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Question #42330

Difficulty: 8/10 | 🟢 Free

The Question

A point charge $+q$ is initially placed at the exact center of a fixed, isolated, uncharged thin conducting spherical shell of inner radius $R$ . The charge is then displaced by a very small vector $\vec{x}$ ( $|\vec{x}| \ll R$ ) from the center. Which of the following represents the approximate electrostatic force $\vec{F}$ acting on the point charge at this displaced position?

Options

A.
$\frac{q^2 \vec{x}}{4\pi\epsilon_0 R^3}$
B.
$-\frac{q^2 \vec{x}}{4\pi\epsilon_0 R^3}$
C.
$\vec{0}$
D.
$\frac{q^2 \vec{x}}{2\pi\epsilon_0 R^3}$

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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

Equilibrium Stability|Coulombic Central Force|Potential of Concentric Shells|Second Derivative Test

Why This Question Is Tricky

Common Mistakes

This question tests the application of the method of images and the physical implications of Earnshaw's theorem in the context of cavity fields.

A common trap is assuming that because the electric field inside an empty conducting shell is zero, a displaced charge will experience zero net force (neutral equilibrium), leading to option C.

Another trap is incorrectly assuming the induced negative charge on the inner surface acts as a restoring force (stable equilibrium), leading to option B.

In reality, the induced charge distribution becomes asymmetric, creating a local field that pulls the charge further from the center, proving that static electrostatic levitation is unstable.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Work, Energy, and Power - Class 11 Part 1
Concept	Equilibrium Stability
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free