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)

198 N

Answer 2

Let two charges are $q_1$ and $q_2$ and r is the distance between them. Then electrical force
F = $\frac{ 1}{ 4 \pi \varepsilon_0 } . \frac{ q_1 q_2}{ r^2 } = 200 \, N$ \hspace25mm ..(i)
If $q_1$ is increased by 10%, then
$q_1' = \frac{ 110}{ 100} q_1$
and $q_2$ is decreased by 10%, then
$q_2' = \frac{ 90}{ 100} q_2$
Then electrical force between them
F' = $\frac{ 1}{ 4 \pi \varepsilon_0 } \frac{ q_1' q_2'}{ r^2 }$
$\Rightarrow F' = \frac{ 1}{ 4 \pi \varepsilon_0 } \frac{ \frac{110}{ 100} q_1 \times \frac{99}{ 100} q_2 }{r^2 }$
$\Rightarrow F' = \frac{ 1}{ 4 \pi \varepsilon_0 } . \frac{ q_1 q_2 }{ r^2 } \times \frac{99}{100}$ \hspace25mm ..(ii)
From Eqs. (i) and (ii)
F'= 200 $\times \frac{99}{100} \Rightarrow F' = 198$ N