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)

-10 N

Answer 2

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, ${{F}_{1}}=-\frac{40}{4}=10\,\,\text{attractive}$