Question

Two point charges repel each other with a force of \(100 \mathrm{~N}\). One of the charges is increased by \(10 \%\) and other is reduced by \(10 \%\). The new force of repulsion at the same distance would be \(\ldots \ldots \mathrm{N}\). (A) 121 (B) 100 (C) 99 (D) 89

Short answer

The new force of repulsion at the same distance after altering the charges is \(99 N\), which corresponds to the answer (C).

Step-by-step solution

Write down the formula for Coulomb's law

The formula for the force between two point charged objects at distance r can be given by Coulomb's Law: \[F = k \frac{q_1 q_2}{r^2}\] Where \(F\) is the force, \(k\) is the Coulomb's constant (\(8.99 \times 10^9 N m^2 C^{-2}\)), \(q_1\) and \(q_2\) are the charges, and \(r\) is the distance between them.

Find the relation between initial charges and force

Let the initial charges be \(q_1\) and \(q_2\), and the force between them is 100 N. According to Coulomb's law, the initial force can be written as: \[F_1 = k \frac{q_1 q_2}{r^2} = 100 N\]

Modify the charges and write the new force equation

Now, we increase one charge by 10% and decrease the other charge by 10%. The new charges will be: \[q_1' = 1.1q_1\] \[q_2' = 0.9q_2\] Now, we can write the equation for the new force between the charges at the same distance \(r\): \[F_2 = k \frac{q_1' q_2'}{r^2}\]

Calculate the new force

Replacing the new charges in the above equation, we get: \[F_2 = k \frac{1.1q_1 * 0.9q_2}{r^2}\] \[F_2 = k \frac{0.99q_1 q_2}{r^2}\] We know that the initial force equation is \(F_1 = k \frac{q_1 q_2}{r^2} = 100 N\). Thus, we can write the new force equation in terms of the initial force equation: \[F_2 = 0.99F_1\] \[F_2 = 0.99(100 N)\]

Find the new force value

Calculating the new force, we get the final value: \[F_2 = 99 N\] So, the new force of repulsion at the same distance after altering charges is 99 N, which corresponds to the answer (C).