Question

$N$ identical spherical drops are charged to the same potential $V$. They combine to form a bigger drop. The potential of the big drop will be:

• A. $V{N}^{1/3}$
• B. $V{N}^{2/3}$
• C. $V$
• D. $VN$

Answer 1

Answer: (B)

Volume of $N$ small drops is equal to volume of single big drop.
Volume of $N$ small drops = volume of one big drop i.e.,
$\left(\frac{4}{3}\pi r^3\right)N=\frac{4}{3}\pi R^3$
$\Rightarrow R=N^{1/3}r$
Also electric potential on big drop is
$V=\frac{1}{4\pi {\epsilon }_0}\frac{Nq}{R}$
$=\frac{1}{4\pi {\epsilon }_0}\frac{Nq}{N^{1/3}r}$
$\Rightarrow V=N^{2/3}V$
$\left(asV=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}\right)$