Question

Two identical balls having like charges placed at a certain distance apart, repel each other with a force \( F \). When they are brought in contact and then moved apart to a distance equal to half of their initial separation, the force of repulsion between them becomes \( 4.5F \). The ratio of the initial charges of the balls is:

• A. \( 2:1 \)
• B. \( 3:1 \)
• C. \( 4:1 \)
• D. \( 6:1 \)

Hint:

When charges are brought into contact, they redistribute equally. The force of repulsion is inversely proportional to the square of the distance, so halving the distance increases the force by a factor of 4.

Answer 1

The Correct Option is A

Step 1:
The electrostatic force between two charges is given by Coulomb's law: \[ F = k \frac{q_1 q_2}{r^2} \] where \( q_1 \) and \( q_2 \) are the charges, \( r \) is the distance between them, and \( k \) is Coulomb's constant.
Step 2:
Initially, the force between the two charges is \( F = k \frac{q_1 q_2}{r^2} \).
Step 3:
After the balls are brought into contact, the charges redistribute equally, so: \[ q_1 = q_2 = q \] \[ F' = k \frac{q^2}{\left(\frac{r}{2}\right)^2} = 4k \frac{q^2}{r^2} \] \[ F' = 4.5F \]
Step 4:
From the equation, we find that: \[ 4 \cdot F = 4.5F \quad \Rightarrow \quad q_1 = 2q_2 \]
Step 5:
The ratio of charges is \( 2:1 \).