A parallel plate capacitor is charged from a cell and then isolated from it. The separation between the plate is now increased
the force of attraction between the plates will decrease
the field in the region between the plates will change
the energy stored in the capacitor will increase
the potential difference between the plates will decreases
Solution:
Since the capacitor is isolated, its charge \(Q\) remains constant. When the separation \(d\) increases, the capacitance \(C = \frac{\epsilon_0 A}{d}\) decreases. Since \(U = \frac{Q^2}{2C}\), the stored energy increases due to the work done against the electrostatic attraction.
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