Question

Relation electric potential at surface and centre of sphere

Answer 1

Answer:

Let the potential of each smaller drop be U, so

U = kq/r

now,

The new potential of the larger drop will be given as

U' = kq'/r'

here

q' = nq

and

as volume V' = nV

r'3 = nr3

or

r' = n1/3r

thus, we have

U' = knq / n1/3r

or

U' = (n/ n1/3).(kq/r)

thus,

U' = n2/3U