Question

'\(n\)' small spherical drops of the same size which are charged to 'V' volt each coalesce to form a single big drop. The potential of a big drop is

• A. \(nV\)
• B. \(\frac{V}{n}\)
• C. \((n^{\frac{1}{3}})V\)
• D. \((n^{\frac{2}{3}})V\)

Hint:

Charge adds up and the radius scales as n^(1/3).

Answer 1

The Correct Option is D

Step 1: Charge and Radius:
Each small drop has charge \(q\) and radius \(r\), with \(V=\dfrac{kq}{r}\). The big drop has charge \(Q=nq\). Volume is conserved: \(\dfrac43\pi R^3=n\cdot\dfrac43\pi r^3\), so \(R=n^{1/3}r\).

Step 2: Potential of the Big Drop:
\[ V'=\frac{kQ}{R}=\frac{k\,nq}{n^{1/3}r}=n^{2/3}\frac{kq}{r}=n^{2/3}V \]

Step 3: Check the Options:
\(nV\) would hold if the radius did not change. \(V/n\) is a decrease and \(n^{1/3}V\) forgets that charge also adds up. So (D) is correct.

Final Answer:
The potential of the big drop is \(n^{2/3}V\), option (D). \[ \boxed{\text{(D) } n^{2/3}V} \]