Stored Energy of Capacitor under Plate Separation – Rankers Physics

Parallel Plate Capacitor: Practice Problem & Solution

A capacitor is connected to a battery. The electric energy stored in it is \(E\). If the separation between the plates is doubled, what will be the energy on the capacitor?
\(0.25 E\)
\(0.50 E\)
\(E\)
\(2 E\)

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Parallel Plate Capacitor. Understanding the underlying formula is key to arriving at the correct answer below:

The energy stored is \(E = \frac{1}{2} C V^2\). With the battery connected, potential \(V\) is constant. Doubling the separation halves the capacitance, so the new capacitance is \(C' = C/2\), and the stored energy becomes \(E' = E/2 = 0.50 E\).

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