Question

The Coulomb force (\( F \)) versus \( \frac{1}{r^2} \) graphs for two pairs of point charges (\( q_1 \) and \( q_2 \)) and (\( q_2 \) and \( q_3 \)) are shown in the figure. The charge \( q_2 \) is positive and has the least magnitude. Then: 

• A. \( q_1>q_2>q_3 \)
• B. \( q_1>q_3>q_2 \)
• C. \( q_3>q_2>q_1 \)
• D. \( q_3>q_1>q_2 \)

Hint:

A steeper slope in the \( F \) vs. \( \frac{1}{r^2} \) graph means a greater product of charge magnitudes.

Answer 1

The Correct Option is D

Understanding Coulomb Force Variation
- According to Coulomb’s Law, the force between two charges is: \[ F = \frac{1}{4\pi \epsilon_0} \frac{|q_1 q_2|}{r^2} \] - The slope of the \( F \) vs. \( \frac{1}{r^2} \) graph represents \( |q_1 q_2| \).
- Since \( q_2 \) has the least magnitude and is positive, comparing slopes gives: \[ q_3>q_1>q_2 \] Thus, the correct order is \( q_3>q_1>q_2 \), matching option (D).