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Q. A capacitor is charged with a battery and energy stored is U. After disconnecting battery another capacitor of same capacity is connected in parallel to the first capacitor. Then energy stored in each capacitor is

AIPMTAIPMT 2000Electrostatic Potential and Capacitance

Solution:

Now charge will flow in such a way that total charge is $Q$ and potential $\frac{ d }{ f }$ across both capacitors remains same.
$
V=\frac{Q}{2 C}
$
Energy in each capacitor $=\frac{1}{2} CV ^{2}=\frac{1}{2} C \left(\frac{ Q }{2 C }\right)^{2}=\frac{1}{4}\left(\frac{1}{2} \cdot\left(\frac{ Q ^{2}}{2 C }\right)^{2}=\frac{ U }{4}\right.$