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Question #42785

Difficulty: 8/10 | 🟢 Free

The Question

A parallel plate capacitor with vertical rectangular plates separated by a distance dd is connected to a constant DC voltage source VV. A dielectric slab of mass mm, thickness dd, and dielectric constant KK is suspended vertically between the plates by an ideal spring of spring constant kk. The slab is shaped as an isosceles triangle of base bb and height hh, and is oriented such that its tip points downwards. The slab is inserted tip-first into the capacitor. Let xx be the depth of the tip inside the capacitor. The spring is unstretched when x=0x = 0. Assuming the slab remains partially inside the capacitor and neglecting fringing fields, determine the equilibrium depth xeqx_{eq} and the angular frequency ω\omega of small vertical oscillations about this equilibrium. (Assume k>λk > \lambda, where λ=ϵ0(K−1)bV22hd\lambda = \frac{\epsilon_0 (K-1) b V^2}{2 h d})

Options

  • A.

    xeq=mgk−λx_{eq} = \frac{mg}{k - \lambda}, ω=k−λm\omega = \sqrt{\frac{k - \lambda}{m}}

  • B.

    xeq=mgk+λx_{eq} = \frac{mg}{k + \lambda}, ω=k+λm\omega = \sqrt{\frac{k + \lambda}{m}}

  • C.

    xeq=mgk−2λx_{eq} = \frac{mg}{k - 2\lambda}, ω=k−2λm\omega = \sqrt{\frac{k - 2\lambda}{m}}

  • D.

    xeq=mgk+2λx_{eq} = \frac{mg}{k + 2\lambda}, ω=k+2λm\omega = \sqrt{\frac{k + 2\lambda}{m}}

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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|Equilibrium Stability|Variable Force Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly apply energy conservation and equilibrium stability principles to a variable capacitor system.

The primary trap lies in the calculation of the electrostatic force.

Students often mistakenly use F=−dUEdxF = -\frac{dU_E}{dx} with UE=12CV2U_E = \frac{1}{2}CV^2, forgetting that the battery does work WB=V2ΔCW_B = V^2 \Delta C.

This omission leads to an incorrect sign for the electrostatic force (Fe=−λxF_e = -\lambda x instead of +λx+\lambda x), resulting in the incorrect equilibrium position and frequency (Option B).

Furthermore, the non-linear area insertion requires careful integration to find C(x)C(x), testing their mathematical formulation of physical systems.

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