Question

A charge of $10^{-9} C$ is placed on each of the $64$ identical drops of radius $2\, cm$. They are then combined to form a bigger drop. Its potential will be

• A. $7.2 \times 10^{3}V$
• B. $7.2 \times 10^{2}V$
• C. $1.44 \times 10^{2}V$
• D. $1.44 \times 10^{3}V$

Answer 1

Answer: (A)

Volume of bigger drop = volume of $64$ small drops
ie., $\frac{4}{3} \pi R^{3}=64 \times \frac{4}{3} \pi r^{3}$
or $R =4r$
Potential on small drop
$V_{1}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}$
Potential on bigger drop
$V_{2} =\frac{1}{4 \pi \varepsilon_{0}} \frac{64 q}{R}$
$V_{2} =\frac{1}{4 \pi \varepsilon_{0}} \frac{64 q}{4 r}$
$=\frac{9 \times 10^{9} \times 64 \times 10^{-9}}{4 \times 2 \times 10^{-2}}$
$=7.2 \times 10^{3} V$