Question

Figure shows variation of Coulomb force (F) acting between two point charges with \( \frac{1}{r^2} \), \( r \) being the separation between the two charges \( (q_1, q_2) \) and \( (q_2, q_3) \). If \( q_2 \) is positive and least in magnitude, then the magnitudes of \( q_1, q_2 \), and \( q_3 \) are such that:

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

Hint:

Remember that the Coulomb force is proportional to the product of the charges and inversely proportional to the square of the distance between them. If one charge is smaller, it produces less force.

Answer 1

The Correct Option is D

Coulomb Force Graph – Step-by-Step Reasoning

Given: The graph shows F (Coulomb force) plotted against 1/r² for two pairs: (q₁,q₂) and (q₂,q₃). Also, q₂ is positive and the smallest in magnitude.

1) Relate slope to charges

F = k (q_i q_j) / r²  ⇒  F vs (1/r²) is a straight line through origin Slope m = k(q_i q_j)

• Positive slope ⇒ product q_i q_j > 0 ⇒ charges have the same sign.
• Negative slope ⇒ product q_i q_j < 0 ⇒ charges have opposite signs.

2) Deduce the signs from the figure

• Line for (q₁,q₂) slopes upward ⇒ q₁ and q₂ have the same sign. With q₂ > 0 ⇒ q₁ > 0.
• Line for (q₂,q₃) slopes downward ⇒ q₂ and q₃ have opposite signs. With q₂ > 0 ⇒ q₃ < 0.

3) Compare magnitudes using steepness

The (q₁,q₂) line is steeper than the (q₂,q₃) line, so

|m₁₂| = k|q₁ q₂|  >  |m₂₃| = k|q₂ q₃|   ⇒  |q₁| > |q₃|  (since q₂ is common and positive)

4) Use the “q₂ is least” condition

Given |q₂| is the smallest: |q₂| < |q₁| and |q₂| < |q₃|.

Final Result

Signs: q₁ > 0, q₂ > 0, q₃ < 0.

Magnitudes (ordering): q₂ < q₃ < q₁.

Correct Option: Option 4 → q₂ < q₃ < q₁