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Simple Circuit

The new poten...

Question

The new potential difference between parallel plate capacitor?

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Solution

The initial parallel plate capacitance is $C_{i} = \frac{A ϵ_{0}}{d}$
and the charge is $Q_{i} = C_{i} V = \frac{A ϵ_{0} V}{d}$
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, $V_{f} = 2 V$

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