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
$C' = \varepsilon_{r} = \left(\frac{\varepsilon_{0}A}{d}\right)$
$ \Rightarrow C' = \varepsilon_{r}C \quad\left[\because C=\frac{\varepsilon_{0} A}{d}\right]$
Which shows that capacitance is increased. Potential across new capacitor,
$ V' = \frac{C'}{Q} = \varepsilon_{r} \frac{C}{Q} = \varepsilon_{r} V \left[\because V = \frac{C}{Q}\right]$
So, potential difference across capacitor increased.
Energy stored in new capacitor and old capacitor.
$E = \frac{1}{2}CV^{2}$ and
$ E' = \frac{1}{2}C'V'^{2} = \frac{1}{2}\left(\varepsilon_{r} C\right)\left(\varepsilon_{r}V\right)^{2}$
$\Rightarrow E' = \frac{1}{2} \varepsilon_{r}^{3}CV^{2}$
Hence, the energy associated with capacitor is increased.