Calculate the work done moving a charge of +3.0 nC through a potential difference of 500 V. [2 marks]

AQA AQA 2019-style practice: An isolated metal sphere of radius 5.0 cm carries a charge of +8.0 nC. Calculate the electric potential at its surface. Take 14π_0 = 8.99 × 10^9 N m^2 C^-2.

Outside (and at the surface of) a charged sphere the field and potential are those of a point charge at its centre, so use V = 14π_0Qr with r the radius. V = (8.99 × 10^9)8.0 × 10^-95.0 × 10^-2 = (8.99 × 10^9)(1.6 × 10^-7). V = 1.4 × 10^3 V, positive because the charge is positive. Markers reward treating the sphere as a point charge at its surface, conversion to SI units, and a positive answer.

EnglandPhysicsSyllabus dot point

How much work is needed to move a charge through an electric field, and how does this compare with gravity?

Absolute electric potential and potential energy in a radial field, the potential gradient, equipotentials, and the work done moving a charge.

A focused answer to AQA A-Level Physics 3.7.3.3, covering absolute electric potential in a radial field, electric potential energy, the potential gradient and its link to field strength, equipotentials, and the work done moving a charge.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Electric potential
Electric potential energy
Potential gradient and field strength
Equipotentials
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What this dot point is asking

AQA specification point 3.7.3.3 wants you to define absolute electric potential, calculate it in a radial field, relate the potential gradient to field strength, describe equipotential surfaces, and calculate the work done moving a charge.

Electric potential

The potential is positive around a positive charge (work must be done against repulsion to bring a positive test charge in) and negative around a negative charge (the test charge is attracted, so work is done by the field). The zero of potential is taken at infinity, where the charges are infinitely far apart and no longer interact. Note the $\dfrac{1}{r}$ dependence: potential falls off more slowly than the field, which falls as $\dfrac{1}{r^2}$ .

Electric potential energy

The potential energy of a charge $q$ at a point of potential $V$ is the energy stored by virtue of its position in the field:

This work is independent of the path taken between the two points (the electric force is conservative), so only the start and end potentials matter.

Potential gradient and field strength

Field strength is the negative potential gradient: the field points in the direction of decreasing potential, from high to low. The steeper the potential changes with distance, the stronger the field. This is why field lines are closely spaced where the potential changes rapidly. In a uniform field the potential changes linearly, giving $E = \dfrac{V}{d}$ as a special case.

Equipotentials

The spacing of equally-spaced equipotentials reveals the field strength: closely spaced equipotentials mean a steep potential gradient and a strong field.

Energy gained by an accelerated electron

An electron ( $q = -1.60 \times 10^{-19} \text{ C}$ ) is accelerated from rest through a potential difference of $2000 \text{ V}$ . Find the kinetic energy it gains and express it in electronvolts.

step 1: Use the work-potential relation

The work done by the field is $W = q\Delta V$ , and this becomes the electron's kinetic energy.

step 2: Compute the magnitude in joules

$|W| = (1.60 \times 10^{-19})(2000) = 3.2 \times 10^{-16} \text{ J}$ .

step 3: Convert to electronvolts

One electronvolt is the energy gained by an electron through $1 \text{ V}$ , so accelerating through $2000 \text{ V}$ gives $2000 \text{ eV} = 2.0 \text{ keV}$ .

step 4: Check the sign

The electron moves to higher potential, and being negative it gains kinetic energy, confirming the energy is positive.

Try this

Q1. State the relationship between electric field strength and electric potential. [1 mark]

Cue. Field strength is the negative potential gradient, $E = -\dfrac{\Delta V}{\Delta r}$ .

Q2. Calculate the work done moving a charge of $+3.0 \text{ nC}$ through a potential difference of $500 \text{ V}$ . [2 marks]

Cue. $W = q\Delta V = 3.0 \times 10^{-9} \times 500 = 1.5 \times 10^{-6} \text{ J}$ .

Q3. State where the zero of electric potential is taken. [1 mark]

Cue. At infinity.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20194 marksAn isolated metal sphere of radius

5.0 \text{ cm}

carries a charge of

+8.0 \text{ nC}

. Calculate the electric potential at its surface. Take

\dfrac{1}{4\pi\varepsilon_0} = 8.99 \times 10^9 \text{ N m}^2 \text{ C}^{-2}

Show worked answer →

Outside (and at the surface of) a charged sphere the field and potential are those of a point charge at its centre, so use $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}$ with $r$ the radius.

$V = (8.99 \times 10^9)\dfrac{8.0 \times 10^{-9}}{5.0 \times 10^{-2}} = (8.99 \times 10^9)(1.6 \times 10^{-7})$ .

$V = 1.4 \times 10^{3} \text{ V}$ , positive because the charge is positive.

Markers reward treating the sphere as a point charge at its surface, conversion to SI units, and a positive answer.

AQA 20213 marksDefine an equipotential surface and explain why no work is done when a charge moves along it.

Show worked answer →

An equipotential surface joins all points that are at the same electric potential.

The work done moving a charge between two points is $W = q\Delta V$ . Along an equipotential the potential does not change, so $\Delta V = 0$ and therefore $W = 0$ .

Equivalently, equipotentials are always perpendicular to the field lines, so a charge moving along one has no component of the electric force in its direction of motion.

Markers reward defining the surface by constant potential and linking zero work to $\Delta V = 0$ (or to motion perpendicular to the field).

Related dot points

Sources & how we know this

AQA A-level Physics (7408) specification — AQA (2017)