Question

A charge \( -6 \mu C \) is placed at the center B of a semicircle of radius 5 cm, as shown in the figure. An equal and opposite charge is placed at point D at a distance of 10 cm from B. A charge \( +5 \mu C \) is moved from point ‘C’ to point ‘A’ along the circumference. Calculate the work done on the charge.

Hint:

Work done in moving a charge along an equipotential surface is always zero because there is no change in electric potential.

Answer 1

Work done in moving a charge in an electrostatic field depends only on the potential difference between the initial and final positions. The work done is given by: \[ W = q \Delta V \] where: - \( W \) is the work done, - \( q = 5 \mu C = 5 \times 10^{-6} C \), - \( \Delta V = V_A - V_C \) (potential difference between points A and C). Since both C and A are on the same equipotential surface (same radial distance from the center B), their potential is the same: \[ V_A = V_C \] Thus, the potential difference: \[ \Delta V = V_A - V_C = 0 \] \[ W = (5 \times 10^{-6}) \times 0 = 0 \] Thus, no work is done on the charge.