Question

Two point charges $+3\mu C$ and $+8\mu C$ repel each other with a force of $40N$. If a charge of $-5\mu C$ is added to each of them, the force between them will become:

• A. $-10N$
• B. $10N$
• C. $20N$
• D. $-20N$

Answer 1

The correct option is A $-10N$

Redistribution of charges takes place.
Charge $q_1=3\mu C$
Charge $q_2=8\mu C$
When third charge $q_3=-5\mu C$ is added to each, then new charges on $q_1$ and $q_2$ will be
$q_1^{\prime }=3-5=-2\mu C$
and, $q_2^{\prime }=8-5=3\mu C$
Now,
In first case,
$40=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{3\times 8}{r^2}$
In second case,
$F=\frac{1}{4\pi {\epsilon }_0}\times \frac{(-2\times 3)}{r^2}$
$\therefore \frac{F}{40}=\frac{-2\times 3}{3\times 8}$
or $F=-10N$