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Question #42315

Difficulty: 8/10 | 🟢 Free

The Question

A parallel plate capacitor with plates of length LL, width ww, and separation dd is connected to a battery of constant voltage VV. A solid dielectric slab of dielectric constant KK and dimensions L×w×dL \times w \times d is inserted a distance xx into the capacitor along its length. The +x+x direction is defined along the inward direction of insertion, with unit vector i^\hat{i}. Which of the following correctly represents the electrostatic force F⃗e\vec{F}_e exerted by the capacitor on the slab, and the external work WextW_{ext} required to slowly pull the slab completely out from a fully inserted state?

Options

  • A.

    F⃗e=ϵ0wV22d(K−1)i^,Wext=ϵ0LwV22d(K−1)\vec{F}_e = \frac{\epsilon_0 w V^2}{2d}(K-1) \hat{i}, \quad W_{ext} = \frac{\epsilon_0 L w V^2}{2d}(K-1)

  • B.

    F⃗e=−ϵ0wV22d(K−1)i^,Wext=−ϵ0LwV22d(K−1)\vec{F}_e = -\frac{\epsilon_0 w V^2}{2d}(K-1) \hat{i}, \quad W_{ext} = -\frac{\epsilon_0 L w V^2}{2d}(K-1)

  • C.

    F⃗e=ϵ0wV2d(K−1)i^,Wext=ϵ0LwV2d(K−1)\vec{F}_e = \frac{\epsilon_0 w V^2}{d}(K-1) \hat{i}, \quad W_{ext} = \frac{\epsilon_0 L w V^2}{d}(K-1)

  • D.

    F⃗e=−ϵ0wV22d(K−1)i^,Wext=ϵ0LwV22d(K−1)\vec{F}_e = -\frac{\epsilon_0 w V^2}{2d}(K-1) \hat{i}, \quad W_{ext} = \frac{\epsilon_0 L w V^2}{2d}(K-1)

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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

Conservative Force Field|Dielectric Slab Displacement|Work-Energy Theorem|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's understanding of conservative forces in systems connected to an external energy source.

A common lethal mistake is blindly applying the relation F=−dU/dxF = -dU/dx (valid only for isolated, constant-charge systems) to a constant-potential system.

When connected to a battery, the battery does work dWbatt=VdQ=2dUdW_{batt} = V dQ = 2dU.

Consequently, the electrostatic force is F=+dU/dxF = +dU/dx.

Failing to account for the battery's work leads to an incorrect sign for the force and equates the external work directly to the change in potential energy, trapping the student in the incorrect options.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterWork, Energy, and Power - Class 11 Part 1
ConceptConservative Force Field
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree