Question:
512 identical drops of mercury are charged to a potential of $2 \mathrm{~V}$ each. The drops are joined to form a single drop. The potential of this drop is__________ $\mathrm{V}$.
Solution:
$Q=512 q$
Volume $_{\mathrm{i}}=$ Volume $_{\mathrm{f}}$
$512 \times \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi R^{3}$
$2^{9} r^{3}=R^{3}$
$\mathrm{R}=8 \mathrm{r}$
$2=\frac{\mathrm{kq}}{\mathrm{r}}$
$\mathrm{V}=\frac{\mathrm{kQ}}{\mathrm{R}}=\frac{\mathrm{k} 512 \mathrm{q}}{8 \mathrm{r}}$
$\mathrm{V}=128$