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  4. A capacitor of capacitance 2 μ F is charged to a potential difference of 200 volt. After disconnecting from the battery, it is connected in parallel with an another uncharged capacitor. The common potential is 20 volt. The capacitance of the second capacitor is

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Q. A capacitor of capacitance $2\,\mu F$ is charged to a potential difference of 200 volt. After disconnecting from the battery, it is connected in parallel with an another uncharged capacitor. The common potential is 20 volt. The capacitance of the second capacitor is

Electrostatic Potential and Capacitance

Solution:

Let the capacitance of the other capacitor be C $\mu$F.
Total capacitance = (C + 2)$\mu$F.
Initial charge, q =VC = $200 \times 2 \times 10^{-6} = 400 \times 10^{-6}$ C.
Hence common potential 20 = $\frac{q}{(C - 2)} = 400 \times 10^{-6} / (C + 2) \times 10^6$
or 20 (C+2) = 400. This gives C = 18 $\mu$ F.