Question

When the distance between the charges is halved, the force between the charges becomes:

• A. Half
• B. Twice
• C. Four times
• D. None of these

Hint:

The force between two charges is inversely proportional to the square of the distance between them, according to Coulomb’s law.

Answer 1

The Correct Option is B

Step 1: Understanding Coulomb's Law.
According to Coulomb's law, the force \( F \) between two point charges is given by the equation: \[ F = k \frac{q_1 q_2}{r^2} \] where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them.
Step 2: Effect of halving the distance.
If the distance \( r \) is halved, the force becomes: \[ F' = k \frac{q_1 q_2}{\left(\frac{r}{2}\right)^2} = 4 \times \left(k \frac{q_1 q_2}{r^2}\right) = 4F \] Step 3: Conclusion.
Thus, when the distance between the charges is halved, the force between the charges becomes four times larger, corresponding to option (C).