A uniformly charged solid insulating sphere has a radius R and total charge +Q . A narrow, smooth tunnel is drilled through its centre. A point charge -q (where q Q ) of mass m is released from rest at the surface of the sphere and allowed to fall through the tunnel. Its speed when it reaches the centre of the sphere is: (Assume ₀ is the permittivity of vacuum and neglect gravity)

Step-by-step solution

The electric potential at the surface of a uniformly charged solid sphere is: V_i = 1 4 ₀ Q R Initial potential energy of the charge -q at the surface is: U_i = -qV_i = - Qq 4 ₀ R The electric potential at the centre of a uniformly charged solid insulating sphere is 3 2 times the potential at the surface: V_f = 3 2 ( 1 4 ₀ Q R ) = 3Q 8 ₀ R Final potential energy of the charge -q at the centre is: U_f = -qV_f = - 3Qq 8 ₀ R Applying conservation of mechanical energy ( K_i + U_i = K_f + U_f ): 0 - Qq 4 ₀ R = 1 2 mv^2