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Question #43865

Difficulty: 8/10 | 🟢 Free

The Question

A small dielectric cylinder of length 2L2L and cross-sectional area AA is placed along the xx-axis from x=−Lx = -L to x=Lx = L. It is subjected to a uniform external electric field E⃗=E0i^\vec{E} = E_0 \hat{i}. The material's electric susceptibility follows a Curie-like temperature dependence, χe=CT\chi_e = \frac{C}{T}, where CC is a constant and χe≪1\chi_e \ll 1 (so the internal field is approximately E0i^E_0 \hat{i}). A temperature gradient is maintained across the cylinder such that T(x)=T0(1+βx)T(x) = T_0 (1 + \beta x), where βL≪1\beta L \ll 1. Let p0p_0 be the dipole moment of the cylinder if it were at a uniform temperature T0T_0. Find the electric potential V(d)V(d) at a point x=dx = d on the axis (d≫Ld \gg L). Retain the leading order term and the first non-vanishing correction term involving β\beta.

Options

  • A.

    p04πϵ0d2(1−2βL23d)\frac{p_0}{4\pi\epsilon_0 d^2} \left( 1 - \frac{2 \beta L^2}{3 d} \right)

  • B.

    p04πϵ0d2(1−βL2d)\frac{p_0}{4\pi\epsilon_0 d^2} \left( 1 - \frac{\beta L^2}{d} \right)

  • C.

    p04πϵ0d2(1+2βL23d)\frac{p_0}{4\pi\epsilon_0 d^2} \left( 1 + \frac{2 \beta L^2}{3 d} \right)

  • D.

    p04πϵ0d2(1−βL22d)\frac{p_0}{4\pi\epsilon_0 d^2} \left( 1 - \frac{\beta L^2}{2 d} \right)

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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

Electric Field Symmetry|Dipole Potential Angular Dependence|Curie’s Law|Binomial Approximation

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental understanding of bound charges in dielectrics and multipole expansions.

Students often correctly identify the temperature dependence of polarization (Curie's Law) but fall into the trap of only calculating the modified surface bound charges σb=P⃗⋅n^\sigma_b = \vec{P} \cdot \hat{n}.

Because the polarization is non-uniform due to the temperature gradient, a volume bound charge ρb=−∇⋅P⃗\rho_b = -\nabla \cdot \vec{P} is generated.

Ignoring this volume charge leads to an incorrect quadrupole moment (Option B).

Furthermore, it tests the ability to recognize that a spatial gradient in susceptibility breaks the pure dipole symmetry, requiring a binomial expansion to find the far-field potential correction.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectric Charges and Fields - Class 12
ConceptElectric Field Symmetry
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree