Question

Two drops of water which are identical carry equal potential of 1 volt. If these two drops merge into one large drop, what is the potential of the new drop?

Answer 1

Assume the radius of the drops to be R ans say they carry a charge Q (>0) each. Now potential of each is kQ/R.
When they merge, new radius is 2^(1/3) times R [Volume conserved] and the charge carried by it is 2Q [charge conserved]. Now, it's potential is k(2Q)/2^(1/3)R, that is 2^(2/3) times the potential of the original drop, ie 1V. Thus, Answer = 2^(2/3) V.