Question

Two point charges $ 3\times {{10}^{-6}}C $ and $ 8\times {{10}^{-6}}C $ repel each other by a force of $ 6\times {{10}^{-3}}N $ . If each of them is given an additional charge $ -6\times {{10}^{-6}}C $ , the force between them will be:

• A. $ 2.4\times {{10}^{-3}}N $ (repulsive)
• B. $ 2.4\times {{10}^{-3}}N $ (attractive)
• C. $ 1.5\times {{10}^{-3}}N $ (repulsive)
• D. $ 1.5\times {{10}^{-3}}N $ (attractive)

Answer 1

The Correct Option is D

Like charges repel each other while unlike charges attract each other.
From Coulombs law, the force of attraction/repulsion between two point charges $q_{1}$ and $q_{2}$ placed a distance $r$ apart is given by
$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}} N$
When similar charges are taken
$q_{1}=3 \times 10^{-6} C , q_{2}=8 \times 10^{-6} C$
$F =\frac{1}{4 \pi \varepsilon_{0}} \frac{\left(3 \times 10^{-6}\right) \times\left(8 \times 10^{-6}\right)}{r^{2}} ?$ (i) (repulsive)
When additional charge $-6 \times 10^{-6} C$ is given to each charge, then
$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{(3-6) \times 10^{-6} \times(8-6) \times 10^{-6}}{r^{2}}$ (attractive)
$\therefore F=\frac{1}{4 \pi \varepsilon_{0}} \frac{(-3) \times 10^{-6} \times 2 \times 10^{-6}}{r^{2}} N ?$ (ii)
Dividing E (ii) by E (i), we get
$\frac{F}{F}=-\frac{6}{24} \Rightarrow F=-\frac{F}{4}$
$=-\frac{6 \times 10^{-3}}{4}=-1.5 \times 10^{-3} N$
Negative sign indicates, force is attractive.