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 charged 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

Answer: (B)

$\frac{3}{2}C{V}^2$

For capacitance C ,

$Q_1=C\cdot V$

$E=\frac{1}{2}C{V}^2$

For Capacitance 2C,

$Q_2=4CV,E=\frac{1}{2}\left(2C\right)(2V{)}^2$

$E=4C{V}^2$

When connected in parallel with positive to negative charge total charge becomes $Q=4CV-CV$

$Q=3CV$

Let x be the charge on the capacitor of capacitance C and $V_f$ be the final voltage of the system .

$x=C\cdot V_f,Q-x=2C\cdot V_f$

$\frac{x}{Q-x}=\frac{C}{2C}$

$2x=Q-x$

$x=\frac{Q}{3}$

$V_f=\frac{x}{C}$

$V_f=\frac{Q}{3C}$

$V_f=\frac{3CV}{3C}=V$

Hence , Energy of the system becomes

$=\frac{1}{2}C{V}^2+\frac{1}{2}\left(2C\right)V^2$

$E=\frac{3}{2}C{V}^2$