Question

If electrical force between two charges is $200 \, N$ 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
$F=\frac{1}{4 \pi \epsilon _{0}}\cdot \frac{q_{1} q_{2}}{r^{2}}=200 \, N$ ..... (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, the electrical force between them,
$F^{′}=\frac{1}{4 \pi \epsilon _{0}}\times \frac{q_{1}^{′} q_{2}^{′}}{r^{2}}$
$F^{′}=\frac{\frac{1}{4 \pi \epsilon _{0}} \times \frac{110}{100} q_{1} \times \frac{90}{100} q_{2}}{r^{2}}$ .... (ii)
From equations (i) and (ii), we get
$F^{′}=200\times \frac{99}{100}\Rightarrow F^{′}=198 \, N$