Question

If the distance between two charges is doubled, what happens to the Coulomb force?

• A. It becomes double
• B. It becomes four times
• C. It becomes half
• D. It becomes one-fourth

Hint:

According to Coulomb’s law, electrostatic force varies inversely with the square of distance. If distance becomes \(2r\), force becomes \(\frac{1}{4}\). If distance becomes \(3r\), force becomes \( \frac{1}{9} \).

Answer 1

The Correct Option is D

Concept: Coulomb’s law states that the electrostatic force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. ::contentReference[oaicite:0]{index=0}

Step 1: Initial force. Let the initial distance between two charges be \(r\). Then the force is: \[ F = k\frac{q_1 q_2}{r^2} \]

Step 2: When distance is doubled. If the distance becomes \(2r\): \[ F' = k\frac{q_1 q_2}{(2r)^2} \] \[ F' = k\frac{q_1 q_2}{4r^2} \] \[ F' = \frac{F}{4} \]

Step 3: Conclusion. Thus, when the distance between two charges is doubled, the Coulomb force becomes \(\frac{1}{4}\) of its original value.