Skip to main content
EnglandPhysicsSyllabus dot point

How do charged objects exert forces on one another, and how do gravitational and electric fields compare?

Coulomb's law, electric field strength as force per unit charge, the radial field of a point charge, uniform fields between plates, and the motion of charged particles in uniform fields.

A focused answer to AQA A-Level Physics 3.7.3, covering Coulomb's law, electric field strength as force per unit charge, radial and uniform fields, the comparison with gravitational fields, and the motion of charged particles in a uniform field.

Generated by Claude Opus 4.810 min answer

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

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. Coulomb's law
  3. Electric field strength
  4. Uniform fields
  5. Comparing electric and gravitational fields
  6. Charged particles in a uniform field
  7. Try this

What this dot point is asking

AQA specification point 3.7.3 wants you to state and use Coulomb's law, define electric field strength, work with radial fields of point charges and uniform fields between parallel plates, compare electric and gravitational fields, and analyse the motion of a charged particle in a uniform field.

Coulomb's law

The force is repulsive for like charges and attractive for unlike charges, acting along the line joining them. It obeys an inverse square law, so doubling the separation quarters the force. The constant 14πε0≈8.99×109 N m2 C−2\dfrac{1}{4\pi\varepsilon_0} \approx 8.99 \times 10^9 \text{ N m}^2 \text{ C}^{-2} is large, which is why electric forces between charged particles dwarf gravitational forces between the same particles.

Electric field strength

The test charge must be small so that it does not disturb the field it is measuring. For a point charge the field is radial, pointing away from a positive charge and towards a negative one:

Field lines for a point charge are straight and radial; their spacing increases with distance, showing the field weakening as 1r2\dfrac{1}{r^2}.

Uniform fields

Between two parallel plates with a potential difference VV separated by a distance dd, the field is uniform (the same magnitude and direction everywhere except near the edges):

The field lines are straight, parallel and equally spaced. This is the arrangement inside a parallel plate capacitor and in the deflecting plates of an electron beam.

Comparing electric and gravitational fields

Charged particles in a uniform field

Try this

Q1. State two similarities between electric and gravitational fields. [2 marks]

  • Cue. Both obey an inverse square law and both define field strength as force per unit (charge or mass).

Q2. Calculate the field strength between plates 5.0 mm5.0 \text{ mm} apart with a potential difference of 200 V200 \text{ V}. [2 marks]

  • Cue. E=Vd=2005.0×10−3=4.0×104 V m−1E = \dfrac{V}{d} = \dfrac{200}{5.0 \times 10^{-3}} = 4.0 \times 10^4 \text{ V m}^{-1}.

Q3. State the direction of the electric field between two parallel plates. [1 mark]

  • Cue. From the positive plate to the negative plate.

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 20184 marksTwo small charged spheres each carry a charge of +3.0 nC+3.0 \text{ nC} and are held 2.0 cm2.0 \text{ cm} apart in a vacuum. Calculate the magnitude and state the direction of the electric force between them. Take 14πε0=8.99×109 N m2 C−2\dfrac{1}{4\pi\varepsilon_0} = 8.99 \times 10^9 \text{ N m}^2 \text{ C}^{-2}.
Show worked answer →

Apply Coulomb's law F=14πε0Q1Q2r2F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q_1 Q_2}{r^2}, converting to SI units: Q1=Q2=3.0×10−9 CQ_1 = Q_2 = 3.0 \times 10^{-9} \text{ C} and r=2.0×10−2 mr = 2.0 \times 10^{-2} \text{ m}.

F=(8.99×109)(3.0×10−9)2(2.0×10−2)2=(8.99×109)9.0×10−184.0×10−4F = (8.99 \times 10^9)\dfrac{(3.0 \times 10^{-9})^2}{(2.0 \times 10^{-2})^2} = (8.99 \times 10^9)\dfrac{9.0 \times 10^{-18}}{4.0 \times 10^{-4}}.

F=(8.99×109)(2.25×10−14)=2.0×10−4 NF = (8.99 \times 10^9)(2.25 \times 10^{-14}) = 2.0 \times 10^{-4} \text{ N}, and the force is repulsive because both charges are positive.

Markers reward conversion to SI units, correct substitution, and stating the force is repulsive.

AQA 20224 marksState two similarities and two differences between the gravitational field of a point mass and the electric field of a point charge.
Show worked answer →

Similarities: both are radial fields whose field strength obeys an inverse square law with distance, and both have an associated potential that varies as 1r\dfrac{1}{r}. In both cases field strength is defined as force per unit (mass for gravity, positive charge for electric).

Differences: the gravitational force is always attractive, whereas the electric force can be attractive or repulsive depending on the signs of the charges. The electric force between fundamental particles is enormously stronger than the gravitational force, because 14πε0\dfrac{1}{4\pi\varepsilon_0} is large while GG is tiny.

Markers reward two valid similarities (inverse square law, force per unit quantity, 1r\dfrac{1}{r} potential) and two valid differences (attractive only versus both signs, relative strength).

Related dot points

Sources & how we know this