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,
${\text{Q}}^{{}^{\prime }}={\text{Q}}_2-{\text{Q}}_1=4\text{CV}-\text{CV}=3\text{CV }$
The two capacitors will have the same potential, say ${\text{V}}^{{}^{\prime }}$ .
The net capacitance of the parallel combination of the two capacitors will be,
${\text{C}}^{{}^{\prime }}={\text{C}}_1+{\text{C}}_2=\text{C}+2\text{C}=3\text{C}$
The potential difference across the capacitors will be,
${\text{V}}^{{}^{\prime }}=\frac{{\text{Q}}^{{}^{\prime }}}{{\text{C}}^{{}^{\prime }}}=\frac{3\text{CV}}{3\text{C}}=\text{V }$
The electrostatic energy of the capacitors will be,
${\mathrm{U}}^{\prime }=\frac{1}{2}{\mathrm{C}}^{\prime }{\mathrm{V}}^{\prime 2}=\frac{1}{2}(3\mathrm{C}){\mathrm{V}}^2=\frac{3}{2}{\mathrm{CV}}^2$