A2 Daily A Level Physics question
A pair of large parallel plates produces a uniform electric field. Equipotential lines are straight and 1.0 cm apart, with a 100 V step between neighbouring lines. A +2.0 nC bead is moved slowly from a point on the 200 V line to a point on the 300 V line. Two routes are tried: - Route P: straight across, perpendicular to the equipotentials, distance 1.0 cm. - Route Q: a longer 3.0 cm zig‑zag path that mostly runs parallel to the equipotentials but ends on the 300 V line. Which statement must be true about the work that must be done by the student?
Answer
The correct answer is D.
Correct: D — Both routes require the same work, 0.20 µJ, because only the 100 V potential rise matters for a +2.0 nC charge. A path length does not affect work in a conservative field; W depends on q and the potential difference: 2.0 nC × 100 V = 0.20 µJ. A More work is needed on Q because the path is three times longer confuses distance with potential change; in a conservative field, work is path‑independent. B More work is needed on P because it goes straight ‘against’ the field lines is incorrect; direction affects force direction, but the work between fixed potentials depends only on ΔV, not on the route. C No work is needed on either route because the bead is moved slowly confuses speed with energy change; moving a positive charge to higher potential requires positive work even if done quasi‑statically. D Correct as stated: same work on both routes, 0.20 µJ, set by the 100 V increase for a +2.0 nC charge.