A soap bubble of radius Ro is slowly given a charge q. Because of mutual repulsion of the surface charges,the radius increases slightly to R. The air pressure inside the bubble drops, because of the expansion, to p(Vo/V) where p is the atmospheric pressure, Vo is the initial volume, and V is the final volume.Show that
q2 = 32
2
o pR [ R3 - (Ro)3 ].