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Question

Physics Question on electrostatic potential and capacitance

A spherical drop of capacitance 1μF1 \mu \,F is broken into eight drops of equal radius. Then, the capacitance of each small drop is ........

A

18μF\frac{1}{8}\,\mu F

B

8μF8\,\mu F

C

12μF\frac{1}{2}\,\mu F

D

14μF\frac{1}{4}\,\mu F

Answer

12μF\frac{1}{2}\,\mu F

Explanation

Solution

Potential of a charged sphere is given by the formula
V=kQRV =\frac{ kQ }{ R }
Hence, by Q=CVQ = CV,
C=QV=RkC =\frac{ Q }{ V }=\frac{ R }{ k }
As per given information, Rk=1μF\frac{ R }{ k }=1 \mu F
After splitting,
8(43πr3)=43πR38\left(\frac{4}{3} \pi r ^{3}\right)=\frac{4}{3} \pi R ^{3}
r=R2r =\frac{ R }{2}
Thus, capacitance of each small drop is V=rk=12Rk=12μFV =\frac{ r }{ k }=\frac{1}{2} \frac{ R }{ k }=\frac{1}{2} \mu F