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Question #43778

Difficulty: 8/10 | 🟢 Free

The Question

Two identical point charges, each of mass $m$ and charge $q$ , are initially separated by a very large distance in free space. One charge is initially at rest, and the other is projected towards it with an initial speed $v_0$ and an impact parameter $b$ . Given the relation $b = \frac{2 q^2}{4 \pi \epsilon_0 m v_0^2}$ , what is the minimum separation distance between the two charges during their subsequent motion? (Assume all physical quantities are expressed in standard SI units).

Options

A.
$b(\sqrt{2} + 1)$
B.
$2b$
C.
$b$
D.
$b(\sqrt{2} - 1)$

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

Distance of Closest Approach|Relative Velocity in 2D

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to analyze 2D scattering using the Center of Mass frame and exposes the common misconception of treating the closest approach as a 1D head-on collision.

Students frequently assume that at the minimum separation, the relative velocity is zero, leading them to equate the entire initial COM kinetic energy to the electrostatic potential energy (yielding $r_{min} = 2b$ ).

By introducing a non-zero impact parameter, the relative velocity at the closest approach has a non-zero tangential component dictated by the Conservation of Angular Momentum.

Failing to account for this residual kinetic energy is the primary trap.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Atoms - Class 12
Concept	Distance of Closest Approach
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free