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Question #42695

Difficulty: 8/10 | 🟢 Free

The Question

A parallel plate capacitor with plates of length LL, width bb, and separation dd is fixed horizontally. The lower plate is rough with a coefficient of kinetic friction μ\mu. A dielectric slab of mass mm, dielectric constant κ\kappa, and dimensions L×b×dL \times b \times d is placed such that it can slide between the plates. The slab is connected to a fixed wall by a horizontal ideal spring of spring constant kk. Initially, the spring is relaxed and the slab is just at the edge of the capacitor (x=0x=0). A constant voltage VV is applied across the plates. Assume the normal force exerted by the lower plate on the slab is mgmg, the electrostatic force is sufficient to pull the slab in, and the maximum displacement xmax<Lx_{max} < L. What is the total work done by the battery from the instant the slab is released until it reaches its maximum displacement?

Options

  • A.

    ϵ0b(κ−1)V2kd(ϵ0b(κ−1)V2d−2μmg)\frac{\epsilon_0 b (\kappa - 1) V^2}{k d} \left( \frac{\epsilon_0 b (\kappa - 1) V^2}{d} - 2\mu m g \right)

  • B.

    ϵ0b(κ−1)V22kd(ϵ0b(κ−1)V2d−2μmg)\frac{\epsilon_0 b (\kappa - 1) V^2}{2k d} \left( \frac{\epsilon_0 b (\kappa - 1) V^2}{d} - 2\mu m g \right)

  • C.

    ϵ0b(κ−1)V24kd(ϵ0b(κ−1)V2d−2μmg)\frac{\epsilon_0 b (\kappa - 1) V^2}{4k d} \left( \frac{\epsilon_0 b (\kappa - 1) V^2}{d} - 2\mu m g \right)

  • D.

    ϵ0b(κ−1)V2kd(ϵ0b(κ−1)V2d+2μmg)\frac{\epsilon_0 b (\kappa - 1) V^2}{k d} \left( \frac{\epsilon_0 b (\kappa - 1) V^2}{d} + 2\mu m g \right)

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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

Dielectric Slab Displacement|Kinetic Friction Transition|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the rigorous application of the Work-Energy Theorem in an electromechanical system.

A common trap is equating the electrostatic force to the restoring forces (Fe=kx+μmgF_e = kx + \mu mg) to find the maximum displacement, which actually yields the equilibrium position (where velocity is maximum), missing a factor of 2.

Another critical trap is misunderstanding the energy supplied by the battery; students often incorrectly assume Wbatt=ΔUcapW_{batt} = \Delta U_{cap} instead of Wbatt=2ΔUcapW_{batt} = 2\Delta U_{cap}.

Both of these misconceptions independently lead to the exact same incorrect distractor (Option B), making it a highly effective trap for students who rely on partial intuition rather than a strict energy balance.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectrostatic Potential and Capacitance - Class 12
ConceptDielectric Slab Displacement
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree