Q118 / 150 · MHT CET 2021 (24 Sep Shift 1) · Physics
Two, point charges $+3 \mu \mathrm{C}$ and $+8 \mu \mathrm{C}$ repel each other with a force of $40 \mathrm{~N}$. If a charge of $-5 \mu \mathrm{C}$ is added to each of them, then force between them will become
- Correct answerA$-10 \mathrm{~N}$
- B$10 \mathrm{~N}$
- C$20 \mathrm{~N}$
- D$-20 \mathrm{~N}$
Solution
$$
\mathrm{q}_1=3 \mu \mathrm{C} \text { and } \mathrm{q}_2=8 \mu \mathrm{C}
$$
When third charge $q_3=-5 \mu \mathrm{C}$ is added to each, then new charges on $\mathrm{q}_1$ and $\mathrm{q}_2$ will be
$$
\mathrm{q}_1=3-5=-2 \mu \mathrm{C}
$$
and $\mathrm{q}_2=8-5=3 \mu \mathrm{C}$
Now,
Case I: $40=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{3 \times 8}{r^2}$
Case II: $F=\frac{1}{4 \pi \varepsilon_0} \times \frac{(-2 \times 3)}{r^2}$
$$
\begin{aligned}
& \therefore \frac{F}{40}=\frac{-2 \times 3}{3 \times 8} \\
& \Rightarrow F=-10 \mathrm{~N}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.