A parallel plate capacitor of capacity $C_0$ is charged to a potential $V_0$. (i) The energy stored in the…
A parallel plate capacitor of capacity $C_0$ is charged to a potential $V_0$. (i) The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is $E_1$. (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is $E_2$. Then, $E_1 / E_2$ value is:
4
$3 / 2$
2
$1 / 2$
Solution
Capacitance of parallel plate capacitor,
$C_0=\frac{\varepsilon_0 A}{d}$
where, $A=$ area of the plates, $d=$ separation between the plates Charge stored in the capacitor,
$Q=C_0 V_0$
When battery is disconnected, then charge remains same.
So, energy, $E_1=\frac{1}{2} \frac{Q^2}{C}$
$C=$ capacitance when plate separation is doubled
So, $\quad C_1=\frac{C_0}{2}$
$E_1=\frac{1}{2} \frac{Q^2}{C_0 / 2}=\frac{Q^2}{C_0}=\frac{C_0^2 V_0^2}{C_0}=C_0 V_0^2$
When battery is connected, then
Energy, $E_2=\frac{1}{2} C V_0^2$
where, $E_2=\frac{1}{2} \frac{C_0}{2} V_0^2=\frac{1}{4}\left(C_0 V_0^2\right)$
$\therefore \quad \frac{E_1}{E_2}=\frac{C_0 V_0^2}{\frac{1}{4} C_0 V_0^2}=\frac{1}{4}$
$E_1: E_2=4: 1$