Question

When n small drops of a conducting liquid , each of surface charge density $\sigma$ and radius $r$ are made to combine to form a big drop of radius $R$, then:

• A. potential becomes $n^{1/3}$ times original potential and charge density decreases to $n^{1/3}$ times original charge density
• B. potential becomes $n^{2/3}$ times and charge density increases to $n^{1/3}$times original charge density
• C. potential and charge density decrease to $n^{4/3}$ times original values
• D. potential and charge density increase to n times original values

Answer 1

The correct option is B potential becomes $n^{2/3}$ times and charge density increases to $n^{1/3}$times original charge density

Volume of n small drops is equal to that of the big drop
$n\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3\Rightarrow n{r}^3=R^3$
Let each drop has a surface charge density $\sigma$.
Therefore,
$V_{smalldrop}=\frac{1}{\left(4\pi {ϵ}_0\right)}\times \frac{\sigma 4\pi r^2}{r}=\frac{\sigma r}{{ϵ}_0}$.
Now, $n\sigma 4\pi r^2=x4\pi R^2\Rightarrow n\times n^{\frac{-2}{3}}=\frac{x}{\sigma }$
$x=n^{\frac{1}{3}}\sigma$
Also,potential on big drop $=\frac{k\times n\times \sigma 4\pi r^2}{R}=\frac{n}{{ϵ}_0}\times \frac{r^2}{\left(n^{\frac{1}{3}}r\right)}=n^{\frac{2}{3}}V$