Change in Capacitor Properties – Rankers Physics

Parallel Plate Capacitor: Practice Problem & Solution

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 :
capacitance
surface charge density on each plate
stored energy
electric field between the two plates

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 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.

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Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. The electric field between the plates of a parallel-plate capacitor of capacitance \(2.0~\mu\text{F}\) drops to one third of its initial value in \(4.4~\mu\text{s}\) when the plates are connected by a thin wire. Find the resistance of the wire.
2. Two metallic spheres of radii $1 \text{ cm}$ and $2 \text{ cm}$ are given charges $10^{-2} \text{ C}$ and $5 \times 10^{-2} \text{ C}$ respectively. If they are connected by a conducting wire, the final charge on the smaller sphere is (1995)
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