Two parallel plates R and S are 2 mm apart in a vacuum. An electron with charge –1.6 × 10-19 C moves along a straight line in the electric field between the plates.

Question 17

Two parallel plates R and S are 2 mm apart in a vacuum. An electron with charge –1.6 × 10^-19 C moves along a straight line in the electric field between the plates. The graph shows how the potential energy of the electron varies with its distance from plate R.

Which deduction is not correct?

A The electric field between R and S is uniform.

B The electric field strength is 3000 N C^-1.

C The force on the electron is constant.

D The magnitude of the potential difference between R and S is 3 V.

Reference: Past Exam Paper – November 2015 Paper 13 Q29

Solution:

Answer: B.

The electric field between two parallel plates is uniform.

Electric field strength, E = F / q

Force on the charge: F = Eq

Since the electric field strength is constant, the force on the electron is also constant.

(Potential energy) Work done = Force × distance, s

Work done = (Eq) × s

From the graph, when distance s = 2 mm, potential energy (work done) = 4.8×10^-19 J

Work done = (Eq) × s

4.8×10^-19 = E × 1.6×10^-19 × 2×10^-3

Electric field strength, E = (4.8×10^-19) / (1.6×10^-19 × 2×10^-3) = 1500 N C^-1

So, the electric field strength E is 1500 N C^-1, not 3000 N C^-1. Deduction B is NOT correct. [Choice B is the answer]

For parallel plates,

E = V / d

Potential difference, V = Ed = 1500 × 2×10^-3 = 3 V