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).