Question

A parallel plate capacitor of capacitance $C$ is connected to a battery and is charge to a potential difference $V$. Another capacitor of capacitor $2C$ 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. $\left(3/2\right)C{V}^2$
• C. $25/6C{V}^2$
• D. $\left(9/2\right)C{V}^2$

Answer 1

The correct option is B $\left(3/2\right)C{V}^2$

Initial charges are given in this diagram now after joining, charges will flow due to potential difference.

Finally, the potential difference across the capacitors will be same as they are connected in parallel.

[ Potentntial difference across a capacitor $=\frac{Q}{C}$ ]

$\Rightarrow \frac{CV-q}{C}=\frac{-4CV+q}{2C}$

$\Rightarrow 2CV-2q=-4CV+q$

$\Rightarrow 6CV=3q$

$\Rightarrow q=2CV$

$\Rightarrow$ Total energy $=\frac{{\left(CV\right)}^2}{2C}+\frac{{\left(2CV\right)}^2}{2C}$ [ Energy of a capactor $=\frac{Q^2}{2C}$ ]

$=\frac{C^2{V}^2}{2C}+\frac{4C^2{V}^2}{2\left(2C\right)}$

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

$=\frac{3C{V}^2}{2}$ $\left(B\right)$

Hence, the answer is $\frac{3C{V}^2}{2}.$