Question

If $n$ drops of potential $V$ merge, find new potential on the big drop.

• A. $n^{2/3}V$
• B. $n^{1/3}V$
• C. $nV$
• D. $V^{n/3}$

Answer 1

Answer: (A)

Volume of big drop $=n\times$ volume of a small drop
$\Rightarrow \frac{4}{3}\pi R^3=n\frac{4}{3}\pi r^3$
$\Rightarrow R=n^{1/3}r\dots$ (i)
Potential of small drop $V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$
$\Rightarrow q=V\times 4\pi {\epsilon }_{0}r\dots$(ii)
So, potential of big drop $=\frac{1}{4\pi {\epsilon }_0}\times \frac{nq}{R}$
$\Rightarrow V^{\prime }=\frac{4\pi {\epsilon }_0\times V\times r}{4\pi {\epsilon }_0\times n^{1/3}r}=n^{2/3}V$