Question

If electrical force between two charges is $200N$ and we increase $10\%$ charge on one of the charges and decrease $10\%$ charge on the other, then the electrical force between them for the same distance becomes

• A. $200N$
• B. $202N$
• C. $198N$
• D. $199N$

Answer 1

Answer: (C)

Let two charges are $q_{1}and{q}_2$ and r is the distance between them, then electrical force
$F=\frac{1}{4\pi {ϵ}_0}\cdot \frac{q_1{q}_2}{r^2}=200N$ ..... (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, the electrical force between them,
$F^{\prime }=\frac{1}{4\pi {ϵ}_0}\times \frac{q_1^{\prime }{q}_2^{\prime }}{r^2}$
$F^{\prime }=\frac{\frac{1}{4\pi {ϵ}_0}\times \frac{110}{100}{q}_1\times \frac{90}{100}{q}_2}{r^2}$ .... (ii)
From equations (i) and (ii), we get
$F^{\prime }=200\times \frac{99}{100}\Rightarrow F^{\prime }=198N$