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Question #42018

Difficulty: 8/10 | 🟢 Free

The Question

A non-uniform line charge is placed along the positive x-axis from $x = 0$ to $x = a$ . The linear charge density is given by $\lambda = c x^2$ , where $c$ is a positive constant. A point charge $q_0$ is placed on the y-axis at $y = a$ . Which of the following expressions correctly represents the net electric force vector $\vec{F}$ exerted on $q_0$ by the line charge?

Options

A.
$\frac{q_0 c a}{4\pi\epsilon_0} \left[ \left( 2 - \frac{3}{\sqrt{2}} \right) \hat{i} + \left( \ln(\sqrt{2} + 1) - \frac{1}{\sqrt{2}} \right) \hat{j} \right]$
B.
$\frac{q_0 c a}{4\pi\epsilon_0} \left[ \left( \frac{3}{\sqrt{2}} - 2 \right) \hat{i} + \left( \ln(\sqrt{2} + 1) - \frac{1}{\sqrt{2}} \right) \hat{j} \right]$
C.
$\frac{q_0 c a}{4\pi\epsilon_0} \left[ \left( 2 - \frac{3}{\sqrt{2}} \right) \hat{i} + \left( \ln(\sqrt{2} + 1) + \frac{1}{\sqrt{2}} \right) \hat{j} \right]$
D.
$\frac{q_0 c a}{4\pi\epsilon_0} \left[ \left( \frac{3}{\sqrt{2}} - 2 \right) \hat{i} - \left( \ln(\sqrt{2} + 1) - \frac{1}{\sqrt{2}} \right) \hat{j} \right]$

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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

Dimensional Independence vs. Scalar/Vector Nature|Coulomb's Law Vector Form|Integration

Why This Question Is Tricky

Common Mistakes

This problem tests the student's ability to perform rigorous vector integration for a non-uniform spatial charge distribution.

A common trap in JEE Advanced is treating the superposition principle as a scalar summation or failing to correctly assign the direction of the differential position vector $\vec{R} = \vec{r}_{final} - \vec{r}_{initial}$ .

Students frequently miss the negative sign in the x-component because they blindly use $x\hat{i}$ instead of $-x\hat{i}$ for the vector pointing from the source charge on the x-axis to the test charge on the y-axis.

Furthermore, the non-uniform density $\lambda = cx^2$ prevents the use of standard symmetry arguments, forcing a component-by-component integration that strictly evaluates dimensional independence.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Units and Measurements - Class 11 Part 1
Concept	Dimensional Independence vs. Scalar/Vector Nature
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free