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

Answer: (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
$F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}N$
When similar charges are taken
$q_1=3\times 10^{-6}C,q_2=8\times 10^{-6}C$
$F=\frac{1}{4\pi {\epsilon }_0}\frac{\left(3\times 10^{-6}\right)\times \left(8\times 10^{-6}\right)}{r^2}?$ (i) (repulsive)
When additional charge $-6\times 10^{-6}C$ is given to each charge, then
$F=\frac{1}{4\pi {\epsilon }_0}\frac{(3-6)\times 10^{-6}\times (8-6)\times 10^{-6}}{r^2}$ (attractive)
$\therefore F=\frac{1}{4\pi {\epsilon }_0}\frac{(-3)\times 10^{-6}\times 2\times 10^{-6}}{r^2}N?$ (ii)
Dividing Eq. (ii) by Eq. (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.