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Question #42914

Difficulty: 8/10 | 🟢 Free

The Question

A solid non-conducting sphere of radius $R$ has a volume charge density $\rho(r)$ that varies with the radial distance $r$ from the center as $\rho(r) = \rho_0 \frac{r}{R}$ , where $\rho_0$ is a positive constant. If the total charge on the sphere is $Q$ , what is the total electrostatic self-energy of this spherical charge distribution? Assume the permittivity of free space is $\epsilon_0$ .

Options

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

Gauss Law Applicability|Self-Energy of Charged Sphere|Variable Force Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to derive electrostatic self-energy from first principles for a non-uniform charge distribution.

The primary trap is the misapplication of the standard formula for a uniformly charged solid sphere ( $U = 3Q^2 / 20\pi\epsilon_0 R$ ), which is a common rote-memorization error.

Additionally, students might correctly find the internal electric field but fail to account for the external field's energy contribution ( $U_{out}$ ), leading to the trap option $Q^2 / 56\pi\epsilon_0 R$ , or conversely, they might only calculate the external energy, leading to $Q^2 / 8\pi\epsilon_0 R$ .

The problem strictly requires applying Gauss's Law to find the non-linear internal field $E(r) \propto r^2$ and correctly integrating the energy density $\frac{1}{2}\epsilon_0 E^2$ over all space.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electric Charges and Fields - Class 12
Concept	Gauss Law Applicability
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free