Question

The distance between two charges is made half and one of the charges is also halved. The force acting between the two will become as compared to previous value.

• A. half
• B. double
• C. thrice
• D. none of these

Hint:

Force between two charges increases with decreasing distance and is directly proportional to the product of charges.

Answer 1

The Correct Option is B

Step 1: Coulomb's law.
Coulomb’s law for the force between two charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is: \[ F = k \cdot \frac{{q_1 q_2}}{{r^2}} \] where \( k \) is Coulomb’s constant.
Step 2: Applying changes.
When the distance is halved, \( r \rightarrow \frac{r}{2} \). So the force increases by a factor of \( 4 \) (since \( F \propto \frac{1}{r^2} \)). Also, when one charge is halved, \( q_1 \rightarrow \frac{q_1}{2} \), so the force decreases by a factor of \( 2 \).
Step 3: Final force.
Thus, the final force will be: \[ F' = 4F \times \frac{1}{2} = 2F \] Conclusion: The force will become double the original force.