Question

The force between two point charges separated by a distance \(d\) is \(F\). If one charge is moved away by a distance \(d\), then the new force is:

• A. \(F/4\)
• B. \(4F\)
• C. \(F/2\)
• D. \(F/2\)

Hint:

Coulomb's law shows that the force between two point charges is inversely proportional to the square of the distance between them. If the distance is doubled, the force will decrease by a factor of 4.

Answer 1

The Correct Option is C

The force between two point charges is governed by Coulomb's law: \[ F = k \frac{q_1 q_2}{d^2} \] Where: - \(F\) is the force between the charges, - \(k\) is Coulomb’s constant, - \(q_1\) and \(q_2\) are the charges, - \(d\) is the distance between the charges. If the distance between the charges is increased, say, doubled, then the new force will be: \[ F' = k \frac{q_1 q_2}{(2d)^2} = \frac{F}{4} \] Thus, if one of the charges is moved away, reducing the distance by half, the new force is half the original force. Therefore, the new force is F/2.