All subjects · Physics · WPH14 (Unit 4)

Electric fields and Coulomb's law

Charges exert forces on each other without touching. The idea of a field describes this, and Coulomb's law gives the size of the force.

Electric fields

An electric field is a region where a charged particle experiences a force. The electric field strength at a point is the force per unit positive charge:

E = F / Q (unit N/C)

Its direction is the direction of the force on a small positive test charge.

Coulomb's law

Two point charges Q_1 and Q_2 a distance r apart exert forces on each other of

F = Q_1Q_2 / (4πε_0r2) = kQ_1Q_2 / r2

with ε_0 = 8.85 x 10-12 F/m and k = 1/(4πε_0) = 8.99 x 109 N m2 C-2. Like charges repel; unlike charges attract. It is an inverse square law: double r and F falls to a quarter.

Field of a point charge

Putting F = QQ_test / (4πε_0r2) into E = F / Q_test gives the radial field of a point charge:

E = Q / (4πε_0r2)

The field lines point radially outwards from a positive charge and inwards towards a negative charge. A charged sphere behaves, outside itself, like a point charge at its centre.

Worked example

Find the electric field strength 0.10 m from a +4.0 nC point charge.

  1. E = kQ / r2 = 8.99 x 109 x 4.0 x 10-9 / 0.102
  2. = 35.96 / 0.010

Answer: E = 3600 N/C (2 s.f.), directed away from the charge

Key idea: E = F/Q; F = kQ_1Q_2/r2 with k = 8.99 x 109 N m2 C-2; E = kQ/r2 for a point charge; inverse square.

Check yourself

  1. A charge of 2.0 µC feels a force of 0.050 N. Find the electric field strength in N/C.
    Show answer

    25000 N/C. E = F / Q = 0.050 / 2.0 x 10-6 = 25000 N/C.

  2. Find the force between two protons 1.0 x 10-15 m apart, in N, to 3 s.f.
    Show answer

    230 N. F = 8.99 x 109 x (1.60 x 10-19)2 / (1.0 x 10-15)2 = 230 N.

  3. Charges of +3.0 µC and -5.0 µC are 0.20 m apart. Find the size of the force between them in N, to 3 s.f.
    Show answer

    3.37 N. F = 8.99 x 109 x 3.0 x 10-6 x 5.0 x 10-6 / 0.040 = 3.37 N (attractive).

  4. Find the field strength 0.050 m from a 6.0 nC point charge, in kN/C, to 3 s.f.
    Show answer

    21.6 kN/C. E = 8.99 x 109 x 6.0 x 10-9 / 0.0025 = 21600 N/C = 21.6 kN/C.

  5. An electron is in a uniform field of 5000 N/C. Find its acceleration, in units of 1014 m/s2, to 3 s.f.
    Show answer

    8.78 x 1014 m/s2. F = eE = 1.60 x 10-19 x 5000 = 8.0 x 10-16 N; a = F / m = 8.0 x 10-16 / 9.11 x 10-31 = 8.78 x 1014 m/s2.

  6. The distance between two point charges is doubled. What happens to the force between them?
    • It halves
    • It becomes a quarter as large
    • It doubles
    • It stays the same
    Show answer

    It becomes a quarter as large. F is proportional to 1/r2, and (1/2)2 = 1/4.

  7. How is the direction of an electric field at a point defined?
    • The direction of the force on an electron placed there
    • Always towards the nearest charge
    • At right angles to the force on a positive charge
    • The direction of the force on a small positive charge placed there
    Show answer

    The direction of the force on a small positive charge placed there. The field direction is defined using a positive test charge.

  8. Which describes the field lines around an isolated positive point charge?
    • Radial lines pointing inwards
    • Parallel, equally spaced lines
    • Radial lines pointing outwards
    • Circles centred on the charge
    Show answer

    Radial lines pointing outwards. A positive charge pushes a positive test charge directly away from it.

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Last updated 7 November 2026. Spotted a mistake? Tell us.