IMAT Physics · Electricity

Electric Potential, the Electronvolt and Capacitors

Potential difference is the energy given to each coulomb of charge; a capacitor stores charge, and energy, in proportion to the voltage across it.

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Why this matters

Only one past question sits here: a 2012 question ranking three arrangements of identical capacitors. Potential, the work W = qV and the electronvolt are in the syllabus but not yet asked, and they are quick marks when they come.

Concept 1 of 2: Potential difference, the work W = qV and the electronvolt

Lifting a mass gives it gravitational energy; moving a charge against an electric field gives it electrical energy. The potential difference between two points tells you how much energy each coulomb gains or loses going from one to the other. Multiply by the charge and you have the energy.

Definition

The potential difference (p.d., voltage) between two points is the work done per unit charge moving a charge between them: $V = W/q$ , in volts ( $1\ \text{V} = 1\ \text{J/C}$ ).

The work done on a charge $q$ crossing a p.d. $V$ is $W = qV$ . A charge released in a field turns that work into kinetic energy.
The electronvolt is a unit of energy: the energy one elementary charge gains crossing 1 V. $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$ .
In a uniform field between parallel plates a distance $d$ apart, $E = V/d$ .
Moving a charge along an equipotential (a line of equal potential, always at right angles to field lines) takes no work.

Work on a charge, and the uniform field

W = qV \qquad E = \frac{V}{d}

$W$ work done on the charge (energy gained), in J
$q$ charge, in C
$V$ potential difference crossed, in V
$d$ separation of the plates, in m

Worked example

An electron starts at rest and is accelerated between two plates 4.0 mm apart with a p.d. of 2000 V between them. Find the field between the plates and the kinetic energy the electron gains, in eV and in joules.

Practice this conceptself-check · 4 quick reps

The electronvolt is energy, not voltage

Despite its name, the eV measures energy:

1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}

. A particle with charge

2e

crossing 500 V gains 1000 eV, not 500 eV. Writing an eV value in joules without converting gives an answer about

10^{19}

times too big.

Plate spacing goes in metres

E = V/d

the spacing must be in metres. Using millimetres directly makes the field 1000 times too small.

Concept 2 of 2: Capacitors: C = Q/V, stored energy, series and parallel

A capacitor is two conducting plates separated by an insulator. Connect it to a battery and electrons pile up on one plate and leave the other, until the voltage across the plates matches the battery. A bigger capacitor holds more charge for the same voltage, and the energy stored is what it took to push that charge on.

Definition

Capacitance is the charge stored per volt: $C = Q/V$ , in farads (F). Real capacitors are usually $\mu\text{F}$ or nF.

The plates carry $+Q$ and $-Q$ ; $Q$ is the charge on one plate.
For parallel plates, $C$ grows with the plate area and falls as the gap widens: $C = \varepsilon A/d$ .
Stored energy: $U = \tfrac{1}{2}QV = \tfrac{1}{2}CV^2$ .
Parallel: capacitances add, $C = C_1 + C_2 + \dots$ (more plate area).
Series: reciprocals add, $1/C = 1/C_1 + 1/C_2 + \dots$ (the total is smaller than the smallest).
These rules are the reverse of the rules for resistors.

Capacitance, energy and combinations

C = \frac{Q}{V} \qquad U = \tfrac{1}{2}CV^2 \qquad C_{\parallel} = C_1 + C_2 \qquad \frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2}

$C$ capacitance, in F
$Q$ charge on one plate, in C
$V$ p.d. across the capacitor, in V
$U$ energy stored, in J

Worked example

50\ \mu\text{F}

capacitor is charged to 20 V. Find the charge on its plates and the energy it stores.

Practice this conceptself-check · 3 quick reps

Capacitors combine the opposite way to resistors

Capacitors in parallel add directly; in series you add reciprocals. Resistors do the reverse. Swapping the rules is the most common wrong answer in capacitor network questions.

Stored energy is half QV

The energy in a charged capacitor is

\tfrac{1}{2}QV

, not

QV

. The voltage builds up from zero while charging, so on average each bit of charge was pushed on at half the final voltage.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Potential difference, the work W = qV and the electronvolt
Work on a charge, and the uniform field
$W = qV \qquad E = \frac{V}{d}$
Capacitors: C = Q/V, stored energy, series and parallel
Capacitance, energy and combinations
$C = \frac{Q}{V} \qquad U = \tfrac{1}{2}CV^2 \qquad C_{\parallel} = C_1 + C_2 \qquad \frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2}$

Watch out for (4)

The electronvolt is energy, not voltage
→ Potential difference, the work W = qV and the electronvolt
Plate spacing goes in metres
→ Potential difference, the work W = qV and the electronvolt
Capacitors combine the opposite way to resistors
→ Capacitors: C = Q/V, stored energy, series and parallel
Stored energy is half QV
→ Capacitors: C = Q/V, stored energy, series and parallel

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