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Q. Statement I A charged capacitor is disconnected from a battery. Now if its plates are separated further, the potential energy will fall.
Statement II Energy stored in a capacitor is equal to the work done in charging it.

Electrostatic Potential and Capacitance

Solution:

Battery is disconnected from the capacitor,
So, $Q=$ constant
Energy $=\frac{Q^{2}}{2 C}=\frac{Q^{2} d}{2 \varepsilon_{0} A}$
$\Rightarrow$ Energy $\propto d$