Question

A fully charged parallel plate capacitor is connected across an uncharged identical capacitor . Show that the energy stored in the combination is less than stored initially in the single capacitor

Answer 1

Initially we consider a charge capacitor; then its charge would be

$Q=cv$

$U_1=\frac{1}{2}C{V}^2$ __(I)

Now the uncharged capacitor is connected to charged capacitor

Then charge flows from 1st capacitor to other capacitor unless both capacitor attain common potential

$Q_1=C{V}_1$ and $Q_2=C{V}_2$

Applying conservation of charge

$Q=Q_1+Q_2\Rightarrow CV=C{V}_1+C{V}_2$

$\Rightarrow V=V_1+V_2\Rightarrow V=2V_1\Rightarrow V_1=\frac{V}{2}$

Total energy stored $U_2=\frac{1}{2}C{V}_1^2+\frac{1}{2}C{V}_2^2$

$U_2=\frac{1}{4}C{V}^2$ __(II)

eq (II) < eq (I)

So $U_2<U_1$