UPSEE 2019 Solved Paper

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 $C'=\varepsilon_{r}\left(\frac{\varepsilon_{0}A}{d}\right)$ ⇒ $C'=\varepsilon_{r}C\left[\because C=\frac{\varepsilon_{0}A}{d}\right]$ 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. $E=\frac{1}{2}CV^{2}\text{ and }$ $E'=\frac{1}{2}C'V'^{2}=\frac{1}{2}(\varepsilon_{r}C)(\varepsilon_{r}V)^{2}$ $\Rightarrow E'=\frac{1}{2}\varepsilon_{r}^{3}CV^{2}$ Hence, the energy associated with capacitor is increased.