Question:
Two charges \( q_1 \) and \( q_2 \) are placed at a distance \( r \). If the distance between them is doubled, the electrostatic force between them becomes:
Two charges \( q_1 \) and \( q_2 \) are placed at a distance \( r \). If the distance between them is doubled, the electrostatic force between them becomes:
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In inverse-square laws (like gravity or electrostatics), doubling the distance reduces the force by a factor of 4.
Updated On: May 29, 2026
- One-fourth of the original force
- Half of the original force
- Four times the original force
- Twice the original force
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The Correct Option is A
Solution and Explanation
Concept: Coulomb's Law describes the electrostatic force ($F$) between two point charges: \[ F = k \frac{q_1 q_2}{r^2} \] where $k$ is Coulomb's constant, $q_1, q_2$ are the magnitudes of the charges, and $r$ is the distance between them.
Step 1: Identify the relationship between Force and Distance. From the formula, $F$ is inversely proportional to the square of the distance between the charges: \[ F \propto \frac{1}{r^2} \]
Step 2: Calculate the effect of doubling the distance. Let the original force be $F_1 = \frac{k q_1 q_2}{r^2}$. When the distance is doubled, $r' = 2r$. The new force $F_2$ is: \[ F_2 = \frac{k q_1 q_2}{(2r)^2} = \frac{k q_1 q_2}{4r^2} \]
Step 3: Compare the new force to the original. \[ F_2 = \frac{1}{4} \left( \frac{k q_1 q_2}{r^2} \right) = \frac{1}{4} F_1 \]
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