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

Updated On: Apr 5, 2022
  • $7.2 \times 10^{3}V$
  • $7.2 \times 10^{2}V$
  • $1.44 \times 10^{2}V$
  • $1.44 \times 10^{3}V$
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The Correct Option is A

Solution and Explanation

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 $V_{2} =\frac{1}{4 \pi \varepsilon_{0}} \frac{64 q}{R}$ $V_{2} =\frac{1}{4 \pi \varepsilon_{0}} \frac{64 q}{4 r}$ $=\frac{9 \times 10^{9} \times 64 \times 10^{-9}}{4 \times 2 \times 10^{-2}}$ $=7.2 \times 10^{3} V$
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