Question

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

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

Answer 1

The correct option is A -10 N

In second case, charges will be $-2\mu C$ and $+3\mu C$
Since $F\propto Q_1{Q}_{2}i.e.\frac{F}{F^{\prime }}=\frac{Q_1{Q}_2}{Q_1^{\prime }{Q}_2^{\prime }}$
$\therefore \frac{40}{F^{\prime }}=\frac{3\times 8}{-2\times 3}=-4\Rightarrow F^{\prime }=10N$ (Attractive)