Question

A parallel plate capacitor is charged and the charging battery is then disconnected. A dielectric slab is now introduced between the plates of the capacitor. Which of the following is correct?

• A. potential difference across capacitor remains constant
• B. capacitance of the capacitor remains constant.
• C. energy associated with capacitor increases.
• D. energy associated with capacitor decreases.

Answer 1

Answer: (C)

Initially, a capacitor of capacitance $C$ is connected to a battery of potential $V$, then the charge stored
$Q=\frac{C}{V}$
If battery is removed then stored charge remains constant. i.e.
$Q=$ constant ($\because$ conservation of charge)
As, new capacitance of the capacitor
$C^{\prime }={\epsilon }_r=\left(\frac{{\epsilon }_{0}A}{d}\right)$
$\Rightarrow C^{\prime }={\epsilon }_{r}C\left[\because C=\frac{{\epsilon }_{0}A}{d}\right]$
Which shows that capacitance is increased. Potential across new capacitor,
$V^{\prime }=\frac{C^{\prime }}{Q}={\epsilon }_r\frac{C}{Q}={\epsilon }_{r}V\left[\because V=\frac{C}{Q}\right]$
So, potential difference across capacitor increased.
Energy stored in new capacitor and old capacitor.
$E=\frac{1}{2}C{V}^2$ and
$E^{\prime }=\frac{1}{2}{C}^{\prime }{V}^{\prime 2}=\frac{1}{2}\left({\epsilon }_{r}C\right){\left({\epsilon }_{r}V\right)}^2$
$\Rightarrow E^{\prime }=\frac{1}{2}{\epsilon }_r^{3}C{V}^2$
Hence, the energy associated with capacitor is increased.