Question

A parallel plate capacitor of capacitance $C$ is connected to a potential difference $V$. Another capacitor of capacitance $2C$ is connected to another battery and is charged to potential difference $2V$. The charging batteries are 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 one is connected to that negative terminal of the other.The final energy of the configuration is

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

Answer 1

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

The net charges stored between the two capacitors

$Q^{\prime }=Q_1-Q_2=C_2{V}_2-C_1{V}_1$

$\left(20\right)\left(2V\right)-CV=3CV$

When they are connected in parallel

$C_{net}=C+2C=3C$

$V^{\prime }=\frac{Q^{\prime }}{C_{net}}=\frac{3CV}{3C}=V$

The Energy of capacitor

$E^{\prime }=\frac{1}{2}{C}^{\prime }{V}^2=\frac{1}{2}\left(3C\right)V^2$