Energy of Isolated Parallel Plate Capacitor – Rankers Physics

Parallel Plate Capacitor: Practice Problem & Solution

Assertion (A): If separation between plates of a parallel plate isolated charged capacitor is increased, its energy stored will be increased. Reason (R): Work done to separate the plates get converted in electrostatic potential energy.  
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(A) is true but (R) is false
Both (A) and (R) are false

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:

For an isolated capacitor, charge (Q) is constant. Energy stored is \(U = \frac{Q^2}{2C}\). If separation (d) increases, capacitance \(C = \frac{\epsilon_0 A}{d}\) decreases. Therefore, (U) increases. This increase in energy comes from the work done by an external agent to separate the plates against attractive electrostatic forces.

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