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Three charges  $$\mathrm{q}_{1}=3\ \mathrm{m}\mathrm{C},\mathrm{q}_{2}=-3\ \mathrm{m}\mathrm{C}$$ and $$q_3$$ are kept at the vertices of a triangle as shown in the figure. lf the net force acting on $$\mathrm{q}_{1}$$ is $$\overline{\mathrm{F}}$$, the charge $$\mathrm{q}_{3}$$ would have the magnitude $$($$1 +$$\displaystyle \frac{1}{\mathrm{n}})^{2}\mathrm{m}\mathrm{C}$$. So, $$n$$ is:

44237.jpg


A
8
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B
4
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C
1
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D
16
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Solution

The correct option is A $$8$$
Let $$\vec{F_{1}}$$ be the force on $$q_{1}$$ due to $$q_{2}$$ and $$\vec{F_{2}}$$ be that due to $$q_{3}$$
As, $$q_{1} > 0,q_{2} <0,$$ the charge $$q_{3}$$ should be + ve to get the net  force $$\vec{F}$$.
$$F_{1}=\dfrac{1}{4\pi \varepsilon _{0}} \dfrac{q_{1}q_{2}}{4^2},F_{2}=\dfrac{1}{4\pi \varepsilon _{0}}\dfrac{q_{1}q_{3}}{3^2}$$
From the figure, $$\tan\theta=\dfrac{F_2}{F_1}$$
$$\therefore \dfrac{3}{4}=\dfrac{q_{3}}{9}\times \dfrac{16}{q_{2}}$$
$$\therefore q_{3}=\dfrac{27}{64}\times q_{2}=\dfrac{81}{64}=\left ( \dfrac{9}{8} \right )^{2}mC$$
$$=\left ( 1+\dfrac{1}{8} \right )^{2}mC$$
Hence, $$n=8$$
122092_44237_ans.png

Physics

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