A unit positive point charge is taken slowly through an infinitesimally thin tube inside a charged dielectric sphere of radius R having uniform positive charge density ρ. The initial and final positions A and B are at distances 2R and 3R from the centre. In this process the magnitude of the total work done on the point charge is (ρ R²)/(nε₀). The value of n is: (ε₀ = permittivity of vacuum)

Solution & Explanation

Both points A (2R) and B (3R) lie OUTSIDE the sphere, so the field there is that of a point charge Q at the centre. Total charge: Q = ρ·(4/3)π R³. Potential outside at distance r: V = kQ/r, with k = 1/(4πε₀). Work by external agent (slow move) = q(V_B − V_A), with q = +1: W = kQ[1/(3R) − 1/(2R)] = kQ·(2 − 3)/(6R) = −kQ/(6R). Magnitude: |W| = kQ/(6R) = (1/(4πε₀))·(1/(6R))·ρ·(4/3)π R³. Simplify the constants: (1/(4π))·(1/6)·(4/3)π = (4π)/(72π) = 1/18. So |W| = ρR²/(18ε₀). Comparing with ρR²/(nε₀) gives n = 18, option D. Trap: since both points are outside, the uniform-density interior field is never used — treat it as a point charge.