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Question

Physics Question on electrostatic potential and capacitance

A charge of 109C10^{-9} C is placed on each of the 6464 identical drops of radius 2cm2\, cm. They are then combined to form a bigger drop. Its potential will be

A

7.2×103V7.2 \times 10^{3}V

B

7.2×102V7.2 \times 10^{2}V

C

1.44×102V1.44 \times 10^{2}V

D

1.44×103V1.44 \times 10^{3}V

Answer

7.2×103V7.2 \times 10^{3}V

Explanation

Solution

Volume of bigger drop = volume of 6464 small drops ie., 43πR3=64×43πr3\frac{4}{3} \pi R^{3}=64 \times \frac{4}{3} \pi r^{3} or R=4rR =4r Potential on small drop V1=14πε0qrV_{1}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r} Potential on bigger drop V2=14πε064qRV_{2} =\frac{1}{4 \pi \varepsilon_{0}} \frac{64 q}{R} V2=14πε064q4rV_{2} =\frac{1}{4 \pi \varepsilon_{0}} \frac{64 q}{4 r} =9×109×64×1094×2×102=\frac{9 \times 10^{9} \times 64 \times 10^{-9}}{4 \times 2 \times 10^{-2}} =7.2×103V=7.2 \times 10^{3} V