Three charges +5 q, Q and -2 q are kept along | vedclass

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(A) Let the positions of the charges be $x_1 = 0$ for $+5q$,$x_2 = 2r/3$ for $Q$,and $x_3 = r$ for $-2q$. The distance between $+5q$ and $-2q$ is $r$,

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$+\frac{5}{9} q$

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(A) Let the positions of the charges be $x_1 = 0$ for $+5q$,$x_2 = 2r/3$ for $Q$,and $x_3 = r$ for $-2q$. The distance between $+5q$ and $-2q$ is $r$,and the distance between $Q$ and $-2q$ is $r/3$.For the net force on charge $-2q$ to be zero,the forces exerted by $+5q$ and $Q$ on $-2q$ must be equal in magnitude and opposite in direction.Let $F_1$ be the force due to $+5q$ on $-2q$ and $F_2$ be the force due to $Q$ on $-2q$.$F_1 = \frac{k |5q| |-2q|}{r^2} = \frac{10kq^2}{r^2}$ (Attractive,towards left).For the net force to be zero,$F_2$ must be repulsive (towards right),so $Q$ must be positive.$F_2 = \frac{k |Q| |-2q|}{(r/3)^2} = \frac{k |Q| 2q}{r^2/9} = \frac{18k|Q|q}{r^2}$.Equating the magnitudes: $\frac{10kq^2}{r^2} = \frac{18kQq}{r^2}$.$10q = 18Q \Rightarrow Q = \frac{10}{18}q = \frac{5}{9}q$.

Three charges $+5 q, Q$ and $-2 q$ are kept along a straight line in the same order such that $+5 q$ and $-2 q$ charges are at a distance of $\frac{2 r}{3}$ and $\frac{r}{3}$ from the charge $Q$ respectively. If the net force on the charge $-2 q$ is zero,then $Q$ is

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