Question

$64$ droplets of mercury each of radius $r$ and carrying charge $q$, coalesce to form a big drop. Compare the surface density of charge of each drop with is that of the big drop.

Answer 1

Given that,

Droplets of mercury = $64$

Density of small droplet $={\sigma }_{small}$

Density of big droplet $={\sigma }_{big}$

We know that,

$\frac{{\sigma }_{small}}{{\sigma }_{big}}=\frac{q}{Q}\times \frac{R^2}{r^2}$

$\frac{{\sigma }_{small}}{{\sigma }_{big}}=\frac{q}{\left(nq\right)}\times \frac{\left(n^{\frac{1}{3}}r\right)}{r^2}$

$\frac{{\sigma }_{small}}{{\sigma }_{big}}=n^{-\frac{1}{3}}$

Now, put the value of $n$

$\frac{{\sigma }_{small}}{{\sigma }_{big}}={\left(64\right)}^{-\frac{1}{3}}$

$\frac{{\sigma }_{small}}{{\sigma }_{big}}=\frac{1}{4}$

Hence, the ratio is $\frac{1}{4}$