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?
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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