a small ball of mass 1kg and a charge 2/3 uC is placed at the center of a uniformly charged sphere of radius 1m and charge 1/3 mC. a narrow smooth horizontal groove is made in the sphere from centre to surface as shown in figure. the sphere is made to rotate about its vertical diameter at a constant rate of 1/2pi revolutions per second. find the spedd w.r.t ground with which the ball slide out from the groove. neglect any magnetic force acting on ball?

Joshi sir comment

Potential difference from centre to surface = 3V/2 - V = V/2 here V = kq/r (q is charge of sphere) mvr2/2 = VQ/2 (Q is charge of particle) vr is radial velocity Besides tangential velocity due to rotation of sphere = rω=r2πn now solve  

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