Question

Eight small drops, each of radius $r$ and having same charge $q$ are combined to form a big drop. The ratio of potential of the bigger drop to the smaller drop is

• A. $8:1$
• B. $4:1$
• C. $2:1$
• D. $1:8$

Answer 1

The correct option is B $4:1$

When $8$ small drops are combined to a single drop,
let $Q$ be charge and $R$ be radius of drop after coalesce.
Net charge $Q=8q$
Final volume $=8\times$ initial volume.
$4\pi R^3=8\times 4\pi r^3$
$R^3=8r^3$
$R=2r$

Potential of a sphere of charge $Q$ and radius $R$ is $V=\frac{Q}{4\pi {ϵ}_{0}R}$

Ratio of potentials $\frac{V_{big}}{V_{small}}=\frac{Q}{R}\times \frac{r}{q}$
$\frac{V_{big}}{V_{small}}=\frac{8q}{2r}\times \frac{r}{q}=4$