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Question #42327

Difficulty: 8/10 | 🟢 Free

The Question

A parallel plate capacitor with plates of width bb, length LL, and separation dd is connected to a constant voltage source VV. A dielectric slab of mass mm, dielectric constant KK, and dimensions matching the capacitor's interior is initially placed just at the edge of the plates (x=0x=0). The slab is pulled into the capacitor by electrostatic forces. As it enters, it experiences a position-dependent resistive friction force f=αxf = \alpha x, where α\alpha is a positive constant and xx is the length of the slab inside the capacitor. Assuming gravity is negligible and the plates are long enough, what is the maximum kinetic energy acquired by the dielectric slab?

Options

  • A.

    ϵ02b2(K−1)2V48d2α\frac{\epsilon_0^2 b^2 (K-1)^2 V^4}{8 d^2 \alpha}

  • B.

    ϵ02b2(K−1)2V42d2α\frac{\epsilon_0^2 b^2 (K-1)^2 V^4}{2 d^2 \alpha}

  • C.

    ϵ02b2(K−1)2V44d2α\frac{\epsilon_0^2 b^2 (K-1)^2 V^4}{4 d^2 \alpha}

  • D.

    ϵ02b2(K−1)2V4d2α\frac{\epsilon_0^2 b^2 (K-1)^2 V^4}{d^2 \alpha}

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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

Mechanical Energy Conservation|Dielectric Slab Displacement|Vector Cross-Products

Why This Question Is Tricky

Common Mistakes

This question tests the application of the Work-Energy Theorem in an electro-mechanical system with non-uniform acceleration.

A common lethal trap for JEE Advanced students is incorrectly equating the electrostatic force to the spatial derivative of the battery's work (F=dWbatt/dxF = dW_{batt}/dx), leading to a force twice as large (F=2FeF = 2F_e).

Another trap is assuming simple energy conservation (ΔK=ΔU\Delta K = \Delta U) without accounting for the work done by the battery and the non-conservative friction force.

By requiring the integration of a position-dependent net force, students must correctly identify Fe=12V2dCdxF_e = \frac{1}{2}V^2 \frac{dC}{dx} and apply ΔK=∫Fnetdx\Delta K = \int F_{net} dx, avoiding the false assumption of constant acceleration or isolated system energy balances.

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Question Information

SubjectPhysics
ChapterWork, Energy, and Power - Class 11 Part 1
ConceptMechanical Energy Conservation
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree