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Question #42131

Difficulty: 8/10 | 🟢 Free

The Question

A thick uncharged conducting spherical shell has an inner radius R1R_1 and outer radius R2R_2. The spherical cavity (0≤r<R10 \le r < R_1) is completely filled with a uniformly charged compressible fluid of total charge QQ which always fills the cavity. The shell is made of a material whose coefficient of linear thermal expansion varies radially as α(r)=cr2\alpha(r) = \frac{c}{r^2} for R1≤r≤R2R_1 \le r \le R_2, where cc is a positive constant. The outer surface of the shell (r=R2r = R_2) is mechanically fixed such that it cannot move. The system is heated, increasing its temperature at a constant rate dTdt=β\frac{dT}{dt} = \beta. Assuming the radial displacement u(r)u(r) of the shell material is purely kinematic and governed by ∂u∂r=α(r)ΔT\frac{\partial u}{\partial r} = \alpha(r) \Delta T, find the rate of change of the electric field magnitude ∂E∂t\frac{\partial E}{\partial t} at a fixed distance rr (r<R1r < R_1) from the center at t=0t=0.

Options

  • A.

    3Qrcβ(R2−R1)4πϵ0R15R2\frac{3 Q r c \beta (R_2 - R_1)}{4\pi\epsilon_0 R_1^5 R_2}

  • B.

    −3Qrcβ(R2−R1)4πϵ0R15R2-\frac{3 Q r c \beta (R_2 - R_1)}{4\pi\epsilon_0 R_1^5 R_2}

  • C.

    Qcβ(R2−R1)2πϵ0R14R2\frac{Q c \beta (R_2 - R_1)}{2\pi\epsilon_0 R_1^4 R_2}

  • D.

    −3Qrcβ4πϵ0R15-\frac{3 Q r c \beta}{4\pi\epsilon_0 R_1^5}

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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

Expansion of Cavities in Solids|Cavity Field Shielding|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to couple Gauss's Law for a volumetric charge distribution with the kinematics of non-uniform thermal expansion.

The primary trap lies in the boundary conditions: students intuitively assume cavities expand outward.

However, because the outer boundary is mechanically fixed, the non-uniform expansion forces the inner boundary to move inward, shrinking the cavity and yielding a negative dR_1/dt.

Furthermore, students must correctly apply the chain rule to the time-dependent volume of the charge distribution to find the rate of change of the internal electric field, avoiding the mistake of treating the field as that of a point charge or surface charge.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterThermal Properties of Matter - Class 11 Part 2
ConceptExpansion of Cavities in Solids
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree