Two point charges \(+q_1\) and \(q_2\) repel each other with a force of \(100 \mathrm{~N} \cdot \mathrm{q}_1\) is increased by \(10 \%\) and \(\mathrm{q}_2\) is decreased by \(10 \%\). If they are kept at their original positions the change in the force of repulsion between them is

\(\mathrm{F}_{\text {net }}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}^2}\)
\(\mathrm{q}_1\) and \(\mathrm{q}_2\) are increased and decreased respectively by \(10 \%\)
\(\begin{aligned}
& \mathrm{F}_{\text {nett }}=\frac{1}{4 \pi \varepsilon_0} \frac{1}{\mathrm{r}^2}\left(\mathrm{q}_1+\mathrm{q}_1 \frac{10}{100}\right)\left(\mathrm{q}_2-\mathrm{q}_2 \frac{10}{100}\right) \\
\quad & \mathrm{F}_{\text {net' }}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}^2}\left(\frac{110}{1.00}\right)\left(\frac{90}{100}\right) \\
& \text { i.e., } \frac{99}{100} \operatorname{times} \mathrm{~F}_{\text {net }} \\
\therefore \quad & \mathrm{F}_{\text {net' }}=\frac{99}{100} \times \mathrm{F}_{\text {net }}=\frac{99}{100} \times 100=99 \mathrm{~N} \\
& \mathrm{~F}_{\text {net }}-\mathrm{F}_{\text {net }}=100-99=1 \mathrm{~N}
\end{aligned}\)
\(\therefore \quad\) The net force decreases by 1 N