Electrostatics • Topic 3 of 3

Capacitance & Capacitors

A capacitor is a device that stores electric charge and energy. It is two conductors separated by an insulator; when connected to a battery they acquire equal and opposite charges $\pm Q$. The capacitance measures how much charge the device holds per unit potential difference:

$$C=\frac{Q}{V}$$ The SI unit is the farad (F = C/V). A farad is very large, so practical values are in $\mu\text{F}$, $\text{nF}$ and $\text{pF}$. Capacitance depends only on the geometry of the conductors and the medium between them — not on the charge or voltage applied.

For a parallel-plate capacitor of plate area $A$ and separation $d$ with vacuum between the plates,

$$C=\frac{\epsilon_0 A}{d}$$ Larger plates and a smaller gap give a larger capacitance. The uniform field between the plates is $E=\frac{\sigma}{\epsilon_0}=\frac{V}{d}$.

Capacitors are combined in two ways:

  • Series: the same charge sits on each capacitor and voltages add, so $\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\dots$ — the net capacitance is less than the smallest member.
  • Parallel: the voltage is common and charges add, so $C_p=C_1+C_2+\dots$ — the net capacitance is the sum.

Inserting a dielectric (an insulator such as glass, mica or oil) of dielectric constant $K$ multiplies the capacitance: $C=K\frac{\epsilon_0 A}{d}=KC_0$. The dielectric reduces the effective field because its molecules polarise and set up an opposing field, allowing the capacitor to store more charge at the same voltage. This is why real capacitors are filled with dielectric.

The energy stored in a charged capacitor, equal to the work done to charge it, is

$$U=\frac12 CV^2=\frac12 QV=\frac{Q^2}{2C}$$ This energy resides in the electric field between the plates; the energy density (energy per unit volume) is $u=\frac12\epsilon_0 E^2$. Capacitors are everywhere — in power supplies, camera flashes, defibrillators and tuning circuits — wherever rapid storage and release of energy is needed.

Parallel-plate capacitor with dielectric between the platesdielectric K+-Vplate separation darea A
1
Worked Example
A capacitor stores a charge of $6\times10^{-6}\ \text{C}$ when connected to a 12 V battery. Find its capacitance.
Solution
  1. Use $C=\frac{Q}{V}$.
  2. $C=\frac{6\times10^{-6}}{12}$.
  3. $C=5\times10^{-7}\ \text{F}=0.5\ \mu\text{F}$.

Answer: $C=0.5\ \mu\text{F}$.

2
Worked Example
A parallel-plate capacitor has plates of area $0.02\ \text{m}^2$ separated by $1\ \text{mm}$ in vacuum. Find its capacitance.
Solution
  1. Use $C=\frac{\epsilon_0 A}{d}$.
  2. $C=\frac{8.85\times10^{-12}\times0.02}{1\times10^{-3}}$.
  3. $=\frac{1.77\times10^{-13}}{10^{-3}}$.
  4. $C=1.77\times10^{-10}\ \text{F}=177\ \text{pF}$.

Answer: $C\approx177\ \text{pF}$.

3
Worked Example
Three capacitors of $2\ \mu\text{F}$, $3\ \mu\text{F}$ and $6\ \mu\text{F}$ are connected in series. Find the equivalent capacitance.
Solution
  1. Use $\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$.
  2. $\frac{1}{C_s}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3+2+1}{6}=1$.
  3. $C_s=1\ \mu\text{F}$.

Answer: $C_s=1\ \mu\text{F}$.

4
Worked Example
The same three capacitors ($2$, $3$, $6\ \mu\text{F}$) are now connected in parallel. Find the equivalent capacitance.
Solution
  1. Use $C_p=C_1+C_2+C_3$.
  2. $C_p=2+3+6$.
  3. $C_p=11\ \mu\text{F}$.

Answer: $C_p=11\ \mu\text{F}$.

5
Worked Example
A $4\ \mu\text{F}$ capacitor is charged to 200 V. Find the energy stored.
Solution
  1. Use $U=\frac12 CV^2$.
  2. $U=\frac12\times4\times10^{-6}\times(200)^2$.
  3. $=\frac12\times4\times10^{-6}\times4\times10^{4}$.
  4. $U=0.08\ \text{J}$.

Answer: $U=0.08\ \text{J}$.

6
Worked Example
A parallel-plate capacitor of capacitance $5\ \mu\text{F}$ in air is filled with a dielectric of $K=4$. Find the new capacitance.
Solution
  1. A dielectric multiplies capacitance: $C=KC_0$.
  2. $C=4\times5\ \mu\text{F}$.
  3. $C=20\ \mu\text{F}$.

Answer: $C=20\ \mu\text{F}$.

Key Points

  • Capacitance $C=\frac{Q}{V}$ (unit farad); it depends only on geometry and the medium, not on $Q$ or $V$.
  • Parallel-plate capacitor: $C=\frac{\epsilon_0 A}{d}$; field between plates $E=\frac{V}{d}=\frac{\sigma}{\epsilon_0}$.
  • Series: $\frac{1}{C_s}=\sum\frac{1}{C_i}$ (less than the smallest). Parallel: $C_p=\sum C_i$ (the sum).
  • A dielectric of constant $K$ raises capacitance to $C=KC_0$ by polarising and weakening the internal field.
  • Energy stored $U=\frac12 CV^2=\frac12 QV=\frac{Q^2}{2C}$; energy density $u=\frac12\epsilon_0 E^2$.
Tap an option to check your answer0 / 4
Q1.The SI unit of capacitance is the:
Explanation: $C=\frac{Q}{V}$ has unit coulomb/volt = farad (F).
Q2.If the separation between the plates of a parallel-plate capacitor is halved, its capacitance:
Explanation: $C=\frac{\epsilon_0 A}{d}\propto\frac{1}{d}$, so halving $d$ doubles $C$.
Q3.Capacitors in series have an equivalent capacitance that is:
Explanation: $\frac{1}{C_s}=\sum\frac{1}{C_i}$, so $C_s$ is smaller than every individual capacitor.
Q4.Inserting a dielectric of constant $K$ between the plates changes the capacitance to:
Explanation: The dielectric multiplies the capacitance: $C=KC_0$.