A fixed uniformly charged insulating sphere of radius R has total charge +Q. A point charge -q (q Q, mass m) is released from rest at a distance 3R from the centre. When it reaches the surface of the sphere, its speed is: (ε₀ = permittivity of vacuum; neglect gravity)

Solution & Explanation

Both the starting point (3R) and the surface (R) are outside or on the sphere, so the uniformly charged sphere acts like a point charge +Q at the centre, with potential V = (1/4πε₀)(Q/r). The charge −q is released from rest; use energy conservation: KE_gain = PE_loss = (−q)(V_surface − V_start)... more cleanly, ½mv² = q·Q/(4πε₀)·(1/R − 1/3R). Compute the bracket: 1/R − 1/3R = (3 − 1)/3R = 2/(3R). So ½mv² = (1/4πε₀)·(Qq)·(2/3R) = Qq/(6πε₀R). Then v² = 2·Qq/(6πε₀mR) = Qq/(3πε₀mR). Taking the square root: v = √(Qq/(3πε₀mR)). Key point: the negative charge is attracted toward +Q, so it speeds up; the magnitude of the energy released gives answer (C).