Question

Two point charges $A$ and $B$, having charges $+ Q$ and $- Q$ respectively, are placed at certain distance apart and force acting between them is $F$. If $25\%$ charge of $A$ is transferred to $B$. then force between the charges become

• A. $\frac{16F}{9}$
• B. $\frac{4F}{3}$
• C. F
• D. $\frac{9F}{16}$

Answer 1

The Correct Option is D

To solve this problem, let's analyze the situation before and after 25% of the charge from charge \(A\) is transferred to charge \(B\).

1. Initial Situation: The initial charges on points \(A\) and \(B\) are \(+Q\) and \(-Q\), respectively. According to Coulomb's Law, the force \(F\) between two point charges is given by: F = \frac{k \cdot |Q_1 \cdot Q_2|}{r^2} Here, \(Q_1 = +Q\), \(Q_2 = -Q\), and the distance \(r\) between charges is constant. Therefore, the initial force is: F = \frac{k \cdot Q^2}{r^2}
2. Charge Transfer: When 25% of charge \(Q\) from \(A\) is transferred to \(B\):
   • Charge on \(A\) becomes \(+0.75Q\).
   • Charge on \(B\) becomes \(-Q + 0.25Q = -0.75Q\).
3. New Force Calculation: Using Coulomb's Law again for the new configuration: F' = \frac{k \cdot |0.75Q \cdot (-0.75Q)|}{r^2} F' = \frac{k \cdot 0.5625Q^2}{r^2} Given the initial force: F = \frac{k \cdot Q^2}{r^2} Comparing both, \(F'\) becomes: F' = 0.5625F
4. Simplification: Converting 0.5625 into fraction form gives us: F' = \frac{9}{16}F
5. Conclusion: After transferring 25% of the charge from \(A\) to \(B\), the force between the charges becomes: \frac{9F}{16}

Thus, the correct answer is: \frac{9F}{16}