Question:medium

Two point charges placed a distance d apart in vacuum exert a force of magnitude F on each other. One of the two charges is doubled. To keep the magnitude of force same the separation between the charges should be changed to

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In problems involving ratios and proportions like Coulomb's law (\(F \propto \frac{q_1 q_2}{d^2}\)), you can set up a ratio. If \(F\) is to remain constant, then the ratio \(\frac{q_1 q_2}{d^2}\) must also be constant. So, \(\frac{q_1 q_2}{d_{initial}^2} = \frac{(2q_1) q_2}{d_{final}^2}\), which quickly leads to \(d_{final}^2 = 2d_{initial}^2\).
Updated On: May 26, 2026
  • 2d
  • d/2
  • \(\sqrt{2}\) d
  • d/\(\sqrt{2}\)
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The Correct Option is C

Solution and Explanation


Step 1: Conceptual Foundation:
This problem utilizes Coulomb's Law, which quantifies the electrostatic force between two point charges. This force is directly proportional to the product of the charges and inversely proportional to the square of the distance separating them.

Step 2: Governing Equation:
Coulomb's Law is expressed as: \[ F = k \frac{q_1 q_2}{d^2} \] where \(F\) is the force, \(q_1\) and \(q_2\) are the charges, \(d\) is the distance between them, and \(k\) is Coulomb's constant. This formula will be applied to compare the initial and final scenarios.

Step 3: Analytical Breakdown:
Initial State:
Charges: \(q_1\), \(q_2\). Distance: \(d\). Force: \(F\). \[ F = k \frac{q_1 q_2}{d^2} \cdots (1) \] Final State:
One charge is doubled, resulting in new charges \(q'_1 = 2q_1\) and \(q'_2 = q_2\). The new distance is denoted as \(d'\). The force \(F'\) must remain equal to the initial force \(F\). \[ F' = k \frac{(2q_1) q_2}{(d')^2} \cdots (2) \] Force Equivalence:
Given that \(F' = F\), we equate equation (1) and equation (2): \[ k \frac{q_1 q_2}{d^2} = k \frac{2q_1 q_2}{(d')^2} \] The constants \(k\), \(q_1\), and \(q_2\) cancel from both sides: \[ \frac{1}{d^2} = \frac{2}{(d')^2} \] Solving for the new distance \(d'\): \[ (d')^2 = 2d^2 \] \[ d' = \sqrt{2d^2} = \sqrt{2} d \]

Step 4: Conclusion:
To maintain the original force after doubling one charge, the separation distance must be increased to \(\sqrt{2} d\).

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