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Question #42318

Difficulty: 8/10 | 🟢 Free

The Question

A solid non-conducting sphere of radius RR has a spherically symmetric volume charge density given by ρ(r)=ρ0rR\rho(r) = \rho_0 \frac{r}{R} for r≤Rr \le R, where rr is the radial distance from the center and ρ0\rho_0 is a positive constant. The electrostatic potential at the surface of the sphere is explicitly defined to be zero (V(R)=0V(R) = 0). Which of the following correctly gives the electrostatic potential at the center of the sphere (VcV_c) and at a radial distance r=2Rr = 2R (V2RV_{2R})? Let ϵ0\epsilon_0 be the permittivity of free space.

Options

  • A.

    Vc=ρ0R212ϵ0,V2R=−ρ0R28ϵ0V_c = \frac{\rho_0 R^2}{12 \epsilon_0}, \quad V_{2R} = -\frac{\rho_0 R^2}{8 \epsilon_0}

  • B.

    Vc=ρ0R23ϵ0,V2R=ρ0R28ϵ0V_c = \frac{\rho_0 R^2}{3 \epsilon_0}, \quad V_{2R} = \frac{\rho_0 R^2}{8 \epsilon_0}

  • C.

    Vc=ρ0R24ϵ0,V2R=−ρ0R24ϵ0V_c = \frac{\rho_0 R^2}{4 \epsilon_0}, \quad V_{2R} = -\frac{\rho_0 R^2}{4 \epsilon_0}

  • D.

    Vc=ρ0R212ϵ0,V2R=ρ0R28ϵ0V_c = \frac{\rho_0 R^2}{12 \epsilon_0}, \quad V_{2R} = \frac{\rho_0 R^2}{8 \epsilon_0}

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Correct Answer

Step-by-Step Solution

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

Potential Energy Reference|Reference Potential Convention|Charged Conducting Sphere

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental definition of electrostatic potential as the line integral of the electric field, specifically targeting the trap of the reference potential convention.

Students often blindly memorize formulas like V=kQ/rV = kQ/r for regions outside a sphere, which implicitly assumes V(∞)=0V(\infty) = 0.

By explicitly setting the boundary condition V(R)=0V(R) = 0, the problem forces students to abandon memorized formulas and integrate E⃗⋅dr⃗\vec{E} \cdot d\vec{r} from the new reference point.

If a student incorrectly assumes V(∞)=0V(\infty) = 0, they will calculate V2R=ρ0R28ϵ0V_{2R} = \frac{\rho_0 R^2}{8 \epsilon_0} and Vc=ρ0R23ϵ0V_c = \frac{\rho_0 R^2}{3 \epsilon_0}, leading directly to the primary distractor.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterWork, Energy, and Power - Class 11 Part 1
ConceptPotential Energy Reference
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree