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Question #42658

Difficulty: 8/10 | 🟢 Free

The Question

A thick non-conducting spherical shell has an inner radius $R$ and an outer radius $2R$ . The shell carries a non-uniform volume charge density given by $\rho(r) = \alpha r$ for $R \le r \le 2R$ , where $\alpha$ is a positive constant with units of $\text{C m}^{-4}$ , and $r$ is the radial distance from the center of the shell. The regions $r < R$ and $r > 2R$ contain no charge. Assuming the electric potential at infinity is zero, what is the electric potential at the center of the shell ( $r = 0$ )?

Options

A.
$\frac{7 \alpha R^3}{3 \epsilon_0}$
B.
$\frac{15 \alpha R^3}{8 \epsilon_0}$
C.
$\frac{15 \alpha R^3}{4 \epsilon_0}$
D.
$\frac{11 \alpha R^3}{24 \epsilon_0}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Reference Potential Convention|Shell Theorem (Internal vs External)|Principle of Superposition|Vector Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to calculate electric potential for non-uniform, non-conducting charge distributions, bypassing the standard Shell Theorem shortcut ( $V = kQ/R$ ).

A common trap is calculating the total charge $Q = 15\pi\alpha R^4$ and incorrectly assuming the potential inside the cavity is uniform and equal to the surface potential $V = kQ/(2R)$ (Option B), which is only true for conductors.

Another trap is integrating the electric field across the shell but forgetting to add the reference potential from infinity to the outer surface, leading to Option D.

The most robust method requires applying the superposition principle $V = \int k dq / r$ directly, testing fundamental field theory concepts over rote formulas.

Quick Facts About This Question

Question Information

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