Question

Two point charges M and N having charges +q and -q respectively are placed at a distance apart. Force acting between them is F. If 30% of charge of N is transferred to M, then the force between the charges becomes:

• A. \( \frac{100}{49} F \)
• B. \( \frac{49}{100} F \)
• C. \( \frac{49}{16} F \)
• D. \( \frac{9}{16} F \)

Hint:

When transferring charge between two point charges, the resulting force can be calculated by adjusting the magnitude of the charges according to Coulomb’s law.

Answer 1

The Correct Option is C

The force between two point charges is given by Coulomb's Law: \[ F = k \frac{|q_1 q_2|}{r^2} \] Where: 
- \( F \) is the force between the charges, 
- \( k \) is Coulomb's constant, - \( q_1 \) and \( q_2 \) are the magnitudes of the two charges, 
- \( r \) is the distance between the charges. Initially, the charges are \( +q \) and \( -q \). 
The force between them is: \[ F = k \frac{|q \times (-q)|}{r^2} = k \frac{q^2}{r^2} \] 
Now, if 30% of the charge from \( N \) (which is \( -q \)) is transferred to \( M \), the new charge on \( N \) becomes \( -0.7q \) and the new charge on \( M \) becomes \( +1.3q \). The new force \( F' \) between the charges is: \[ F' = k \frac{|(1.3q) \times (-0.7q)|}{r^2} = k \frac{(1.3 \times 0.7) q^2}{r^2} \] Thus: \[ F' = k \frac{0.91 q^2}{r^2} \] Comparing the new force \( F' \) with the initial force \( F \), we have: \[ F' = \frac{0.91}{1} F = \frac{49}{16} F \] 
Thus, the force between the charges becomes \( \frac{49}{16} F \).