Two parallel plates 0.020\ m apart have a potential difference of 300\ V. Find the field strength between them. [2 marks]

Find the force on a charge of 2.0\ nC placed at a point where the electric field strength is 5.0 × 10^4\ V m^-1. [2 marks]

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How do electric charges interact through fields, and how do those fields compare with gravity?

Coulomb's law, electric field strength and electric potential, uniform and radial fields, and the parallels and differences between electric and gravitational fields.

A CCEA A-Level Physics answer on Coulomb's law, electric field strength and electric potential, uniform fields between parallel plates and radial fields around point charges, and the parallels with gravitational fields.

Generated by Claude Opus 4.811 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
The answer
Examples in context
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What this dot point is asking

CCEA wants you to state and apply Coulomb's law, define electric field strength and electric potential, describe uniform and radial fields, and compare electric fields with gravitational fields. Expect a Coulomb-law calculation and a structured comparison with gravity.

The answer

Coulomb's law

Field strength and potential

Around a positive charge the potential is positive and falls to zero at infinity; around a negative charge it is negative. The field strength is the negative potential gradient, $E = -\frac{\mathrm{d}V}{\mathrm{d}r}$ , so field lines run from high to low potential, perpendicular to the equipotentials.

Comparing electric and gravitational fields

The two field types are mathematically similar: both forces follow an inverse-square law, both field strengths are the force per unit "charge" (mass or charge), and both potentials follow an inverse law with distance. The key difference is that gravity is always attractive, so masses cannot be shielded, whereas electric forces can be attractive or repulsive, which is why a conductor can screen an external field (a Faraday cage).

Worked example: accelerating an electron

Energy gained crossing a uniform field

An electron ( $q = 1.6 \times 10^{-19}\ \text{C}$ , $m = 9.11 \times 10^{-31}\ \text{kg}$ ) is accelerated from rest across a uniform field between plates $0.050\ \text{m}$ apart with a potential difference of $2000\ \text{V}$ . Find the field strength, the force on the electron, and its final speed.

step Find the field strength

E = \frac{V}{d} = \frac{2000}{0.050} = 4.0 \times 10^{4}\ \text{V m}^{-1}.

step Find the force

F = qE = 1.6 \times 10^{-19} \times 4.0 \times 10^{4} = 6.4 \times 10^{-15}\ \text{N}.

step Use energy conservation

The work done on the electron equals $qV$ , which becomes kinetic energy:

qV = \tfrac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2qV}{m}} = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 2000}{9.11 \times 10^{-31}}} = \sqrt{7.02 \times 10^{14}} = 2.6 \times 10^{7}\ \text{m s}^{-1}.

Examples in context

Example 1. An inkjet printer. Tiny ink droplets are given a controlled charge and then deflected by a uniform field between parallel plates, $E = V/d$ , steering each drop to the right place on the page. Varying the plate voltage changes the deflection, building up the printed image dot by dot.

Example 2. A lightning conductor. The strong radial field near a sharp point (small $r$ gives large $E$ ) ionises the surrounding air, allowing charge to leak away gradually and providing a low-resistance path to earth. This bleeds off the charge build-up that would otherwise discharge violently through a building.

Try this

Q1. Two parallel plates $0.020\ \text{m}$ apart have a potential difference of $300\ \text{V}$ . Find the field strength between them. [2 marks]

Cue. $E = \frac{V}{d} = \frac{300}{0.020} = 1.5 \times 10^{4}\ \text{V m}^{-1}$ .

Q2. State one similarity and one difference between electric and gravitational fields. [2 marks]

Cue. Both follow an inverse-square force law; gravity is always attractive, but electric forces can attract or repel.

Q3. Find the force on a charge of $2.0\ \text{nC}$ placed at a point where the electric field strength is $5.0 \times 10^{4}\ \text{V m}^{-1}$ . [2 marks]

Cue. $F = qE = 2.0 \times 10^{-9} \times 5.0 \times 10^{4} = 1.0 \times 10^{-4}\ \text{N}$ .

Exam-style practice questions

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

CCEA 20206 marksTwo small charged spheres carry charges of plus 3.0 nC and minus 5.0 nC and are held 4.0 cm apart in a vacuum. Calculate the magnitude of the electrostatic force between them and state its nature. Determine the electric field strength at the midpoint of the line joining the spheres. Take the permittivity constant 1 over 4 pi epsilon nought as 8.99 times 10 to the 9.

Show worked answer →

The force from Coulomb's law:

$F = \frac{Qq}{4\pi\varepsilon_0 r^2} = 8.99 \times 10^{9} \times \frac{(3.0 \times 10^{-9})(5.0 \times 10^{-9})}{(0.040)^2} = 8.99 \times 10^{9} \times \frac{1.5 \times 10^{-17}}{1.6 \times 10^{-3}} = 8.4 \times 10^{-5}$ N.

The force is attractive because the charges are of opposite sign.

At the midpoint each charge is 0.020 m away. Both fields point from the positive sphere towards the negative sphere (a positive test charge is pushed away from the plus and pulled towards the minus), so they add:

$E_+ = 8.99 \times 10^{9} \times \frac{3.0 \times 10^{-9}}{(0.020)^2} = 6.7 \times 10^{4}$ V per metre.

$E_- = 8.99 \times 10^{9} \times \frac{5.0 \times 10^{-9}}{(0.020)^2} = 1.12 \times 10^{5}$ V per metre.

$E_{\text{mid}} = E_+ + E_- = 1.8 \times 10^{5}$ V per metre.

Markers reward the correct force with "attractive", and recognising the two field contributions add at the midpoint.

CCEA 20184 marksState two similarities and two differences between an electric field around a point charge and a gravitational field around a point mass. Explain why a charged object can be shielded from an external electric field but a mass cannot be shielded from gravity.

Show worked answer →

Similarities: both field strengths obey an inverse-square law with distance from the source; both potentials obey an inverse law with distance; both are radial fields with the field strength being the force per unit charge or per unit mass.

Differences: gravity is always attractive whereas electric forces can be attractive or repulsive depending on the signs of the charges; the gravitational field depends on mass while the electric field depends on charge (and the constants differ).

A conductor can be shielded because mobile charges rearrange on its surface so the field inside a hollow conductor is zero (a Faraday cage). There are no negative masses to rearrange and cancel a gravitational field, so gravity cannot be screened.

Markers reward two valid similarities, two valid differences, and the Faraday-cage reasoning contrasted with the absence of negative mass.

Related dot points

Sources & how we know this

CCEA GCE Physics specification — CCEA (2016)