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Question #78389

Difficulty: 8/10 | 🟢 Free

The Question

A thin, non-conducting spherical shell of radius $R$ is centered at the origin of a coordinate system. The surface charge density at any point on the shell is given by the spatial function $\sigma(r) = \sigma_0 \frac{\sin(kr)}{r}$ , where $r$ is the radial distance of the point from the origin, $\sigma_0$ is a constant with appropriate SI units to ensure $\sigma(r)$ has units of $\text{C}\cdot\text{m}^{-2}$ , and $k$ is a positive constant. If $\epsilon_0$ is the vacuum permittivity, which of the following expressions correctly represents the total electric flux through a concentric imaginary spherical surface of radius $2R$ ?

Options

A.
$\frac{4\pi \sigma_0 R \sin(kR)}{\epsilon_0}$
B.
$\frac{4\pi \sigma_0}{\epsilon_0} \left( \frac{\sin(kR)}{k^2} - \frac{R \cos(kR)}{k} \right)$
C.
$\frac{2\pi \sigma_0}{\epsilon_0 k} (1 - \cos(kR))$
D.
$\frac{4\pi \sigma_0 R^2 \sin(kR)}{\epsilon_0}$

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

Dimensional Consistency and Transcendental Functions|Continuous Charge Distribution|Electric Charges and Fields

Why This Question Is Tricky

Examiner Levers

Visualization Stress

Requires mental rotation or 3D mapping (Optics/Geometry/Vectors).

Difficulty Boost : +4

Common Mistakes

This question tests the student's ability to synthesize dimensional analysis with Gauss's Law while navigating a deceptive coordinate-dependent charge distribution.

The primary trap lies in the function $\sigma(r) = \sigma_0 \sin(kr)/r$ .

Students often reflexively integrate this expression over a volume (Option B) or a flat disk (Option C), ignoring the physical constraint that for a surface charge on a spherical shell, the radial coordinate $r$ is fixed at $R$ .

Consequently, the surface charge density is actually uniform.

Furthermore, the problem serves as a rigorous test of dimensional consistency: because the argument $kr$ must be dimensionless, $[k] = [L]^{-1}$ , which forces $[\sigma_0] = [C][L]^{-1}$ .

Distractors B and D can be eliminated purely because they yield dimensionally incorrect expressions for enclosed charge, rewarding students who verify dimensional homogeneity before performing complex integrations.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Units and Measurements - Class 11 Part 1
Concept	Dimensional Consistency and Transcendental Functions
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free