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Question #41995

Difficulty: 8/10 | 🟢 Free

The Question

A spherically symmetric volume charge distribution in free space is given by $\rho(r) = \frac{\rho_0}{r} \sin(kr)$ for $0 < r \le R$ , and $\rho(r) = 0$ for $r > R$ , where $\rho_0$ is a dimensional constant. The constant $k$ is chosen such that the charge density becomes zero for the first time at the surface of the sphere $r = R$ . What is the total electric flux through a concentric spherical surface of radius $2R$ ?

Options

A.
$\frac{4 \rho_0 R^2}{\epsilon_0}$
B.
$\frac{4 \pi \rho_0 R^2}{\epsilon_0}$
C.
$\frac{\rho_0 R^2}{\pi \epsilon_0}$
D.
$\frac{2 \rho_0 R^2}{\epsilon_0}$

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

Common Mistakes

This question tests the student's ability to apply dimensional consistency and Gauss's Law to a continuous, non-uniform charge distribution.

The primary trap lies in the transcendental function's argument: students often fail to recognize that 'kr' must be dimensionless, leading to incorrect assumptions about 'k'.

$Furthermore, the condition that the density vanishes for the first time at r=R requires setting kR = \pi.$

$A common mistake is ignoring the spherical volume element 4\pi r^2 dr and integrating \rho(r) directly, or failing to cancel the \pi factor during integration, which leads to dimensionally inconsistent or numerically incorrect flux expressions.$

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