Small drops of the same size are charged to v volt each. If such drops coalesce to form a single large drop, its potential will be
Answers
Answered by
11
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
Similar questions