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Question #43687

Difficulty: 8/10 | 🟢 Free

The Question

A fixed ideal electric dipole of dipole moment p⃗=pi^\vec{p} = p \hat{i} is located at the origin of a coordinate system. A stationary radioactive nucleus is located at point AA with polar coordinates (r,π)(r, \pi) in the xyxy-plane, where the polar angle θ\theta is measured from the positive xx-axis. The nucleus undergoes a reaction, emitting a charged fragment of mass mm and quantized charge q=+neq = +ne (where ee is the elementary charge and nn is a positive integer). The heavy residual nucleus remains fixed at AA and is uncharged. The QQ-value of the reaction is entirely converted into the initial kinetic energy of the emitted fragment. The fragment is constrained to move along a fixed, frictionless, non-conducting circular track of radius rr centered at the origin. It just reaches point BB with polar coordinates (r,π/3)(r, \pi/3) in the xyxy-plane, where its kinetic energy momentarily becomes zero. Assuming the only force doing work on the fragment is the electrostatic force from the dipole, which of the following expressions gives the correct value of the integer nn?

Options

  • A.

    8πϵ0r2Q3ep\frac{8\pi\epsilon_0 r^2 Q}{3 e p}

  • B.

    8πϵ0r2Qep\frac{8\pi\epsilon_0 r^2 Q}{e p}

  • C.

    8πϵ0r2Q3ep\frac{8\pi\epsilon_0 r^2 Q}{\sqrt{3} e p}

  • D.

    4πϵ0r2Qep\frac{4\pi\epsilon_0 r^2 Q}{e p}

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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

Q-value Calculation|Quantization of Charge|Dipole Potential Angular Dependence

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to bridge microscopic nuclear energetics (QQ-value) with macroscopic classical electrostatics (dipole field) using the Work-Energy Theorem.

The primary trap lies in the angular dependence of the dipole potential.

Students frequently misapply the potential formula by either ignoring the cos⁡θ\cos\theta term (treating it like a monopole V∝1/rV \propto 1/r), using the wrong reference axis (leading to a sin⁡θ\sin\theta error), or making a sign error when calculating the potential difference between an obtuse and an acute angle.

By constraining the motion to a circular track, the problem isolates the conservative nature of the electrostatic field, ensuring that only the initial and final potentials dictate the required energy, while rigorously testing the exact spatial variation of the dipole potential.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterNuclei - Class 12
ConceptQ-value Calculation
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree