Question

22.A capacitor is connected with a battery and stores energy U.After removing the battery, it is connected with another similar capacitor in parallel.The new stored energy in each capacitor will be

Answer 1

Let C be the capacitance of the capacitors and Q be the charge developed in the charged capacitor. The energy stored in it is given by

$U=\frac{1}{2}\frac{Q^2}{C}$

Now, when this capacitor is connected to another similar capacitor, charge will flow from the charged capacitor to the uncharged capacitor till they develop a common potential V, given by

$V=\frac{\mathrm{total}\mathrm{charge}}{\mathrm{total}\mathrm{capacitance}}=\frac{Q}{2C}$
Energy stored in each capacitor

$$\begin{gathered} U\text{'}=\frac{1}{2}C{V}^2=\frac{1}{2}C{\left(\frac{Q}{2C}\right)}^2=\frac{1}{4}\left(\frac{Q^2}{2C}\right) \\ \Rightarrow U\text{'}=\frac{U}{4} \end{gathered}$$