Relation electric potential at surface and centre of sphere

1.	Relation electric potential at surface and centre of sphere
Answer» 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

Answer»

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

r' = n1/3r

thus, we have

U' = knq / n1/3r

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

thus,

U' = n2/3U

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Relation electric potential at surface and centre of sphere

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