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Question #43080

Difficulty: 8/10 | 🟢 Free

The Question

A parallel plate capacitor consists of two rectangular plates of length LL, width ww, and separation dd. A dielectric slab of identical dimensions is partially inserted between the plates to a depth xx. The dielectric constant of the slab varies along its length as κ(y)=1+αy\kappa(y) = 1 + \alpha y, where yy is the distance measured from its leading edge (the edge deepest inside the capacitor). An alternating voltage V(t)=V0sin⁡(ωt)V(t) = V_0 \sin(\omega t) is applied across the capacitor plates.

An external mechanical agent forces the slab to oscillate such that its insertion depth is given by x(t)=x0+Asin⁡(2ωt)x(t) = x_0 + A \sin(2\omega t), where A<x0A < x_0 and x0+A<Lx_0 + A < L. During its motion, the slab experiences a viscous drag force Fdrag=−bvF_{\text{drag}} = -b v, where vv is its instantaneous velocity and bb is a positive constant. Assuming edge effects are negligible, which of the following expressions represents the average mechanical power ⟨Pext⟩\langle P_{\text{ext}} \rangle that the external agent must supply to maintain this steady oscillation over one complete cycle?

Options

  • A.

    ϵ0wαV02Aωx04d+2bA2ω2\frac{\epsilon_0 w \alpha V_0^2 A \omega x_0}{4 d} + 2 b A^2 \omega^2

  • B.

    2 b A^2 \omega^2

  • C.

    2bA2ω2−ϵ0wαV02Aωx04d2 b A^2 \omega^2 - \frac{\epsilon_0 w \alpha V_0^2 A \omega x_0}{4 d}

  • D.

    ϵ0wαV02Aωx08d+bA2ω2\frac{\epsilon_0 w \alpha V_0^2 A \omega x_0}{8 d} + b A^2 \omega^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

Average Power over Cycle|Dielectric Slab Displacement|Variable Force Integration

Why This Question Is Tricky

Common Mistakes

This question tests the application of the Work-Energy Theorem in a system with parametric excitation, specifically targeting the misconception that an AC electrical force does zero net work over a cycle if its time-average is considered independently of the object's motion.

Students often incorrectly assume that they can use the RMS voltage to find an average force and then calculate work, which yields zero net electrical work (Trap B).

However, because the dielectric's position oscillates at twice the AC frequency (2ω2\omega), it couples non-linearly with the V2∝1−cos⁡(2ωt)V^2 \propto 1 - \cos(2\omega t) term.

This phase relationship means the electrical field acts as a damper, doing net negative work.

The external agent must therefore supply energy not only to overcome mechanical viscous drag but also to pump the system against the reactive electrical forces.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterAlternating Current - Class 12
ConceptAverage Power over Cycle
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree