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A2 Daily A Level Physics question

2026-05-21 OCR A Electric fields: Coulomb; uniform fields; E = V/d; potential & equipotentials; motion of charges (qual.) 6.2.2

Two parallel horizontal conducting plates are separated by a distance d in a vacuum. A constant potential difference V is applied across the plates, creating a uniform electric field. A small sphere with a fixed positive charge is held stationary between the plates and experiences an electrostatic force F. If the potential difference is doubled and the plate separation is halved, what is the new electrostatic force experienced by the sphere?

  1. A It remains at F because the two changes cancel each other out.
  2. B It increases to 2F because the force is only proportional to the potential difference.
  3. C It increases to 4F because the field strength is proportional to potential difference and inversely proportional to plate separation. (correct)
  4. D It decreases to 0.5F because halving the distance reduces the strength of the field at that point.

Answer

The correct answer is C.

Correct: C — It increases to 4F because the field strength is proportional to potential difference and inversely proportional to plate separation. A is incorrect because it assumes the doubled potential difference and halved separation cancel out, whereas they both act to increase the field strength. B is incorrect because it ignores the effect of the plate separation change, which also increases the field strength. C is correct because the uniform electric field strength is given by potential difference divided by plate separation; doubling the potential difference and halving the separation increases the field strength by a factor of four, thereby quadrupling the electrostatic force on the charge. D is incorrect because it assumes that decreasing the plate separation decreases the field strength, whereas it actually increases it.