A large charged spherical liquid drop has an electric potential of 320 V at its surface. It breaks into N identical smaller spherical droplets. If the electric potential at the surface of each small droplet is 20 V , find the value of N . (Assume no loss of charge or volume during the process).

Step-by-step solution

Let the radius of the large drop be R and its charge be Q . The electric potential at the surface of the large drop is V = kQ R = 320 V . Let the radius of each small droplet be r and its charge be q . By conservation of charge, Q = Nq q = Q N . By conservation of volume, 4 3 R^3 = N ( 4 3 r^3 ) r = R N^ 1/3 . The electric potential at the surface of each small droplet is v = kq r . Substituting the values of q and r , we get: v = k(Q/N) R/N^ 1/3 = kQ R 1 N^ 2/3 = V N^ 2/3 Given v = 20 V and V = 320 V , we have: 20