Question

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

• A. 198 N
• B. 100 N
• C. 200 N
• D. 99 N

Answer 1

Answer: (A)

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' = $\frac{1}{4\pi {\epsilon }_0}\frac{q_1^{\prime }{q}_2^{\prime }}{r^2}$
$\Rightarrow F^{\prime }=\frac{1}{4\pi {\epsilon }_0}\frac{\frac{110}{100}{q}_1\times \frac{99}{100}{q}_2}{r^2}$ $\Rightarrow F^{\prime }=\frac{1}{4\pi {\epsilon }_0}.\frac{q_1{q}_2}{r^2}\times \frac{99}{100}$ ..(ii)
From Eqs. (i) and (ii)
F'= 200 $\times \frac{99}{100}\Rightarrow F^{\prime }=198$ N