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. $nV^2$
• D. $n^{2/3} V$

Answer 1

Answer: (D)

$n^{2/3} V$

Answer 2

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 $C'$, charge $Q$ and potential $V'$ will become :
Charge on big drop $Q=n q$
Capacitance of big drop $C'=n^{1 / 3} C$
Hence potential of big drop
$V'=\frac{Q}{C'}$
$=\frac{n q}{n^{1 / 3} C}$
$=n^{2 / 3} V$