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The energy required to charge a parallel plate condenser of plate separation $d$ and plate area of cross-section $A$ such that the uniform electric field between the plates is $E$, is :
A
$\varepsilon_0 E^2 Ad$
B
$\dfrac{1}{2}\varepsilon_0 E^2 Ad$
C
$\dfrac{1}{2}\varepsilon_0 E^2 / Ad$
D
$\varepsilon_0 E^2 / Ad$
Detailed Solution
Capacitance of the parallel plate condenser: $C = \dfrac{\varepsilon_0 A}{d}$
Potential difference when the field between the plates is $E$: $V = Ed$
Energy stored in the capacitor: $U = \dfrac{1}{2}CV^2$
$U = \dfrac{1}{2}\left(\dfrac{\varepsilon_0 A}{d}\right)(Ed)^2$
$U = \dfrac{1}{2}\varepsilon_0 E^2 Ad$
While charging, the cell delivers charge $Q = CV$ at a constant potential difference $V$, so the energy given by the cell is $W = QV = CV^2$.
$W = \left(\dfrac{\varepsilon_0 A}{d}\right)(Ed)^2 = \varepsilon_0 E^2 Ad$
Energy given by the cell = energy stored in the capacitor + energy lost in the process of charging, and the energy lost equals the energy stored.
So the energy required $= 2 \times$ energy stored in the capacitor $= \varepsilon_0 E^2 Ad$.
Potential difference when the field between the plates is $E$: $V = Ed$
Energy stored in the capacitor: $U = \dfrac{1}{2}CV^2$
$U = \dfrac{1}{2}\left(\dfrac{\varepsilon_0 A}{d}\right)(Ed)^2$
$U = \dfrac{1}{2}\varepsilon_0 E^2 Ad$
While charging, the cell delivers charge $Q = CV$ at a constant potential difference $V$, so the energy given by the cell is $W = QV = CV^2$.
$W = \left(\dfrac{\varepsilon_0 A}{d}\right)(Ed)^2 = \varepsilon_0 E^2 Ad$
Energy given by the cell = energy stored in the capacitor + energy lost in the process of charging, and the energy lost equals the energy stored.
So the energy required $= 2 \times$ energy stored in the capacitor $= \varepsilon_0 E^2 Ad$.