Question

If electrical force between two charges is $\text{2}00\text{ N}$ and we increase $\text{1}0$ charge on one of the charges and decrease $\text{1}0$ charge on the other, then electrical force between them for the same distance becomes

• A. $\text{2}00\text{ N}$
• B. 202 N
• C. 198 N
• D. 199 N

Answer 1

Answer: (C)

Let two charges are $q_1$ and $q_2$ and r is the distance between them, then electrical force $F=\frac{1}{4\pi {\epsilon }_0}.\frac{q_1{q}_2}{r^2}=200N$ ? (i) If $q_1$ is increased by 10%, then $q_1^{{}^{\prime }}=\frac{110}{100}{q}_1$ and $q_2$ is decreased by 10%, then $q_2^{{}^{\prime }}=\frac{90}{100}{q}_2$ Then, electrical force between them, $F^{\prime }=\frac{1}{4\pi {\epsilon }_0}\times \frac{q_1^{{}^{\prime }}{q}_2^{{}^{\prime }}}{r^2}$ $F^{\prime }=\frac{\frac{1}{4\pi {\epsilon }_0}\times \frac{110}{100}{q}_1\times \frac{90}{100}{q}_2}{r^2}$ ? (ii) From Eqs. (i) and (ii), we get $F^{\prime }=200\times \frac{99}{100}\Rightarrow F^{\prime }=198N$