27 identical drops of water are equally and simillarly charged to potential V.They are then united to form a bigger drop.The potential of the bigger drop is??Thank u!!

1 Answer
Feb 1, 2016

Let me derive the general expressions for this condition.

Let there be n small drops each having a charge q on it and the radius r, V be its potential and let the volume of each be denoted by B.

When these n small drops are coalesced there is a new bigger drop formed.

Let the radius of the bigger drop be R, Q be charge on it, V' be its potential and its volume be B'

The volume of the bigger drop must be equal to the sum of volumes of n individual drops.

implies B'=B+B+B+......+B

There are total n small drops therefore the sum of volumes of all the individual drops must be nB.

implies B'=nB

A drop is spherical in shape. Volume of a sphere is given by 4/3pir^3 where r is its radius.

implies 4/3piR^3=n4/3pir^3

implies R^3=nr^3

Taking third root on both sides.

implies R=n^(1/3)r

Also the charge of the bigger drop must be equal to the sum of charges on the individual drops.

implies Q=nq

The potential of the bigger drop can be given by

V'=(kQ)/R

implies V'=(knq)/(n^(1/3)r)

implies V'=n^(1-1/3)(kq)/r

implies V'=n^(2/3)(kq)/r

Since, kq/r represents the potential of small drop which we have symbolized by V.

Therefore, V'=n^(2/3)V

Now we have found a general equation for this case.

In this case there are 27 identical drops.

implies V'=27^(2/3)V

implies V'=9V

This shows that in your case the potential of the bigger drop is 9 times the potential of the smaller drop.