Start Learning Free

Before You Begin

Question #42909

Difficulty: 8/10 | 🟢 Free

The Question

A solid non-conducting sphere of radius RR has a non-uniform volume charge density given by ρ(r)=ρ0(1−r2R2)\rho(r) = \rho_0 \left( 1 - \frac{r^2}{R^2} \right), where rr is the radial distance from the center of the sphere and ρ0\rho_0 is a positive constant. A spherical cavity of radius a=R4a = \frac{R}{4} is created within the sphere. The center of the cavity is located at a distance r0=R2r_0 = \frac{R}{2} from the center of the main sphere. What is the magnitude of the electric field at the center of the cavity?

Options

  • A.

    21ρ0R160ϵ0\frac{21 \rho_0 R}{160 \epsilon_0}

  • B.

    17ρ0R120ϵ0\frac{17 \rho_0 R}{120 \epsilon_0}

  • C.

    73ρ0R480ϵ0\frac{73 \rho_0 R}{480 \epsilon_0}

  • D.

    11ρ0R80ϵ0\frac{11 \rho_0 R}{80 \epsilon_0}

Check Your Answer

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

Continuous Charge Distribution|Superposition Principle|Nuclear Radius Scaling|Integration

Why This Question Is Tricky

Common Mistakes

The core trap in this problem lies in the misapplication of the Superposition Principle for non-uniform continuous charge distributions.

Students frequently memorize that the electric field at the center of a spherical cavity is simply the field produced by the main sphere at that point, implicitly assuming that the 'removed' charge produces zero field at its own center.

While this is true for uniform charge densities (where the removed charge is spherically symmetric about the cavity center), it fails catastrophically for non-uniform densities.

Because the original charge density ρ(r)\rho(r) has a radial gradient, the removed charge possesses a dipole-like asymmetry (a linear density gradient) relative to the cavity's center.

Consequently, the removed charge generates a non-zero electric field at its own center, which must be rigorously calculated and subtracted from the full sphere's field.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectric Charges and Fields - Class 12
ConceptContinuous Charge Distribution
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree