Question

Two point charges repel each other with a force of $100\text{N}$. One of the charges is increased by $10\%$ and the other is reduced by $10\%$. The new force of repulsion at the same distance would be (in newtons)

Answer 1

Given that,
Repulsion between the charges, $F_e=100\text{N}$
Let the point charges be $q_1$ and $q_2$ and $r$ is distance between the charges.

The Coulombic force of repulsion is given by,
$F_e=\frac{k{q}_1{q}_2}{r^2}=100\text{N}\dots \dots \left(i\right)$
Case I: After one of the charge is increases by $10\%$,
$q_1^{\prime }=q_1+\frac{q_1\times 10}{100}=1.1q_1$

Case II: another charge is reduced by $10\%$,
$q_2^{\prime }=q_2-\frac{q_2\times 10}{100}=0.9q_2$
Thus, the new force of repulsion is,
$F_e^{{}^{\prime }}=\frac{k\left(1.1q_1\right)\left(0.9q_2\right)}{r^2}$
$\Rightarrow F_e^{{}^{\prime }}=\frac{k{q}_1{q}_2}{r^2}\times (1.1\times 0.9)$
Substituting from eq. (i),
$F_e^{{}^{\prime }}=100\times 1.1\times 0.9$
$\therefore F_e^{{}^{\prime }}=99\text{N}$

Accepted answers: $99,99.0,99.00$