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Question #42837

Difficulty: 8/10 | 🟢 Free

The Question

A solid non-conducting sphere of radius RR contains a uniform free volume charge density ρ0\rho_0. The sphere is filled with a non-uniform linear dielectric material whose relative permittivity varies with the radial distance rr from the center as ϵr(r)=Rr\epsilon_r(r) = \frac{R}{r} for 0<r≤R0 < r \le R. Which of the following expressions correctly represents the volume bound charge density ρb(r)\rho_b(r) inside the dielectric at a distance rr from the center?

Options

  • A.

    ρ0(4r3R−1)\rho_0 \left( \frac{4r}{3R} - 1 \right)

  • B.

    ρ0(rR−1)\rho_0 \left( \frac{r}{R} - 1 \right)

  • C.

    ρ0(2r3R−13)\rho_0 \left( \frac{2r}{3R} - \frac{1}{3} \right)

  • D.

    ρ0(1−rR)\rho_0 \left( 1 - \frac{r}{R} \right)

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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

Gauss Law Applicability|Shell Theorem (Internal vs External)|Non-Uniform Dielectric Filling

Why This Question Is Tricky

Common Mistakes

This question tests the application of Gauss's Law in non-uniform dielectrics.

A common student trap is to blindly apply the formula ρb=−ρf(1−1/ϵr)\rho_b = -\rho_f (1 - 1/\epsilon_r), which is only valid for homogeneous linear dielectrics where ∇ϵr=0\nabla \epsilon_r = 0.

In an inhomogeneous medium, the spatial variation of permittivity acts as an additional source of bound charge.

To solve correctly, students must revert to fundamental principles: finding the displacement field D⃗\vec{D} from free charges, determining E⃗\vec{E} and P⃗\vec{P}, and finally applying the differential operator ρb=−∇⋅P⃗\rho_b = -\nabla \cdot \vec{P} in spherical coordinates.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectric Charges and Fields - Class 12
ConceptGauss Law Applicability
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree