points radially outward and depends only on $r$ . A coaxial Gaussian cylinder of radius $r$ ( $a < r < b$ ) and length $L$ encloses the inner conductor's charge $+Q$ ; flux passes only through its curved side:

E(2\pi rL) = \frac{Q}{\varepsilon_0} \quad\Rightarrow\quad E = \frac{Q}{2\pi\varepsilon_0 rL}, \text{ radially outward.}

(b) Integrate across the gap from the inner conductor to the outer tube:

\Delta V = \int_a^b \frac{Q}{2\pi\varepsilon_0 L r}\,dr = \frac{Q}{2\pi\varepsilon_0 L}\ln\frac{b}{a}, \qquad C = \frac{Q}{\Delta V} = \frac{2\pi\varepsilon_0 L}{\ln(b/a)}.

As it must, $Q$ cancels; the result depends only on the dimensions.

C = \frac{2\pi(8.85\times10^{-12})(10)}{1.386} = 4.0\times10^{-10}\ \text{F}, \qquad U_C = \tfrac{1}{2}C(\Delta V)^2 = \tfrac{1}{2}(4.0\times10^{-10})(100)^2 = 2.0\times10^{-6}\ \text{J}.

(d) Less. The new ratio is $\frac{b}{a} = 8$ , and $C \propto \frac{1}{\ln(b/a)}$ , so $\frac{C_2}{C_1} = \frac{\ln 4}{\ln 8} = \frac{2\ln 2}{3\ln 2} = \frac{2}{3}$ . A wider gap means the same charge produces a larger potential difference, so less charge is stored per volt.

Where students lose points

On the exam

The course limits quantitative capacitor work to three shapes (parallel plates, concentric spheres, coaxial cylinders), so expect derivations of $C$ from Gauss's law for any of them, factor-of-change questions on $C$ , $Q$ , $\Delta V$ , $E$ and $U$ , and charged-particle motion between plates solved as projectile motion. Sketches of $\Delta V$ against $Q$ should be a straight line through the origin with slope $\frac{1}{C}$ ; the area under it is the stored energy. A calculator is allowed throughout.

Check yourself

Question 1
The plates of an air-filled parallel-plate capacitor are replaced with plates of twice the area, and the separation is halved. By what factor does the capacitance change?
1. A
  It is unchanged.
2. B
  It doubles.
3. C
  It increases by a factor of 4.
4. D
  It depends on the charge on the plates.
Show answer
C
$C = \frac{\varepsilon_0 A}{d}$ , so doubling $A$ doubles $C$ and halving $d$ doubles it again, a factor of 4. Choice D treats $C = \frac{Q}{\Delta V}$ as if $C$ depended on $Q$ ; it depends only on the geometry and the material.
Question 2
A capacitor holds $2.0\ \mu\text{C}$ on each plate when the potential difference across it is $50\ \text{V}$ . Find its capacitance and the energy it stores.
Show answer
$C = 4.0\times10^{-8}\ \text{F}$ ( $40\ \text{nF}$ ) and $U = 5.0\times10^{-5}\ \text{J}$ .
$C = \frac{Q}{\Delta V} = \frac{2.0\times10^{-6}}{50} = 4.0\times10^{-8}\ \text{F}$ . With $Q$ and $\Delta V$ both given, $U = \frac{1}{2}Q\,\Delta V = \frac{1}{2}(2.0\times10^{-6})(50) = 5.0\times10^{-5}\ \text{J}$ . Leaving out the $\frac{1}{2}$ doubles the answer.
Question 3
A proton is released from rest next to the positive plate of a parallel-plate capacitor held at a fixed potential difference by a battery. The plate separation is then doubled and the experiment is repeated. Compared with the first trial, the proton reaches the negative plate with
1. A
  the same kinetic energy, after a longer time
2. B
  half the kinetic energy, after the same time
3. C
  the same kinetic energy, after the same time
4. D
  twice the kinetic energy, after a longer time
Show answer
A
The kinetic energy gained is $e\,\Delta V$ , and $\Delta V$ is fixed by the battery, so it is unchanged. But the field $\frac{\Delta V}{d}$ halves, so the acceleration halves while the distance doubles; from $d = \frac{1}{2}at^2$ , the time grows by a factor of 2. Choice C misses that the trip takes longer.
Question 4
A spherical capacitor is a metal ball of radius $10\ \text{cm}$ inside a concentric metal shell of inner radius $12\ \text{cm}$ , with air between. Find its capacitance. Use $k = 9.0\times10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$ .
Show answer
$C \approx 6.7\times10^{-11}\ \text{F}$ (about $67\ \text{pF}$ ).
$C = \frac{ab}{k(b - a)} = \frac{(0.10)(0.12)}{(9.0\times10^{9})(0.02)} = 6.7\times10^{-11}\ \text{F}$ . It comes from integrating $E = \frac{kQ}{r^2}$ across the gap to get $\Delta V$ and dividing $Q$ by it.
Question 5
Why is the electric field outside a large, charged parallel-plate capacitor (away from the edges) nearly zero?
1. A
  Each plate's charge is shielded by the other plate's metal.
2. B
  The fields of the two oppositely charged plates are equal in magnitude and opposite in direction there.
3. C
  All the charge is inside the plates, so no field reaches outside.
4. D
  The field outside points along the plates and has no component away from them.
Show answer
B
Each plate alone makes a field $\frac{\sigma}{2\varepsilon_0}$ on both sides. Outside the pair, the positive plate's field points away and the negative plate's field points toward the plates, so they cancel; between the plates they add to $\frac{\sigma}{\varepsilon_0}$ . Choice C is wrong because a conductor's charge sits on its surface, not inside it.

Capacitors

Capacitance

The field between parallel plates

Charges moving between the plates

Spherical and cylindrical capacitors

Energy stored in a capacitor

Worked examples

Where students lose points

On the exam

Check yourself

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