MHT CET · Physics · Electrostatics
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
- A \(-10 \mathrm{~N}\)
- B \(10 \mathrm{~N}\)
- C \(20 \mathrm{~N}\)
- D \(-20 \mathrm{~N}\)
Answer & Solution
Correct Answer
(A) \(-10 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
\(
\mathrm{q}_1=3 \mu \mathrm{C} \text { and } \mathrm{q}_2=8 \mu \mathrm{C}
\)
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: \(F=\frac{1}{4 \pi \varepsilon_0} \times \frac{(-2 \times 3)}{r^2}\)
\(
\begin{aligned}
& \therefore \frac{F}{40}=\frac{-2 \times 3}{3 \times 8} \\
& \Rightarrow F=-10 \mathrm{~N}
\end{aligned}
\)
\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: \(F=\frac{1}{4 \pi \varepsilon_0} \times \frac{(-2 \times 3)}{r^2}\)
\(
\begin{aligned}
& \therefore \frac{F}{40}=\frac{-2 \times 3}{3 \times 8} \\
& \Rightarrow F=-10 \mathrm{~N}
\end{aligned}
\)
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