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Question #42664

Difficulty: 8/10 | 🟢 Free

The Question

A solid non-conducting sphere of radius $R$ has a radially varying volume charge density given by $\rho(r) = \rho_0 \frac{r}{R}$ for $0 \le r \le R$ , where $\rho_0$ is a positive constant and $r$ is the radial distance from the center. If the total charge of the sphere is $Q$ , what is the total electrostatic self-energy of this charge configuration? (Assume the potential at infinity is zero, and $\epsilon_0$ is the permittivity of free space).

Options

A.
$\frac{Q^2}{7\pi\epsilon_0 R}$
B.
$\frac{3Q^2}{20\pi\epsilon_0 R}$
C.
$\frac{Q^2}{8\pi\epsilon_0 R}$
D.
$\frac{Q^2}{56\pi\epsilon_0 R}$

Check Your Answer

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

Potential Gradient Direction|Self-Energy of Charged Sphere|Integration

Why This Question Is Tricky

Common Mistakes

Students often memorize the self-energy of a uniform solid sphere ( $U = \frac{3Q^2}{20\pi\epsilon_0 R}$ ) or a spherical shell ( $U = \frac{Q^2}{8\pi\epsilon_0 R}$ ) and incorrectly apply these to non-uniform charge distributions.

This question tests the fundamental definition of electrostatic self-energy via the energy density of the electric field ( $u = \frac{1}{2}\epsilon_0 E^2$ ).

The trap lies in assuming the internal potential gradient behaves identically to a uniform sphere, or completely neglecting the internal field energy and only calculating the external field energy (which yields the shell result).

To succeed, students must derive the correct radially dependent internal electric field using Gauss's Law and integrate the energy density over all space.

Quick Facts About This Question

Question Information

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