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Question #42905

Difficulty: 8/10 | 🟢 Free

The Question

A rigid electric dipole consists of two point charges +q+q and −q-q (each of mass mm) connected by a massless non-conducting rod of length 2a2a. The dipole is pivoted at its center OO, which is fixed at the coordinates (a,0)(a, 0) in the xyxy-plane. A non-uniform static electric field E⃗=αx2i^\vec{E} = \alpha x^2 \hat{i} is present in the region, where α\alpha is a positive constant in SI units. The dipole is released from rest from a position where it is aligned parallel to the positive yy-axis (with +q+q at y>0y > 0). Assuming no gravity and no friction at the pivot, what is the angular velocity ω\omega of the dipole when it aligns with the positive xx-axis (with +q+q at x>ax > a)?

Options

  • A.

    8qαa3m\sqrt{ \frac{8 q \alpha a}{3 m} }

  • B.

    2qαam\sqrt{ \frac{2 q \alpha a}{m} }

  • C.

    4qαam\sqrt{ \frac{4 q \alpha a}{m} }

  • D.

    16qαa3m\sqrt{ \frac{16 q \alpha a}{3 m} }

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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

Dipole in Non-Uniform Field|Impulsive Torque and Angular Momentum

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle an extended dipole in a non-uniform field with a spatially varying gradient.

The primary trap is the 'Point Dipole Approximation'.

Students commonly misapply the standard formula U=−p⃗⋅E⃗U = -\vec{p} \cdot \vec{E} or τ⃗=p⃗×E⃗\vec{\tau} = \vec{p} \times \vec{E}, evaluating the electric field only at the pivot (a,0)(a,0).

This yields an incorrect work done of pαa2=2qαa3p \alpha a^2 = 2q\alpha a^3.

However, because the field gradient ∂E/∂x\partial E/\partial x is not constant, the force varies non-linearly across the dipole's length.

To find the correct work done, one must either integrate the exact variable torque over the rotation angle or calculate the exact change in potential energy using V(x)=−∫EdxV(x) = -\int E dx for the individual point charges.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectric Charges and Fields - Class 12
ConceptDipole in Non-Uniform Field
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree