If 216 drops of the same size are charged at 200 ~V each and they combine to form a bigger drop, the potential of the bigger drop will be

Step-by-step solution

Let r be the radius of each small drop and q be the charge on each small drop. The potential of a small drop is given by V = kq r = 200 V . When n = 216 drops combine to form a bigger drop of radius R and charge Q , the volume remains conserved. Thus, 4 3 R^3 = n 4 3 r^3 , which implies R = n^ 1/3 r . Substituting n = 216 , we get R = (216)^ 1/3 r = 6r . The total charge on the bigger drop is Q = nq = 216q . The potential of the bigger drop V' is given by V' = kQ R = k(nq) n^ 1/3 r = n^ 2/3 kq r . Substituting the