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Question #43847

Difficulty: 8/10 | 🟢 Free

The Question

A polar molecule with a constant dipole moment p⃗=pcos⁡θi^+psin⁡θj^\vec{p} = p \cos\theta \hat{i} + p \sin\theta \hat{j} is initially located at x=0x = 0 in a depletion region. The electric field in this region is non-uniform and given by E⃗(x)=E0(1−x2W)i^\vec{E}(x) = E_0 \left(1 - \frac{x}{2W}\right) \hat{i} for 0≤x≤W0 \le x \le W. Initially, the dipole is aligned with the electric field (θ=0\theta = 0). An external agent moves the molecule quasi-statically from x=0x = 0 to x=Wx = W. During this translation, the agent simultaneously rotates the dipole such that its angle θ\theta with the +x+x-axis increases linearly with xx, reaching θ=π/2\theta = \pi/2 at x=Wx = W. What is the work done by the electric torque on the dipole during this process?

Options

  • A.

    −pE0(1−1π)-p E_0 \left( 1 - \frac{1}{\pi} \right)

  • B.

    −pE0-p E_0

  • C.

    −34pE0-\frac{3}{4} p E_0

  • D.

    −pE0(1+1π)-p E_0 \left( 1 + \frac{1}{\pi} \right)

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Depletion Region Dynamics|Work Done in Rotating Dipole|Reference Potential Convention|Vector Cross-Products

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to decouple the work done by electric torque from the total electrostatic potential energy change in a non-uniform field.

A common trap is assuming that the work done by the electric torque simply equals the negative change in potential energy (W=−ΔU=−pE0W = -\Delta U = -p E_0), ignoring that ΔU\Delta U accounts for both rotational work and translational work (Wtrans=∫FxdxW_{trans} = \int F_x dx).

Another trap is treating the electric field as uniform or using its average value (−34pE0-\frac{3}{4} p E_0) for the torque integral.

By forcing a specific kinematic coupling between translation and rotation (θ∝x\theta \propto x), the problem requires a rigorous path integral of the torque τ⃗=p⃗×E⃗(x)\vec{\tau} = \vec{p} \times \vec{E}(x), penalizing heuristic shortcuts.

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Question Information

SubjectPhysics
ChapterSemiconductor Electronics: Materials, Devices, and Simple Circuits - Class 12
ConceptDepletion Region Dynamics
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree