A conducting bubble of outer radius a, thickness t <<<a is charged to potential V . The bubble collapses into a droplet . The potential of such a droplet is ???

It should be ( a/t ) V .

q/4 pi epsilon a = V

q = 4 pi epsilon V a

E(r) = q/ 4 pi epsilon r^2 = V a / r^2

so V(r) = -qVa/r

From potential , first find the charge , then this charge remains constant. Now  the new radius is t. Then find the new potential.

first ....

Q= CV...

now thre C = 4 pie epsilon a...

and now whn the bubble is changed into droplet..... u will get one droplet with same charge Q.. as charge wont change... remains const..

now for droplet we hve to find C' .... and put in relation Q' = C' V' = Q....now taking density of bubble and droplet to be const... as of same material....

we equate volumes....

4/3 pie[ (a+t)^3 - a^3]  = 4/3  pie r'^3......... by this way we can find radius of  new droplet ... using r' calculatin C'= 4 pie epsilon  r'....

hehe big calculation.... but yeah... as t<< a... u can binomially expand and reduce calculation... neglecting higher powers of t....

hence put in relation of C' V' = Q...

and find new potential... V'

Cheers!!!!!!!