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. $\frac{25}{6}\text{CV}^{2}$
• C. $\frac{9}{2}\text{CV}^{2}$
• D. $\frac{3}{2}\text{CV}^{2}$

Answer 1

Answer: (D)

The diagramatic representation of given problem is shown in the figure.
The net charge shared between the two capacitors is,
$\text{Q}^{'}=\text{Q}_{2}-\text{Q}_{1}=4\text{CV}-\text{CV}=3\text{CV }$
The two capacitors will have the same potential, say $\text{V}^{'}$ .
The net capacitance of the parallel combination of the two capacitors will be,
$\text{C}^{'}=\text{C}_{1}+\text{C}_{2}=\text{C}+2\text{C}=3\text{C}$
The potential difference across the capacitors will be,
$\text{V}^{'}=\frac{\text{Q}^{'}}{\text{C}^{'}}=\frac{3 \text{CV}}{3 \text{C}}=\text{V }$
The electrostatic energy of the capacitors will be,
$\mathrm{U}^{\prime}=\frac{1}{2} \mathrm{C}^{\prime} \mathrm{V}^{\prime 2}=\frac{1}{2}(3 \mathrm{C}) \mathrm{V}^2=\frac{3}{2} \mathrm{CV}^2$