A large solid sphere with uniformly distributed positive charge has a smooth narrow tunnel through its centre. A small particle with negative charge, initially at rest far from the sphere, approaches it along the line of the tunnel, reaches its surface with a speed v , and passes through the tunnel. Its speed at the centre of the sphere will be

Step-by-step solution

Potential at infinity =V_ =0 Potential at the surface of the sphere, V_s=k Q R Potential at the centre of the sphere, V_c= 3 2 k Q R Let m and -q be the mass and the charge of the particle respectively. Let v₀= speed of the particle at the centre of the sphere. aligned & 1 2 m v^2=-q [V_ -V_s ]=q k Q R & 1 2 m v₀^2=-q [V_ -V_c ]=q 3 2 k Q R aligned Dividing eqn. (ii) by eqn. (i), v₀^2 v^2 = 3 2 =1.5 or v₀= 1.5 v