2. plate 1of a parallel plate capacitor of plate area A has a charge Q while its
other plate 2 has no charge, There is an identical uncharged plate 3 which can
be connected to plate 2 by means of a thin conducting wire and a switch.
Consider the following statements and choose the correct one/(s)
(a) Upon closing the switch, the force of attraction on plate 2 is equal to
Q^2/4 e_oA

(b) Upon closing the switch there will be no change in the electrostatic energy stored in the
intervening space between plates 1 and 2
(c) Plate 3 experiences no force after the switch is closed
(d) Upon closing the switch, the force of attraction on plate 2 is equal to Q^2/8e_oQA

To analyze the situation with the parallel plate capacitor and the effects of connecting an uncharged plate, let's break down each statement and evaluate its validity based on the principles of electrostatics and capacitor behavior.

Understanding the Setup

We have a parallel plate capacitor with two plates: Plate 1, which has a charge Q, and Plate 2, which is initially uncharged. When we introduce Plate 3, which is also uncharged, and connect it to Plate 2 via a conducting wire and a switch, we need to consider how the charges redistribute and the forces that come into play.

Evaluating Each Statement

• (a) Upon closing the switch, the force of attraction on plate 2 is equal to Q²/4ε₀A

This statement is incorrect. When the switch is closed, Plate 2 will acquire a charge of Q/2 due to the influence of Plate 1. The force of attraction between two charged plates can be calculated using the formula: F = (1/2) * (Q²/ε₀A). Therefore, the correct expression for the force on Plate 2 will be Q²/8ε₀A, not Q²/4ε₀A.

• (b) Upon closing the switch, there will be no change in the electrostatic energy stored in the intervening space between plates 1 and 2

This statement is also incorrect. Initially, the energy stored in the capacitor is given by U = (1/2) * (Q²/ε₀A). Once the switch is closed, the charge distribution changes, and the energy will adjust based on the new configuration. The energy will decrease as the system stabilizes with the new charge distribution.

• (c) Plate 3 experiences no force after the switch is closed

This statement is true. After the switch is closed, Plate 3 becomes neutral and does not have any net charge. Since it does not have a charge, it will not experience any electrostatic force. The forces acting on it from the other plates will cancel out, resulting in no net force.

• (d) Upon closing the switch, the force of attraction on plate 2 is equal to Q²/8ε₀QA

This statement is correct. As mentioned earlier, when the switch is closed, Plate 2 acquires a charge of Q/2. The force of attraction can be calculated using the modified charge, leading to the expression F = (1/2) * (Q²/ε₀A), which simplifies to Q²/8ε₀A.

Summary of Findings

From our analysis, the correct statements are (c) and (d). Plate 3 does not experience any force after the switch is closed, and the force of attraction on Plate 2 is indeed equal to Q²/8ε₀A. The other statements regarding the force and energy changes are not accurate based on the principles of electrostatics and capacitor behavior.

Key Takeaways

Understanding how charges redistribute in capacitors and the forces involved is crucial in electrostatics. The behavior of connected plates can lead to significant changes in both charge distribution and energy storage, which are fundamental concepts in electrical engineering and physics.