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A2 · Practice questions · Coulomb’s law

Inverse square, either sign.

Six original Cambridge-style questions on Coulomb's law F = Q₁Q₂ / (4πε₀r²), forces between point charges and charged spheres, and the balance of forces along a line.

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

Product of the charges over the square of the separation.

01
Analysis
[3 marks]

State Coulomb's law for the force between two point charges. Explain why the law can be applied to two charged conducting spheres only when the charge on each remains uniformly distributed, and state when this is a good approximation.

  • The force between two point charges is proportional to the product of the charges and inversely proportional to the square of their separation, F = Q₁Q₂ / (4πε₀r²). ✓
  • It acts along the line joining the charges, repulsive for like charges and attractive for unlike charges. ✓
  • A uniformly charged sphere behaves as a point charge at its centre; if the spheres are close, each charge pulls or pushes the free charge on the other sphere to one side, so the distribution is no longer uniform. The approximation is good when the separation is large compared with the radii. ✓
02
Calculation
[3 marks]

Point charges of +4.0 nC and +9.0 nC are fixed 0.15 m apart in a vacuum. A third small charge is placed on the line joining them. Determine the position at which the resultant force on the third charge is zero.

  • The third charge must lie between the two charges, where the pushes (or pulls) from the two are in opposite directions; at distance x from the 4.0 nC charge, 4.0 / x² = 9.0 / (0.15 − x)² since the third charge and 1/(4πε₀) cancel. ✓
  • Taking square roots, (0.15 − x) / x = 3 / 2, so 0.15 − x = 1.5x and x = 0.060 m. ✓
  • The point is 60 mm from the +4.0 nC charge and 90 mm from the +9.0 nC charge, closer to the smaller charge; the result is the same whatever the sign of the third charge. ✓
03
Calculation
[3 marks]

In a hydrogen atom the electron and the proton are, on average, 5.3 × 10⁻¹¹ m apart. Calculate the electric force between them and the gravitational force between them, and comment on the result.

  • FE = e² / (4πε₀r²) = 8.99 × 10⁹ × (1.60 × 10⁻¹⁹)² / (5.3 × 10⁻¹¹)² = 8.2 × 10⁻⁸ N, attractive. ✓
  • FG = Gmpme / r² = 6.67 × 10⁻¹¹ × 1.67 × 10⁻²⁷ × 9.11 × 10⁻³¹ / (5.3 × 10⁻¹¹)² = 3.6 × 10⁻⁴⁷ N. ✓
  • The electric force is larger by a factor of about 2 × 10³⁹, so gravity plays no part in holding the atom together. ✓
04
Calculation
[3 marks]

Three point charges lie on a straight line in a vacuum: A, of charge +2.0 μC, at x = 0; B, of charge −3.0 μC, at x = 0.30 m; and C, of charge +1.0 μC, at x = 0.50 m. Calculate the resultant force on C and state its direction.

  • Force on C due to A: F = 8.99 × 10⁹ × 2.0 × 10⁻⁶ × 1.0 × 10⁻⁶ / 0.50² = 0.072 N, repulsive, so directed away from A (in the +x direction). ✓
  • Force on C due to B: F = 8.99 × 10⁹ × 3.0 × 10⁻⁶ × 1.0 × 10⁻⁶ / 0.20² = 0.67 N, attractive, so directed towards B (in the −x direction). ✓
  • Resultant = 0.674 − 0.072 = 0.60 N towards B, in the −x direction. ✓
05
Analysis
[2 marks]

Sphere X carries a charge ten times as large as sphere Y, and the two spheres are placed some distance apart. A student claims that X exerts a larger force on Y than Y exerts on X. Explain why the student is wrong, and state what does differ between the two spheres if they are released.

  • Coulomb's law contains the product Q₁Q₂, the same whichever sphere is called 1, and the two forces are a Newton's third law pair: they are equal in magnitude and opposite in direction. ✓
  • What differs is the acceleration: if the spheres have different masses, the lighter one accelerates more under the same force. ✓
06
Calculation
[4 marks]

Two identical small conducting spheres carry charges of +6.0 nC and −2.0 nC. Their centres are 50 mm apart.

(a) Calculate the force between the spheres and state whether it is attractive or repulsive. (b) The spheres are touched together and then returned to the same separation. State the charge on each sphere. (c) Calculate the new force between the spheres and state its nature.

  • (a) F = 8.99 × 10⁹ × 6.0 × 10⁻⁹ × 2.0 × 10⁻⁹ / 0.050² ✓
  • = 4.3 × 10⁻⁵ N, attractive because the charges have opposite signs. ✓
  • (b) Charge is conserved and shared equally between identical spheres: (+6.0 − 2.0) / 2 = +2.0 nC on each. ✓
  • (c) F = 8.99 × 10⁹ × (2.0 × 10⁻⁹)² / 0.050² = 1.4 × 10⁻⁵ N, now repulsive because both charges are positive. ✓

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.