Question

If electrical force between two charges is 200 N and we increase 10% charge on one of the charges and decrease 10% charge on the other, then electrical force between them for the same distance becomes

• A. 200 N
• B. 202 N
• C. 198 N
• D. 19 N

Hint:

When the charges are increased or decreased by a certain percentage, the force changes by the product of the factors corresponding to the charges' changes.

Answer 1

The Correct Option is C

The electrical 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.
If the charges are increased and decreased by 10%, we can calculate the new force by considering the change in charges. Let the initial charges be \( q_1 \) and \( q_2 \), and the initial force is \( F = 200 \, \text{N} \).
When one charge is increased by 10% and the other is decreased by 10%, the new charges become \( 1.1 q_1 \) and \( 0.9 q_2 \).
The new force is: \[ F_{\text{new}} = k \frac{(1.1 q_1)(0.9 q_2)}{r^2} = 1.1 \times 0.9 \times F = 0.99 \times 200 = 198 \, \text{N} \]
Thus, the correct answer is (c).