A parallel plate capacitor of capacitance ' $C$ ' is connected to a battery and charged to a potential…
A parallel plate capacitor of capacitance ' $C$ ' is connected to a battery and charged to a potential difference ' $V$ '. Another capacitor of capacitance 3 C is similarly charged to a potential difference 3 V . The charging battery is then disconnected and capacitors are connected in parallel to each other in such a way that positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is
$\frac{3}{2} \mathrm{CV}^2$
$8 \mathrm{CV}^2$
$\frac{13}{2} \mathrm{CV}^2$
$18 \mathrm{CV}^2$
Solution
As both capacitors are in parallel
$\begin{array}{ll}
\therefore \quad & \mathrm{C}_{\mathrm{eq}}=\mathrm{C}+3 \mathrm{C}=4 \mathrm{C} \\
& \text { Net potential, } \mathrm{V}_{\mathrm{net}}=3 \mathrm{~V}-\mathrm{V}=2 \mathrm{~V} \\
& \mathrm{~V}=\frac{1}{2} \mathrm{C}_{\mathrm{eq}} \mathrm{~V}_{\text {net }}^2=\frac{1}{2} \times 4 \mathrm{C} \times(2 \mathrm{~V})^2 \\
& \mathrm{~V}=8 \mathrm{CV}
\end{array}$