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 $C^{'}$, charge $Q$ and potential $V^{'}$ will be 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^{\prime}}=\frac{n q}{n^{1 / 3} C} $
$=n^{2 / 3} V$