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Question

Two fixed point charges $$-q$$ and $$+2q$$ and are placed at a certain distance apart. Where should a third point charge be placed so that it is in equilibrium?


A
on the line joining the two charges and on the right of +2q
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B
on the line joining the two charges and on the left of q
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C
between q and +2q
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D
at any point on the right angle bisector of the line joining q and +2q 
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Solution

The correct option is A on the line joining the two charges and on the left of $$-q$$
The force on a charge $$Q$$ due to another charge $$q$$ at a distance $$d$$ is given by: $$F=\dfrac{1}{4\pi\epsilon_0}\dfrac{Qq}{d^2}$$
The force is attracting for charges of opposite signs and repelling for same sign.
Since the force is proportional to the charge of other particle, and inversely proportional to square of distance, under equilibrium,
$$\dfrac{1}{4\pi \epsilon_0}\dfrac{Q(-q)}{d_1^2}+\dfrac{1}{4\pi \epsilon_0}\dfrac{Q(+2q)}{d_2^2}=0$$
$$\implies d_2=\sqrt{2}d_1$$
The charge $$Q$$ cannot be between the charges, because otherwise both $$-q$$ and $$+2q$$ would exert forces in same direction.
Hence, the charge lies on line joining two charges and on left of $$-q$$ because the distance of $$Q$$ from $$-q$$, $$d_1$$ is lesser than that from $$+2q$$,  $$d_2$$. 

Physics

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