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Question #42945

Difficulty: 8/10 | 🟢 Free

The Question

A particle of mass $m = 1 \text{ kg}$ and charge $q = 1 \text{ C}$ is projected from the coordinates $(0, 2 \text{ m})$ with an initial velocity $\vec{v}_0 = 10 \hat{i} \text{ m/s}$ . The region contains an electric field corresponding to the electric potential $V(x,y) = 5(x^2 + y^2) \text{ V}$ , where $x$ and $y$ are in meters. Neglecting gravity, what is the $x$ -coordinate of the particle when it crosses the $x$ -axis for the first time?

Options

A.
$\sqrt{10} \text{ m}$
B.
$2\sqrt{5} \text{ m}$
C.
$5\sqrt{2} \text{ m}$
D.
$10 \text{ m}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Potential Gradient Direction|Projectile Path Equation|Vector Cross-Products

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to recognize when standard kinematic equations of motion are invalid.

A common trap in JEE is to evaluate the electric field at the initial position ( $\vec{E} = -20\hat{j} \text{ V/m}$ at $y=2$ ) and incorrectly assume it remains constant, leading to a parabolic projectile path.

By doing so, a student would calculate a constant downward acceleration of $20 \text{ m/s}^2$ , yielding a crossing time of $t = 1/\sqrt{5} \text{ s}$ and an incorrect crossing point of $x = 2\sqrt{5} \text{ m}$ (Option B).

The correct approach requires recognizing that a quadratic potential gradient produces a position-dependent restoring force, necessitating the use of simple harmonic motion differential equations to find the true elliptical trajectory.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electrostatic Potential and Capacitance - Class 12
Concept	Potential Gradient Direction
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free