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Question #42848

Difficulty: 8/10 | 🟢 Free

The Question

A small conducting sphere of mass mm, radius RR, and net charge QQ is attached to one end of a rigid, massless rod of length LL. The other end of the rod is pivoted at the origin (0,0)(0,0) and connected to a torsional spring of restoring constant CC. The system is constrained to rotate in a smooth horizontal xyxy-plane. An external electric field is present in the region, given by E⃗=(E0+E1xL)i^−E1yLj^\vec{E} = \left(E_0 + E_1 \frac{x}{L}\right) \hat{i} - E_1 \frac{y}{L} \hat{j}, where E0E_0 and E1E_1 are positive constants. The rod is initially in equilibrium along the positive xx-axis. If the rod is given a small angular displacement θ\theta and released, what is the angular frequency ω\omega of the resulting simple harmonic motion? (Assume R≪LR \ll L and neglect any radiation or damping effects. The permittivity of free space is ϵ0\epsilon_0.)

Options

  • A.

    C+QL(E0+2E1)+4πϵ0R3E0E1mL2\sqrt{ \frac{C + Q L (E_0 + 2 E_1) + 4\pi\epsilon_0 R^3 E_0 E_1}{m L^2} }

  • B.

    C+QL(E0+2E1)mL2\sqrt{ \frac{C + Q L (E_0 + 2 E_1)}{m L^2} }

  • C.

    C+QE0L+4πϵ0R3E0E1mL2\sqrt{ \frac{C + Q E_0 L + 4\pi\epsilon_0 R^3 E_0 E_1}{m L^2} }

  • D.

    C+QL(E0+2E1)+8πϵ0R3E0E1mL2\sqrt{ \frac{C + Q L (E_0 + 2 E_1) + 8\pi\epsilon_0 R^3 E_0 E_1}{m L^2} }

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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

Charged Conducting Sphere|Torsional Oscillations|Mechanical Energy Conservation|Small Angle Approximation

Why This Question Is Tricky

Common Mistakes

This question tests the subtle interplay between mechanical restoring forces and electrostatic energy in non-uniform fields.

The primary trap lies in treating the conducting sphere merely as a point charge QQ and ignoring its finite size and polarizability.

In a non-uniform electric field, a conducting sphere redistributes its charge, creating an induced dipole moment p⃗∝E⃗\vec{p} \propto \vec{E}.

The interaction of this dynamic dipole with the field gradient results in a polarization potential energy Udipole=−12αeE2U_{dipole} = -\frac{1}{2} \alpha_e E^2.

Students commonly misapply the standard SHM logic by only calculating the torque on the net charge QQ (Option B), failing to account for the additional restoring torque generated by the spatial variation of the field magnitude which couples with the induced dipole.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectric Charges and Fields - Class 12
ConceptCharged Conducting Sphere
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree