Energy Given by Battery on Plate Separation Change – Rankers Physics
Topic: Capacitors
Subtopic: Parallel Plate Capacitor

Energy Given by Battery on Plate Separation Change

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:

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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