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Question #42694

Difficulty: 8/10 | 🟢 Free

The Question

A parallel plate capacitor with plate area AA and separation dd is charged to a total free charge QQ and then electrically isolated. A dielectric slab of the same area AA and thickness dd is slowly inserted between the plates. The dielectric constant of the slab varies with the distance xx from one of its faces as K(x)=K01+αxK(x) = \frac{K_0}{1 + \alpha x}, where K0K_0 and α\alpha are positive constants. Assume K0>1+αd2K_0 > 1 + \frac{\alpha d}{2} and that the electric field is strictly normal to the plates. What is the magnitude of the work done by the external agent to completely insert the slab? (Here, ϵ0\epsilon_0 is the vacuum permittivity).

Options

  • A.

    Q2d2ϵ0A(1−2+αd2K0)\frac{Q^2 d}{2 \epsilon_0 A} \left( 1 - \frac{2 + \alpha d}{2 K_0} \right)

  • B.

    Q2d2ϵ0A(1−αdK0ln⁡(1+αd))\frac{Q^2 d}{2 \epsilon_0 A} \left( 1 - \frac{\alpha d}{K_0 \ln(1+\alpha d)} \right)

  • C.

    Q2d2ϵ0A(1−ln⁡(1+αd)K0αd)\frac{Q^2 d}{2 \epsilon_0 A} \left( 1 - \frac{\ln(1+\alpha d)}{K_0 \alpha d} \right)

  • D.

    Q2dϵ0A(1−2+αd2K0)\frac{Q^2 d}{\epsilon_0 A} \left( 1 - \frac{2 + \alpha d}{2 K_0} \right)

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Dielectric Slab Displacement|Conductor Interior Potential|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the application of Gauss's Law in non-uniform dielectrics and the work-energy theorem.

A common trap is treating the non-uniform dielectric by averaging its dielectric constant K(x)K(x) over the distance, or incorrectly applying the series capacitance formula by integrating K(x)K(x) directly instead of its reciprocal (elastance).

By defining K(x)K(x) such that its reciprocal is linear, students are forced to either correctly integrate the infinitesimal elastances d(1/C)=dx/(ϵ0AK(x))d(1/C) = dx / (\epsilon_0 A K(x)) or use Gauss's Law for the displacement field D⃗\vec{D} to find the true potential difference.

The problem also verifies the correct sign convention for the work done by an external agent against the electrostatic pull.

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