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. +20 N
• C. -20 N
• D. -10 N

Answer 1

Answer: (D)

Two charges $+3\mu C$ and $8\mu C$ are given which repel each other when a charge $-5\mu C$ is added then in second case changes becomes $-2\mu C$ and $3\mu C$ . From the relation $F\propto q_1{q}_2$ Hence $\frac{F_1}{F_1}=\frac{q_1\times q_2}{q_1{q}_2}$ or $\frac{40}{F_1}=\frac{3\times 8}{-2\times 3}=-4$ so, $F_1=-\frac{40}{4}=10\text{attractive}$