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Question #42690

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The Question

A solid isotropic dielectric sphere of outer radius $R_2$ has a concentric spherical cavity of radius $R_1$ . The inner surface of the cavity and the outer surface of the dielectric are coated with infinitesimally thin conducting layers to form a spherical capacitor. The dielectric has a relative permittivity $K$ (assumed independent of temperature) and a coefficient of linear thermal expansion $\alpha$ . The capacitor is connected to a battery that maintains a constant voltage $V$ across the conducting layers. The temperature of the dielectric is then increased by a small amount $\Delta T$ . Assuming the conducting layers expand with the dielectric, what is the mechanical work done by the electrostatic forces during this expansion? (Here $\epsilon_0$ is the permittivity of free space).

Options

A.
$\frac{2\pi K \epsilon_0 R_1 R_2 V^2 \alpha \Delta T}{R_2 - R_1}$
B.
$-\frac{2\pi K \epsilon_0 R_1 R_2 V^2 \alpha \Delta T}{R_2 - R_1}$
C.
$\frac{4\pi K \epsilon_0 R_1 R_2 V^2 \alpha \Delta T}{R_2 - R_1}$
D.
$-\frac{2\pi K \epsilon_0 R_1 R_2 (R_1 + R_2) V^2 \alpha \Delta T}{(R_2 - R_1)^2}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Common Potential Redistribution|Expansion of Cavities in Solids|Work-Energy Theorem for Rotation|Differential Calculus

Why This Question Is Tricky

Common Mistakes

This question tests two major physical traps.

First, it evaluates the student's understanding of thermal expansion in solids with cavities; a common misconception is that the cavity shrinks as the surrounding solid expands, which would lead to an incorrect sign and magnitude for the change in capacitance (Option D).

Second, it tests the application of the Work-Energy Theorem in electrodynamics.

Students often incorrectly assume that the work done by the electrostatic field is simply the negative change in potential energy ( $dW_{\text{elec}} = -dU$ ), leading to Option B.

They fail to account for the work done by the battery ( $dW_{\text{batt}} = V dQ$ ) which maintains the constant potential difference during the expansion, fundamentally altering the energy balance to $dW_{\text{elec}} = dW_{\text{batt}} - dU = +dU$ .

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electrostatic Potential and Capacitance - Class 12
Concept	Common Potential Redistribution
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free