Question

If the distance between two point charges is doubled, the electrostatic force between them becomes—

• A. Double
• B. Half
• C. One-fourth
• D. Four times

Hint:

In Coulomb's law, if distance becomes: Double $\rightarrow$ Force becomes one-fourth Half $\rightarrow$ Force becomes four times

Answer 1

The Correct Option is C

Concept: According to Coulomb's Law, the electrostatic force between two point charges is: \[ F = k \frac{q_1 q_2}{r^2} \] where:
\( F \) = force between charges
\( r \) = distance between charges
Thus, force is inversely proportional to the square of the distance. Step 1: Original force \[ F \propto \frac{1}{r^2} \] Step 2: If distance is doubled New distance = \( 2r \) \[ F' \propto \frac{1}{(2r)^2} = \frac{1}{4r^2} \] Step 3: Compare forces \[ F' = \frac{F}{4} \] So, the force becomes one-fourth of the original value. \[ \therefore \text{Correct option: (c) One-fourth} \]