Question

A parallel plate capacitor of capacitance $C$ is connected to a battery and is charged to a potential difference $V$. Another capacitor of capacitance $2C$ is similarly charge to a potential difference $2V$. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is :

• A. Zero
• B. $\frac{3}{2}C{V}^2$
• C. $\frac{25}{6}C{V}^2$
• D. $\frac{9}{2}C{V}^2$

Answer 1

The correct option is B $\frac{3}{2}C{V}^2$

From the figure.
The net charge shared between the two capacitors
$Q^{\prime }=Q_2-Q_2=4CV-CV=3CV$
The two capacitors will have some potential, say $V^{\prime }$.
The net capacitance of the parallel combination of the two capacitors
$C^{\prime }=C_1+C_2=C+2C+3C$
The potential of the capacitors
$V^{\prime }=\frac{Q^{\prime }}{C^{\prime }}=\frac{3CV}{3C}=V$
The electrostatic energy of the capacitors
$E^{\prime }=\frac{1}{2}{C}^{\prime }{V}^{\prime 2}=\frac{1}{2}\left(3C\right)V^2=\frac{3}{2}C{V}^2$.