A parallel plate capacitor or capacity $C_{0}$ is charged to a potential $V_{0}$. The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is $E_{1}$. 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 $\frac{E_{1}}{E_{2}}$ value is

Question

Dr. Abubacker · Accepted Answer

(a) 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, $C_{1} = \frac{C_{0}}{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ E_{1} = \frac{1}{2}\frac{Q^{2}}{C_{0}/2} = \frac{Q^{2}}{C_{0}} = \frac{C_{0}^{2}V_{0}^{2}}{C_{0}} = C_{0}V_{0}^{2}$
When battery is connected, then
Energy $\ \ \ \ \ \ \ \ \ \ E_{2} = \frac{1}{2}CV_{0}^{2}$
where $\ \ \ \ \ \ \ \ \ \ \ \ E_{2} = \frac{1}{2}\frac{C_{0}}{2}V_{0}^{2} = \frac{1}{4}(C_{0}V_{0}^{2})$
$	herefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{E_{1}}{E_{2}} = \ \frac{C_{0}V_{0}^{2}}{\frac{1}{4}C_{0}V_{0}^{2}} = \frac{1}{4}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E_{1}:E_{2} = 4:1$

Explanation

Reviewed by Dr. Abubacker

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