Question

A capacitor $C$ is charged to a potential difference $V$ and another capacitor of capacity $2C$ is charged to a potential difference $2V$. The charging batteries are disconnected and the the two capacitors are connected to each other in parallel with reverse polarity. The energy lost in this process will be

• A. $\frac{1}{2}C{V}^2$
• B. $\frac{2C{V}^2}{3}$
• C. $\frac{3}{2}C{V}^2$
• D. $3C{V}^2$

Answer 1

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

When the first capacitor is changed, it attains a initial potential energy of :-

$u_1=\frac{1}{2}C{V}^2$

and change in each plate $Q_1=CV$

Similarly, the potential energy gained in $2^{nd}$ capacitor

$u_2=\frac{1}{2}\left(2e\right)(2v{)}^2$

$=4C{V}^2$

change in each plate $Q_2=\left(2c\right)\left(2v\right)=4CV$

When both are connected, in the reverse polarity, first the charges neutralize

Remaining change on plate

$Q^{\prime }=Q_2-Q_1=30CV$

and equivalent capacitance of system in parallel

$C^{\prime }=3C$.

$\therefore$ Final energy on system $u^{\prime }=\frac{Q^{\prime }2}{2C}$

$=\frac{9C^2{V}^2}{2.3C}=\frac{3}{2}C{V}^2$

also, initial energy of system $u_1+u_2=4C{V}^2+\frac{C{V}^2}{2}=\frac{9}{2}C{V}^2$

Thus, energy dissipated in process

$\Delta u=u_i-u_f=\frac{9}{2}C{V}^2-\frac{3}{2}C{V}^2=3C{V}^2$