Learn9702Paper 4 · A Level Structured QuestionsElectric fieldsElectric potential

Electric fields · Subtopic 3 of 3

Electric Potential

Electric potential at a point is the work done per unit positive charge bringing a small test charge from infinity: $V=\dfrac Q{4\pi\varepsilon_0r}$ . Unlike gravitational potential, it can be positive (near a positive charge) or negative. A charge $q$ at that point has potential energy $E_P=qV=\dfrac{Qq}{4\pi\varepsilon_0r}$ . Field strength is minus the potential gradient, $E=-\dfrac{\Delta V}{\Delta r}$ , so field lines cross equipotentials at right angles.

Cambridge 9702 Paper 4Edexcel YPH11 WPH14AQA 7408 Paper 2≈ 40 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Define electric potential and use $V=\dfrac Q{4\pi\varepsilon_0r}$ .
Use $E_P=\dfrac{Qq}{4\pi\varepsilon_0r}$ and energy conservation for charges.
Relate $E$ to the potential gradient and draw equipotentials.
Find the distance of closest approach in α-scattering.

Pause and think

Gravitational potential is always negative. Can electric potential be positive?

Section 2 of 14

Core idea

Start hereElectric hills

A positive charge near another positive charge is like a ball near the top of a hill: let go and it rolls away. Electric potential is the height of that hill, per coulomb.

The big idea

$V=\frac Q{4\pi\varepsilon_0r}$ : work per unit positive charge from infinity. $E=-\frac{dV}{dr}$ . Work $=q\Delta V$ .

1. Potential of a point charge

Potential and field around a positive point charge

The field falls off faster than the potential. At any point the slope of the

V

curve equals

-E

From field to potential

$E=-\frac{dV}{dr}$
Field points downhill in potential.
$\frac{dV}{dr}=-\frac Q{4\pi\varepsilon_0r^2}$
Point-charge field.
$V=\frac Q{4\pi\varepsilon_0r}+C,\quad C=0$
Integrate; $V(\infty)=0$ .

Worked example+2.0 μC charge

$V$ at 0.30 m and 0.60 m; work to bring +1 nC from 0.60 to 0.30 m.

Goal: Potentials; work.
Strategy: $W=q\Delta V$ .

$V_{0.30}=\frac{8.99\times10^9\times2\times10^{-6}}{0.30}=6.0\times10^4\text{ V}$
Formula.
$V_{0.60}=3.0\times10^4\text{ V}$
Double $r$ , half $V$ .
$W=10^{-9}\times3.0\times10^4=3.0\times10^{-5}\text{ J}$
Uphill: work done on the charge.

Answer

6.0\times10^4

3.0\times10^4

3.0\times10^{-5}

✓ Check

V\propto\frac1r

, but

E\propto\frac1{r^2}

✓.

Equipotentials (dashed) are perpendicular to field lines

No work is done moving a charge along an equipotential. They crowd together where the field is strong.

Potentials add as scalars; fields as vectors.
Negative charges give negative potential.

Common misconception $\frac1{r^2}$ for potential Potential uses

\frac1r

Pause and think

Work moving along an equipotential?

Summary

$V=Q/(4\pi\varepsilon_0r)$ .
$E=-dV/dr$ ; $W=q\Delta V$ .

Equipotential: Surface of constant potential.

HistoryRutherford's closest approach. Using energy conservation, Rutherford calculated that

5

MeV α-particles came within about

4\times10^{-14}

m of a gold nucleus before turning back. The nucleus had to be smaller than that: about

10\,000

times smaller than the atom. It was the first measurement of nuclear size.

Fun factBirds on wires again. A bird sitting on a

132

kV power line is at a potential of

132\,000

V, but because both feet are on the same equipotential, no charge flows through it. Potential differences, not potentials, drive currents.

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