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Question #42665

Difficulty: 8/10 | 🟢 Free

The Question

A solid non-conducting sphere of radius $R$ is centered at the origin. The volume charge density within the sphere varies with the radial distance $r$ as $\rho(r) = \rho_0 \left(\frac{r}{R}\right)^2$ , where $\rho_0$ is a positive constant. Assuming the electric potential at infinity is zero, what is the electric potential at a distance $r = \frac{R}{2}$ from the center of the sphere?

Options

A.
$\frac{79 \rho_0 R^2}{320 \epsilon_0}$
B.
$\frac{15 \rho_0 R^2}{320 \epsilon_0}$
C.
$\frac{11 \rho_0 R^2}{40 \epsilon_0}$
D.
$\frac{\rho_0 R^2}{320 \epsilon_0}$

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

Equipotential Surface Geometry|Non-conducting Charged Sphere|Gauss Law Applicability|Integration

Why This Question Is Tricky

Common Mistakes

Students often memorize the potential formula for a uniformly charged sphere or incorrectly assume that the potential inside a non-uniform sphere can be found by merely integrating the internal electric field from the point of interest to the center.

The core misconception tested here is the failure to recognize that absolute potential $V(r)$ requires integrating from infinity.

This integral splits into the potential at the surface $V(R)$ (acting as a boundary condition) plus the integral of the internal electric field from $r$ to $R$ .

Ignoring $V(R)$ leads to the trap option B, while integrating from the center leads to option D.

Furthermore, assuming a uniform density with the same total charge leads to option C.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electrostatic Potential and Capacitance - Class 12
Concept	Equipotential Surface Geometry
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free