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A 10 $μ$ F capacitor is charged to a potential difference of 50 V and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes 20 volts. The capacitance of second capacitor is

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detailed solution

Correct option is A

Here, q1=10×50=500μCC1=10μF,C2=? and q2=0Now V=q1+q2C1+C2=500+0C1+C2or C1+C2=50020=25μF∴ C2=25−C1=25−10=15μF

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A 10 μF capacitor is charged to a potential difference of 50 V and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes 20 volts. The capacitance of second capacitor is

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A 10 $μ$ F capacitor is charged to a potential difference of 50 V and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes 20 volts. The capacitance of second capacitor is