# 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!!

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.

$\implies B ' = B + B + B + \ldots \ldots + B$

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

$\implies B ' = n B$

A drop is spherical in shape. Volume of a sphere is given by $\frac{4}{3} \pi {r}^{3}$ where $r$ is its radius.

$\implies \frac{4}{3} \pi {R}^{3} = n \frac{4}{3} \pi {r}^{3}$

$\implies {R}^{3} = n {r}^{3}$

Taking third root on both sides.

$\implies R = {n}^{\frac{1}{3}} r$

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

$\implies Q = n q$

The potential of the bigger drop can be given by

$V ' = \frac{k Q}{R}$

$\implies V ' = \frac{k n q}{{n}^{\frac{1}{3}} r}$

$\implies V ' = {n}^{1 - \frac{1}{3}} \frac{k q}{r}$

$\implies V ' = {n}^{\frac{2}{3}} \frac{k q}{r}$

Since, $k \frac{q}{r}$ represents the potential of small drop which we have symbolized by $V$.

Therefore, $V ' = {n}^{\frac{2}{3}} V$

Now we have found a general equation for this case.

In this case there are $27$ identical drops.

$\implies V ' = {27}^{\frac{2}{3}} V$

$\implies V ' = 9 V$

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