Solution:
When a capacitor is disconnected, charge (Q) remains constant. Capacitance \(C = \frac{\epsilon_0 A}{d}\). If (d) doubles, (C) halves. Stored energy \(U = \frac{Q^2}{2C}\). Since (Q) is constant and (C) halves, (U) doubles.
A battery is used to charge a parallel-plate capacitor, after which it is disconnected. Then the plates are pulled apart to twice their original separation. This process will double the :
Leave a Reply