Electrical force between two point charges is 200 N . If 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 _________.

(C)
Suppose \(q_1\) and \(q_2\) are two point charges. They are separated by distance \(r\).
\(F =\frac{k q_1 q_2}{r^2}=200 N\)
→ charge \(q_1\) increases by \(10 \%\)
\(\begin{aligned} \therefore q_1^{\prime} & =q_1+10 \% q_1 \\ q_1 & =\frac{110}{100} q_1=1.1 q_1\end{aligned}\)
→ and charge \(q_2\) decreases by \(10 \%\)
\(\begin{aligned} \therefore q_2^{\prime} & =q_2-10 \% q_2 \\ & =\frac{90}{100} \quad q_2=0.9 q_2\end{aligned}\)
New electrical force
\(\begin{array}{l} F ^{\prime}=\frac{k q_1^{\prime} q_2^{\prime}}{r^2} \\ F^{\prime}=\frac{k \times \frac{110}{100} q_1 \times \frac{90}{100} q_2}{r^2} \\ F^{\prime}=\frac{k q_1 q_2}{r^2} \times \frac{110}{100} \times \frac{90}{100}\end{array}\)
\(\begin{array}{l}\therefore F^{\prime}=200 \times \frac{110}{100} \times \frac{90}{100} \\ \therefore F^{\prime}=198 N\end{array}\)