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Question #42042

Difficulty: 8/10 | 🟢 Free

The Question

Two fixed point charges, each of magnitude $+Q$ , are located on the vertical $y$ -axis at $y = +d$ and $y = -d$ . A particle of mass $m$ and charge $+q$ is released from rest at $y = +d/2$ . The region has a uniform downward gravitational field of acceleration $g$ . Assuming no air resistance, what is the speed of the particle when it reaches the origin ( $y = 0$ )?

Options

A.
$\sqrt{gd + \frac{8qQ}{9\pi\epsilon_0 md}}$
B.
$\sqrt{gd + \frac{qQ}{2\pi\epsilon_0 md}}$
C.
$\sqrt{gd + \frac{qQ}{3\pi\epsilon_0 md}}$
D.
$\sqrt{gd - \frac{qQ}{3\pi\epsilon_0 md}}$

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

Kinematics of Vertical Free Fall|Superposition of Potentials|Conservative Field Path Independence

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to recognize when standard kinematic equations fail.

A common trap is calculating the initial net acceleration at $y = d/2$ and incorrectly assuming it remains constant to find the final velocity using $v^2 = u^2 + 2as$ .

Because the particle moves through a non-linear field gradient created by the superposition of two $1/r$ potentials, the acceleration is highly variable.

Students must abandon constant- $g$ assumptions and strictly apply the conservation of mechanical energy, carefully accounting for the potential contributions of both fixed charges at the initial and final positions.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Motion in a Straight Line - Class 11 Part 1
Concept	Kinematics of Vertical Free Fall
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free