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(B) Step $1$: Equilibrium of the central charge $Q$.For the system to be in equilibrium,the net force on each charge must be zero.By symmetry,the two

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$-Q/4$

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(B) Step $1$: Equilibrium of the central charge $Q$.For the system to be in equilibrium,the net force on each charge must be zero.By symmetry,the two outer charges $q$ exert equal and opposite forces on $Q$,which cancel each other out. Thus,the net force on $Q$ is zero regardless of the value of $q$.Step $2$: Equilibrium of the outer charge $q$.Let the distance between the two charges $q$ be $2x$. The distance from $q$ to $Q$ is $x$.The net force on the charge $q$ at the right must be zero:$F = \frac{kQq}{x^2} + \frac{kq^2}{(2x)^2} = 0$Dividing by $k/x^2$ (assuming $x 
eq 0$):$Qq + \frac{q^2}{4} = 0$$Qq = -\frac{q^2}{4}$$Q = -\frac{q}{4}$Therefore,$q = -4Q$.

$A$ charge $Q$ is placed at the midpoint of the line joining two identical point charges $q$. For what value of $q$ will the system be in equilibrium?

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