Home · A-Level 9702 · Electric fields · Electric field of a point charge · Practice
A2 · Practice questions · Electric field of a point charge

Radial, and inverse square.

Six original Cambridge-style questions on the field E = Q / (4πε₀r²) of a point charge or charged sphere, its inverse-square shape, and adding the fields of several charges as vectors.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

E is a vector: add directions, not just sizes.

01
Calculation
[2 marks]

Calculate the electric field strength at a distance of 25 mm from a point charge of +8.0 nC in a vacuum, and state its direction.

  • E = Q / (4πε₀r²) = 8.99 × 10⁹ × 8.0 × 10⁻⁹ / 0.025² = 1.2 × 10⁵ V m⁻¹. ✓
  • The field is directed radially outward, away from the charge, because the charge is positive. ✓
02
Analysis
[2 marks]

Use the idea of field lines to explain why the electric field strength of an isolated point charge is proportional to 1 / r².

  • A fixed number of field lines leaves the charge and spreads out radially, so at distance r the same lines are spread over the surface of a sphere of area 4πr². ✓
  • The field strength is shown by the number of lines crossing unit area, so it falls in proportion to 1 / (4πr²), that is as 1 / r²; the 4π is the same factor that appears in E = Q / (4πε₀r²). ✓
03
Calculation
[3 marks]

Point charges of +3.0 nC at A and −5.0 nC at B are fixed 80 mm apart in a vacuum. Point P lies on the line AB extended, 40 mm beyond B on the side away from A. Calculate the resultant electric field strength at P and state its direction.

  • Due to A (r = 0.120 m): E = 8.99 × 10⁹ × 3.0 × 10⁻⁹ / 0.120² = 1.9 × 10³ V m⁻¹, directed away from A, that is from B towards P. ✓
  • Due to B (r = 0.040 m): E = 8.99 × 10⁹ × 5.0 × 10⁻⁹ / 0.040² = 2.8 × 10⁴ V m⁻¹, directed towards the negative charge B, that is from P towards B. ✓
  • The two fields are in opposite directions, so the resultant is 2.81 × 10⁴ − 0.19 × 10⁴ = 2.6 × 10⁴ V m⁻¹ directed from P towards B. ✓
04
Calculation
[3 marks]

Two point charges, each +4.0 nC, are fixed 60 mm apart in a vacuum. Point P lies on the perpendicular bisector of the line joining them, 40 mm from its midpoint, so that each charge is 50 mm from P. Calculate the resultant electric field strength at P and state its direction.

  • Each charge gives a field of size E = 8.99 × 10⁹ × 4.0 × 10⁻⁹ / 0.050² = 1.44 × 10⁴ V m⁻¹ at P, directed away from that charge. ✓
  • The components parallel to the line joining the charges are equal and opposite and cancel; the components along the bisector add, each being E cos θ with cos θ = 40 / 50 = 0.80. ✓
  • Resultant = 2 × 1.44 × 10⁴ × 0.80 = 2.3 × 10⁴ V m⁻¹, directed along the bisector away from the line joining the charges. ✓
05
Analysis
[3 marks]

Compare the electric field of an isolated point charge with the gravitational field of an isolated point mass. State two similarities and one difference.

  • Both are radial fields, directed along lines through the point, and both obey an inverse-square law: E = Q / (4πε₀r²) and g = GM / r² have the same form. ✓
  • Both field strengths are defined as a force per unit property of a test object: force per unit positive charge and force per unit mass. ✓
  • A gravitational field is always directed towards the mass, because gravity is only attractive, whereas an electric field points away from a positive charge and towards a negative one, so electric forces can repel as well as attract. ✓
06
Calculation
[4 marks]

A metal sphere of radius 20 mm carries a charge of +12 nC, uniformly spread over its surface, and is isolated in a vacuum.

(a) Calculate the electric field strength at the surface of the sphere. (b) Determine the distance from the centre of the sphere at which the field strength has fallen to 3.0 × 10⁴ V m⁻¹. (c) State the electric field strength inside the metal of the sphere.

  • (a) Outside the sphere it acts as a point charge at its centre: E = 8.99 × 10⁹ × 12 × 10⁻⁹ / 0.020² = 2.7 × 10⁵ V m⁻¹. ✓
  • (b) r² = Q / (4πε₀E) = 8.99 × 10⁹ × 12 × 10⁻⁹ / 3.0 × 10⁴ = 3.6 × 10⁻³ m². ✓
  • r = 0.060 m, that is 60 mm from the centre or 40 mm above the surface; a check is that the field at three times the radius is one ninth of the surface value, 2.7 × 10⁵ / 9 = 3.0 × 10⁴ V m⁻¹. ✓
  • (c) Zero: all the charge sits on the surface and the free electrons in the metal would move if there were any field inside. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.