Help me details mathematically A soap bubble of radius R0 is

A soap bubble of radius R0 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(V_0/V), where p is the atmospheric pressure, V0isthe initial volume, and V is the final volume. Show that q^2 = 32 pi^2 epsilon pR(R^3 - R^3_o).

Solution

initially :

Vo = 4 pi Ro^3 / 3

and PV =constant

p [ 4 pi Ro^3 / 3 ] = p\' [ 4 pi R^3 / 3 ]

p\' = p (Ro /R)^3   ........Final pressure

electric force = kq/R^2

prssurae change due to this force = Fe/ 4 pi R^2

hence this pressure change due electric force= initial - final pressure

[ k q / R^2 ] ( 4 pi R^2 ) = p - p\'

(kq) / (4 pi R^4) = p - ( p (Ro /R)^3

kq / 4pi R^4 = p (R^3 - Ro^3) / R^3

k q = 4 pi R p (R^3 - Ro^3)

and k = 1 / 8 pi e0

q = (4 pi e0 ) (4 pi R p ) ( R^3 - Ro^3)

q = 32 pi^2 e0 p R ( R^3 - Ro^3)