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Energy of a Capacitor

Question

The initial and final stored energy in the capacitor, $U_{i}$ and $U_{f}$ ?

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Solution

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The initial parallel plate capacitance is $C_{i} = \frac{A ϵ_{0}}{d}$

and the charge is $Q_{i} = C_{i} V_{i} = \frac{A ϵ_{0} V}{d}$

The final parallel plate capacitance is $C_{f} = \frac{A ϵ_{0}}{2 d}$
and charge is $Q_{f} = C_{f} V_{f} = \frac{A ϵ_{0} V_{f}}{2 d}$

As the battery is disconnected so the charge will remain constant so,
$Q_{i} = Q_{f}$

or $\frac{A ϵ_{0} V}{d} = \frac{A ϵ_{0} V_{f}}{2 d}$

or, $V_{f} = 2 V$

Thus, initial energy stored, $U_{i} = \frac{1}{2} C_{i} V_{i}^{2} = \frac{A ϵ_{0} V^{2}}{2 d}$ and final energy stored, $U_{f} = \frac{1}{2} C_{f} V_{f}^{2} = \frac{A ϵ_{0} (2 V)^{2}}{4 d} = \frac{A ϵ_{0}}{d} V^{2}$

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