Total number of drops , N=512
Potential of the each small drop , Vs=14πϵ0 qr ......(1)
Potential of the big drop , Vb=14πϵ0 QR=14πϵ0 NqR .....(2)
We know that, Volume of bigger drop is N times the volume of each small drop.
43πR3=N(43 πr3)
⇒R=N13 r
Equation (2) becomes
Vb=14πϵ0 Nq⎛⎜⎝N13r⎞⎟⎠=N23[14πϵ0 qr]
=N23 Vs
Potential of the big drop Vb=(512)23 (2)
=(83)23 2=64×2=128 V