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Question #88664

Difficulty: 5/10 | 🟢 Free

The Question

A hollow, uncharged spherical conductor has an inner radius $R_1$ and an outer radius $R_2$ . A permanently polarized dielectric spherical shell of inner radius $a$ and outer radius $R_1$ is placed exactly inside the cavity, fitting perfectly against the conductor's inner surface. The polarization vector of the dielectric is given by $\mathbf{P} = \alpha r \hat{r}$ , where $\alpha$ is a positive constant, $r$ is the radial distance from the center, and $\hat{r}$ is the unit radial vector. The region $0 \le r < a$ is vacuum. What is the electric potential at the center of the cavity ( $r=0$ )? Assume the electric potential at infinity is zero.

Options

A.
$\frac{\alpha}{2\epsilon_0} (a^2 - R_1^2)$
B.
$0$
C.
$\frac{\alpha}{2\epsilon_0} (R_1^2 - a^2)$
D.
$\frac{\alpha}{\epsilon_0} \left( \frac{3a^2 - R_1^2}{2} - \frac{a^3}{R_1} \right)$

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

Non-Uniform Dielectric Filling|Boundary Condition Reflection|Cavity Field Shielding

Why This Question Is Tricky

Examiner Levers

Dimensional Analysis

Tests if the student verifies units/dimensions before selecting.

Difficulty Boost : +2

Common Mistakes

This question tests the conflation of Faraday Cage shielding with the behavior of bound charges inside a cavity.

Students commonly assume that because the dielectric is enclosed within an uncharged conducting shell, the electric field inside the cavity must be zero ( $E=0$ ), leading them to incorrectly choose $0$ .

Furthermore, students who attempt to use Gauss's Law for the electric field often ignore the bound polarization charges (both volume $\rho_b = -\nabla \cdot \mathbf{P}$ and surface $\sigma_b = \mathbf{P} \cdot \hat{n}$ ) because there is no free charge.

By using a permanently polarized dielectric, the problem forces the realization that while the displacement field $\mathbf{D}=0$ , the non-zero polarization creates a real internal electric field ( $\mathbf{E} = -\mathbf{P}/\epsilon_0$ ) that modifies the cavity's potential, effectively nullifying the 'complete shielding' shortcut.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electrostatic Potential and Capacitance - Class 12
Concept	Non-Uniform Dielectric Filling
Difficulty	5
Cognitive Load	6
Target Solve Time	140 sec
Access	Free