Energy Given by Battery on Plate Separation Change – Rankers Physics

Parallel Plate Capacitor: Practice Problem & Solution

A capacitor of capacity \(C\) is connected with a battery of potential \(V\) in parallel. The distance between its plates is reduced to half at once, assuming that the charge remains the same. Then to charge the capacitance upto the potential \(V\) again, the energy given by the battery will be
\(CV^2 / 4\)
\(CV^2 / 2\)
\(3CV^2 / 4\)
\(CV^2\)

Solution Explained:

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:

When the plate separation is halved, capacitance becomes \(2C\). Keeping charge constant at \(Q = CV\), to recharge it back to potential \(V\), the final charge is \(Q' = 2CV\). The charge flowing from the battery is \(\Delta Q = 2CV - CV = CV\). The energy supplied by the battery is \(W_b = \Delta Q \cdot V = CV^2\).

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