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Question #42677

Difficulty: 8/10 | 🟢 Free

The Question

A parallel plate capacitor consists of a fixed left plate and a movable right plate, each of area $A$ . The initial separation between the plates is $d$ . The right plate has mass $m$ , rests on a rough horizontal surface with a coefficient of friction $\mu$ , and is connected to a relaxed ideal spring of spring constant $k$ . The left plate is rigidly attached to a wall via an insulating cylindrical rod of cross-sectional area $S$ and breaking stress $\sigma_b$ . A constant voltage $V$ is suddenly applied across the plates. The right plate moves towards the left plate and momentarily comes to rest at a separation of $d/2$ . At this exact instant, the insulating rod holding the left plate breaks. Assuming the coefficient of static friction is equal to the kinetic friction, find the breaking stress $\sigma_b$ of the rod.

Options

A.
$\frac{kd + 4\mu mg}{2S}$
B.
$\frac{kd + 2\mu mg}{2S}$
C.
$\frac{kd + 4\mu mg}{S}$
D.
$\frac{kd + 4\mu mg}{4S}$

Check Your Answer

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

Capacitance Independence|Work Done by Friction|Breaking Stress Concept

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to integrate electrostatics with mechanics, specifically targeting three common traps.

Trap 1 (Option B): Students incorrectly apply force equilibrium ( $F_e = F_s + F_f$ ) instead of the Work-Energy theorem for a dynamic process where the plate momentarily stops (velocity is zero, but acceleration is not).

Trap 2 (Option C): Students use the correct energy formula but incorrectly use $F = Q^2/(\epsilon_0 A)$ for the force, missing the $1/2$ factor derived from energy minimization.

Trap 3 (Option D): Students incorrectly assume the charge remains constant after the voltage is applied, leading to a constant electrostatic force and a completely different work-energy balance.

The correct approach requires recognizing that at constant voltage, the electrostatic force increases as the plates move closer, necessitating integration for work and evaluating the maximum force at the closest approach.

Quick Facts About This Question

Question Information

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