Capacitance: parallel-plate, cylindrical, spherical, and series-parallel combinations

The capacitor is the simplest circuit element that stores electric charge and therefore energy. This week you learn two jobs: computing the capacitance of a geometry, and reducing several capacitors to a single equivalent capacitor. The typical exam question is: “What is the equivalent capacitance of this network, and what is the charge on $C_2$?” This type is worth +5 points and is mechanical for someone who knows the method.

Golden rules:

1. $C = Q/\Delta V$. Capacitance is the ratio of charge to voltage. Here Q is the magnitude of the charge on one plate (the other plate has $-Q$).
2. C is geometry. If you double Q, V doubles too, and C does not change.
3. Parallel: same voltage, capacitances add. Series: same charge, reciprocals add.
4. If the battery is connected, V is constant; if the battery is removed, Q is constant.
5. Convert units to meters. Errors with cm, mm and cm² cost the most points.

Definition of the capacitor and capacitance

A capacitor consists of two conductors with an insulator (or vacuum) between them. When you connect it to a battery, the battery pulls electrons from one conductor and pushes them onto the other. The result is $+Q$ on one conductor and $-Q$ on the other. The total charge is zero, but a potential difference $\Delta V$ appears between them.

Experiments show that $Q$ and $\Delta V$ are proportional. The constant of proportionality is called the capacitance:

$C = \frac{Q}{\Delta V}$

Its unit is the farad (F): $1\ \text{F} = 1\ \text{C/V}$. The farad is a very large unit. In practice you see $\mu\text{F} = 10^{-6}\ \text{F}$, $\text{nF} = 10^{-9}\ \text{F}$, $\text{pF} = 10^{-12}\ \text{F}$.

TIP · C is always positive. $Q$ and $\Delta V$ are taken as magnitudes. If you got something like “negative capacitance”, you mixed up a sign.

The recipe for computing capacitance

The same four steps work for every geometry:

1. Put charge $+Q$ and $-Q$ on the conductors.
2. Find the field $E$ in between with Gauss’s law.
3. Compute the potential difference with $\Delta V = \int_a^b E\,dr$ (as a magnitude).
4. $C = Q/\Delta V$. Q cancels and only geometry remains.

The cancellation of Q is the mathematical proof that C does not depend on the charge.

Parallel-plate capacitor

Two parallel plates of area $A$ with a distance $d$ between them. If the plates are large and close, edge effects are neglected and the field in between is uniform. The surface charge density is $\sigma = Q/A$. From Gauss’s law (the conductor-surface result you saw in earlier weeks):

$E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}, \qquad \Delta V = E d = \frac{Q d}{\varepsilon_0 A}$

$C = \frac{Q}{\Delta V} = \frac{\varepsilon_0 A}{d}$

Intuition: a larger area holds more charge, and closer plates give a lower voltage for the same charge, so C grows.

Example 1. Plate area $A = 4.0\times10^{-3}\ \text{m}^2$, separation $d = 1.5\ \text{mm}$, air. It is connected to a 12 V battery. What are C, Q and E?

C = ε0 A / d = (8.85e-12)(4.0e-3) / (1.5e-3)
  = 2.36e-11 F = 23.6 pF
Q = C ΔV = (23.6 pF)(12 V) = 283 pC = 2.83e-10 C
E = ΔV / d = 12 / 1.5e-3 = 8.0e3 V/m

Cylindrical capacitor

An inner conducting cylinder of radius $a$, an outer conducting shell of radius $b$, length $\ell$. This is the coaxial cable. The charge per unit length is $\lambda = Q/\ell$. Choose a cylinder with $a < r < b$ as the Gaussian surface:

$E = \frac{2k\lambda}{r}$

$\Delta V = \int_a^b \frac{2k\lambda}{r}\,dr = 2k\lambda\ln\frac{b}{a}$

$C = \frac{Q}{\Delta V} = \frac{\ell}{2k\ln(b/a)} = \frac{2\pi\varepsilon_0\,\ell}{\ln(b/a)}$

C is proportional to the length. That is why for cables you are usually asked for the “capacitance per unit length” $C/\ell$.

Example 2. $a = 0.50\ \text{mm}$, $b = 2.0\ \text{mm}$, $\ell = 0.80\ \text{m}$.

ln(b/a) = ln 4 = 1.386
C = 0.80 / (2 · 8.99e9 · 1.386) = 0.80 / 2.49e10
  = 3.21e-11 F = 32.1 pF

PITFALL · Mixing up ln and log. On a calculator the log key is usually base 10. The $\ln$ in the formula is the natural logarithm. If you use $\log_{10} 4 = 0.602$ you get a C that is 2.3 times too large, and that is a distractor waiting ready in the answer choices.

