Question

There are $27$ drops of a conducting fluid. Each drop has radius r, and each of them is charged to the same potential $V_1$.They are then combined to form a bigger drop.The potential of the bigger drop is $V_2$ .Find the ratio $V_2/V_1$. Ignore the change in density of the fluid on combining the drops.

Answer 1

If $q$ be the charge on each drop, charge on big drop is $Q=27q$.
If $R$ be the radius of bigger drop, $\frac{4}{3}\pi R^3=27\left(\frac{4}{3}\pi r^3\right)\Rightarrow R=3r$
Thus, $V_1=k\frac{q}{r}$
and $V_2=k\frac{Q}{R}=k\frac{27q}{3r}=9k\frac{q}{r}=9V_1$
$\therefore \frac{V_2}{V_1}=9$