Capacitors and capacitance: WASSCE Physics

A camera flash charges up with a whine and then releases its energy in a split second. Inside is a capacitor: a device that stores electric charge, and with it energy, ready to be released when needed.

Learning objectives

By the end of this lesson you should be able to:

describe a parallel-plate capacitor and define capacitance
use $Q = CV$ and $C = \dfrac{\varepsilon A}{d}$
state the factors affecting the capacitance of a parallel-plate capacitor
describe charging and discharging of a capacitor

The capacitor

Definition

A capacitor is a device that stores electric charge. The simplest form is two parallel metal plates separated by an insulator called the dielectric (air, paper, mica, plastic or ceramic).

When a capacitor is connected to a battery, electrons flow onto one plate (making it negative) and away from the other (leaving it positive). The plates gain equal and opposite charges, $+Q$ and $-Q$ . No charge passes through the dielectric.

Capacitance

Definition

The capacitance $C$ of a capacitor is the ratio of the charge on either plate to the potential difference between the plates:

C = \dfrac{Q}{V}

Its unit is the farad (F): 1 F = 1 C V⁻¹. Practical capacitors are measured in microfarads (μF = $10^{-6}$ F), nanofarads (nF) and picofarads (pF = $10^{-12}$ F).

Worked example

A 50 μF capacitor is connected to a 12 V battery. How much charge does it store?

Q = CV = 50 \times 10^{-6} \times 12 = 6.0 \times 10^{-4}\ \text{C}

\begin{aligned}Q &= CV \\ &= 50 \times 10^{-6} \times 12 \\ &= 6.0 \times 10^{-4}\ \text{C}\end{aligned}

Capacitance of a parallel-plate capacitor

C = \dfrac{\varepsilon A}{d} = \dfrac{\varepsilon_r \varepsilon_0 A}{d}

\begin{aligned}C &= \dfrac{\varepsilon A}{d} \\ &= \dfrac{\varepsilon_r \varepsilon_0 A}{d}\end{aligned}

where $A$ = area of overlap of the plates, $d$ = separation, $\varepsilon_0 = 8.85 \times 10^{-12}$ F m⁻¹ and $\varepsilon_r$ = relative permittivity (dielectric constant) of the material between the plates.

Factors affecting capacitance

Change	Effect on capacitance
Larger area of overlap $A$	increases
Smaller separation $d$	increases
Dielectric with larger $\varepsilon_r$ (e.g. mica instead of air)	increases

Worked example

Two plates, each of area 0.02 m², are 1 mm apart in air. Find the capacitance. (ε₀ = 8.85 × 10⁻¹² F m⁻¹)

C = \dfrac{8.85 \times 10^{-12} \times 0.02}{1 \times 10^{-3}} = 1.77 \times 10^{-10}\ \text{F} = 177\ \text{pF}

\begin{aligned}C &= \dfrac{8.85 \times 10^{-12} \times 0.02}{1 \times 10^{-3}} \\ &= 1.77 \times 10^{-10}\ \text{F} \\ &= 177\ \text{pF}\end{aligned}

Filling the gap with mica ( $\varepsilon_r = 6$ ) would raise it to about 1.06 nF.

Common mistake

Convert mm to m and μF to F before calculating. A separation of 1 mm is $1 \times 10^{-3}$ m; forgetting this makes the answer 1000 times too small.

Charging and discharging

Charging (capacitor and resistor in series with a battery):

current is largest at the start and falls to zero
p.d. across the capacitor rises from zero to the battery voltage
charging is quicker with a smaller resistance or smaller capacitance

Discharging (charged capacitor connected across a resistor):

current flows the opposite way, largest at first, falling to zero
the p.d. and charge fall gradually

The time scale is set by the product $RC$ (the time constant).

Uses of capacitors

Camera flash units: store charge and release it quickly through the flash tube.
Smoothing the output of rectifiers in power supplies (phone chargers).
Tuning circuits in radios (variable capacitor with the coil selects a frequency).
Timing circuits (the RC time constant), for example in indicator flashers.
Blocking d.c. while passing a.c. in amplifiers.
Reducing sparking at switch contacts and in car ignition systems.

In Sierra Leone

Open an old phone charger or radio and you will see cylindrical electrolytic capacitors marked with values such as "470 μF 16 V". The voltage marked is the maximum it can safely take; exceeding it can make the capacitor burst.

Key points

A capacitor stores charge on two plates separated by a dielectric.
$C = Q/V$ , in farads; practical values in μF, nF, pF.
Parallel-plate: $C = \dfrac{\varepsilon A}{d}$ ; larger area, smaller gap and better dielectric increase $C$ .
Charging and discharging currents are largest at first and fall to zero.

WASSCE examination focus

WAEC tests this

Define capacitance and the farad, use $Q = CV$ , state the factors affecting the capacitance of a parallel-plate capacitor, and give uses of capacitors. Watch the unit prefixes in calculations.

Quick check

What is the unit of capacitance?
A capacitor stores 2 × 10⁻⁴ C at 20 V. What is its capacitance?
How does halving the separation of the plates affect the capacitance?

Answers: 1. The farad (F). 2. $C = \dfrac{2 \times 10^{-4}}{20} = 1 \times 10^{-5}$ F = 10 μF. 3. It doubles.

Capacitors and capacitance