Spherical capacitor

An inner sphere of radius $a$ and an outer shell of inner radius $b$. In between, the field is like that of a point charge: $E = kQ/r^2$.

$\Delta V = \int_a^b \frac{kQ}{r^2}\,dr = kQ\left(\frac{1}{a} - \frac{1}{b}\right) = kQ\,\frac{b-a}{ab}$

$C = \frac{ab}{k(b-a)}$

A single (isolated) sphere. If you take $b \to \infty$ you get $C = a/k = 4\pi\varepsilon_0 a$. So even a single conducting sphere has a capacitance; the “other plate” is at infinity. For a sphere of radius 20 cm this is only 22 pF.

Example 3. $a = 4.0\ \text{cm}$, $b = 6.0\ \text{cm}$.

C = ab / (k(b-a)) = (0.04)(0.06) / (8.99e9 · 0.02)
  = 2.4e-3 / 1.798e8 = 1.34e-11 F = 13.4 pF

Connecting capacitors

In a circuit diagram a capacitor is drawn as two short parallel lines. You can reduce several capacitors to a single equivalent capacitance.

Parallel combination

Both ends of each capacitor are connected to the same two points. Therefore:

• The voltages are the same: $\Delta V_1 = \Delta V_2 = \Delta V$.
• The charges add: $Q = Q_1 + Q_2$.

$C_{eq}\,\Delta V = C_1\Delta V + C_2\Delta V \quad\Rightarrow\quad C_{eq} = C_1 + C_2 + \dots$

Intuition: connecting in parallel is like increasing the plate area.

Series combination

The capacitors are connected end to end with no other path between them. Seen from the battery, charge arrives only at the outermost plates. The middle plates form one isolated piece of conductor whose total charge stays zero, so by induction the same $Q$ appears on each:

• The charges are the same: $Q_1 = Q_2 = Q$.
• The voltages add: $\Delta V = \Delta V_1 + \Delta V_2$.

$\frac{Q}{C_{eq}} = \frac{Q}{C_1} + \frac{Q}{C_2} \quad\Rightarrow\quad \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots$

Shortcut for two capacitors: $C_{eq} = \dfrac{C_1C_2}{C_1+C_2}$. $n$ identical capacitors in series: $C/n$.

PITFALL · Opposite of resistors. For capacitors, parallel adds directly and series adds reciprocals. For resistors it is exactly the reverse. Don’t mix them up when you get to resistors in week 13.

TIP · Quick check. The series equivalent must be smaller than the smallest capacitor, and the parallel equivalent larger than the largest. Immediately eliminate any answer choice that breaks this rule.

A worked network example

Example 4. $C_2 = 2.0\ \mu\text{F}$ and $C_3 = 6.0\ \mu\text{F}$ are in parallel. This pair is in series with $C_1 = 4.0\ \mu\text{F}$. 12 V is applied across the ends. What are the charge and voltage of each capacitor?

Method: reduce from the inside out, then expand back from the outside in.

1) Parallel pair:   C23 = 2.0 + 6.0 = 8.0 μF
2) Series:          Ceq = (4.0 · 8.0)/(4.0 + 8.0) = 2.67 μF
3) Total charge:    Q = Ceq · V = 2.67 · 12 = 32 μC
4) Series elements carry the same Q:  Q1 = Q23 = 32 μC
5) Voltages:        V1 = 32/4.0 = 8.0 V,   V23 = 32/8.0 = 4.0 V
   check:           8.0 + 4.0 = 12 V
6) Same V inside the parallel pair:  Q2 = 2.0 · 4.0 = 8.0 μC,  Q3 = 6.0 · 4.0 = 24 μC
   check:           8.0 + 24 = 32 μC

We did a check at the end of each step: in series the voltages add up to the source voltage, in parallel the charges add up to the incoming charge. On the exam these two checks are the fastest way to find a mistake.

Battery connected or removed?

In questions that change the geometry of a capacitor (pulling the plates apart, inserting a material between them), the first job is to decide which case you are in:

Charge sharing

If you connect a charged capacitor that has been disconnected from the battery to an empty capacitor, the charge is shared between the two capacitors until the voltages are equal. The total charge is conserved:

$V_{final} = \frac{Q_0}{C_1 + C_2}$

Energy is not conserved in this process; part of it is lost as heat and radiation in the wires. We will compute the energy in week 9.

Common mistakes