# n identical droplets are charged to v volt each. If they coalesce to form a single drop, then its potential will be (A) n2/3v (B) n1/3v (C) nv (D) v/n

Arun
25757 Points
3 years ago
Dear Aniket

Let potential f 1 drop = Kq/r =V

Volume of n drops = volume of bigger drop

n  4/3 pi (r)^3  = 4/3 pi (R)^3

Therefore ,  R = ( (n)^1/3) r

also charge remains conserved

charge on 1 drop = q

charge on big drop containing n drops = (nq)

Therefore potential on big drop = K (nq)/ ((n)^1/3)r

=(( n)^2/3) Kq/r

= (( n)^2/3) V

Hope it helps

Vikas TU
14149 Points
3 years ago
Let r, q and v be the radius, charge the potential of a small drop.
The total charge on the bigger drop is the sum of all charge on small drop s
Q = Nq
The volume of N small drops = Nx 4/3 πr cube
And for the bigger drop = 4/3 πR cube
Hence,
4/3 πR cube = N x 4/3 π r cube
R = N raise to power 1/3 r
So the potential on bigger drop
V = 1/ 4π Ɛo Q/R = 1/4πƐ0 Nq/N raise to power 1/3r = N raise to power 2/3 x 1/ 4πƐo q/r
V = N raise to power 2/3 x v
And the capacitance  C = 4πƐoR
C = 4πƐoN raise to power 1/3 r = N raise to power 1/3 (4πƐor)
C = N raise to power 1/3c