Electric potential, uniform fields and equipotentials

Electric potential

The electric potential V at a point is the work done per unit charge in bringing a small positive charge from infinity to that point. For a point charge:

V = Q / (4πε_0r) = kQ / r

V is a scalar: positive near a positive charge, negative near a negative charge, and zero at infinity. Moving a charge Q through a p.d. ΔV takes work W = QΔV.

Field and potential

The field strength is the potential gradient: E = -ΔV / Δr. The field points from high to low potential, and is strongest where the potential changes fastest with distance. Between parallel plates with p.d. V and separation d, the field is uniform:

E = V / d (so N/C = V/m)

Field lines and equipotentials

• Equipotentials join points at the same potential. No work is done moving a charge along one.
• Equipotentials are always at right angles to field lines.
• Uniform field: parallel, equally spaced field lines; equipotentials are parallel, equally spaced planes.
• Radial field: equipotentials are concentric spheres (circles on paper) that get further apart for equal steps in V as r increases.

Worked example

Two parallel plates 5.0 cm apart have a p.d. of 2000 V. Find the field strength and the force on an electron between them.

1. E = V / d = 2000 / 0.050 = 40000 V/m
2. F = eE = 1.60 x 10^-19 x 40000

Answer: E = 4.0 x 10⁴ V/m; F = 6.4 x 10^-15 N

Check yourself

1. Two plates 3.0 mm apart have a p.d. of 1500 V. Find the field strength, in units of 10⁵ V/m.
   Show answer

   5 x 10⁵ V/m. E = V / d = 1500 / 3.0 x 10^-3 = 5.0 x 10⁵ V/m.

2. Find the potential 0.30 m from a +2.0 nC point charge, in V, to 3 s.f.
   Show answer

   59.9 V. V = kQ / r = 8.99 x 10⁹ x 2.0 x 10^-9 / 0.30 = 59.9 V.

3. How much work is done moving a 3.0 µC charge through a p.d. of 200 V? Answer in mJ.
   Show answer

   0.6 mJ. W = QΔV = 3.0 x 10^-6 x 200 = 6.0 x 10^-4 J = 0.60 mJ.

4. Find the p.d. between points 0.10 m and 0.20 m from a +5.0 nC point charge, in V, to 3 s.f.
   Show answer

   225 V. V_0.10 = 449.5 V and V_0.20 = 224.75 V, so the p.d. is 224.75 V, which is 225 V to 3 s.f.

5. An electron is accelerated from rest through 2000 V. Find its speed, in units of 10⁷ m/s, to 3 s.f.
   Show answer

   2.65 x 10⁷ m/s. eV = (1/2)mv²: v = sqrt(2 x 1.60 x 10^-19 x 2000 / 9.11 x 10^-31) = 2.65 x 10⁷ m/s.

6. What do the equipotentials look like in the uniform field between two parallel plates?
   • Parallel, equally spaced planes parallel to the plates
   • Concentric circles
   • Lines joining the two plates
   • Planes that get closer together near one plate
   Show answer

   Parallel, equally spaced planes parallel to the plates. They are perpendicular to the parallel field lines and equally spaced because the field is uniform.

7. How much work is done moving a charge along an equipotential?
   • QV
   • It depends on the path
   • None
   • Q / V
   Show answer

   None. ΔV = 0 along an equipotential, so W = QΔV = 0.

8. How is electric field strength related to electric potential?
   • Field strength equals the potential
   • Field strength is the area under a potential-distance graph
   • They are not related
   • Field strength is the negative of the potential gradient
   Show answer

   Field strength is the negative of the potential gradient. E = -ΔV / Δr: the field points down the potential slope.

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Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics specification (XPH11/YPH11), Issue 3
• Pearson qualifications: International Advanced Levels Physics (2018)
• Pearson Edexcel IAL Physics Unit 4 (WPH14) question paper, June 2024: list of data, formulae and relationships

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