Question

$n$ identical mercury droplets charged to the same potential $V$ coalesce to form a single bigger drop. The potential of new drop will be

• A. $\frac{V}{n}$
• B. $nV$
• C. $n{V}^2$
• D. $n^{2/3}V$

Answer 1

Answer: (D)

Suppose we have $n$ identical drops each having radius $r$, capacitance $C$, charge $q$ and potential $V$. If these drops are combined to form a big drop of radius $R$, capacitance $C^{{}^{\prime }}$, charge $Q$ and potential $V^{{}^{\prime }}$ will be become :
Charge on big drop $Q=nq$
Capacitance of big drop $C^{{}^{\prime }}=n^{1/3}C$
Hence potential of big drop
$V^{\prime }=\frac{Q}{C^{\prime }}=\frac{nq}{n^{1/3}C}$
$=n^{2/3}V$