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Question

Physics Question on Electrostatic potential

Twenty seven drops of same size are charged at 220V220\, V each. They combine to form a bigger drop. Calculate the potential of the bigger drop.

A

660 V

B

1320 V

C

1520 V

D

1980 V

Answer

1980 V

Explanation

Solution

If each drop has a charge ' qq ' and radius ' rr '. Then from conservation of charge, charge on the big drop is nq=27q(n=27)n q=27 q(n=27) from conservation of volume 43πr3n=43πR3\frac{4}{3} \pi r^{3} n=\frac{4}{3} \pi R^{3} R=n1/3rR=n^{1 / 3} r Now potential of the small drop V=q4πϵ0r=220VV=\frac{q}{4 \pi \epsilon_{0} r}=220 V Potential of the big drop, V=nq4π0R=nq4π0n1/3r=n2/3q4π0rV=\frac{n q}{4 \pi \in_{0} R}=\frac{n q}{4 \pi \in_{0} n^{1 / 3} r}=n^{2 / 3} \frac{q}{4 \pi \in_{0} r} V=(27)2/3×220VV=(27)^{2 / 3} \times 220 V =9×220=1980V=9 \times 220=1980 V