Question

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

• A. U/2
• B. U/4
• C. 4U
• D. 2U

Answer 1

Answer: (B)

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