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(B) Let the two charges $Q$ be at points $A$ and $B$,separated by a distance $x$. The charge $q$ is placed at point $C$,the midpoint of $AB$,so $AC =

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$ - \frac{Q}{4} $

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(B) Let the two charges $Q$ be at points $A$ and $B$,separated by a distance $x$. The charge $q$ is placed at point $C$,the midpoint of $AB$,so $AC = CB = x/2$.For the system to be in equilibrium,the net force on each charge must be zero.Consider the equilibrium of charge $Q$ at point $A$. The force exerted by charge $Q$ at $B$ on $A$ is $F_{BA} = \frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{x^2}$ (repulsive).The force exerted by charge $q$ at $C$ on $A$ is $F_{CA} = \frac{1}{4 \pi \varepsilon_0} \frac{qQ}{(x/2)^2}$ (attractive,so $q$ must be negative).For equilibrium,$F_{BA} + F_{CA} = 0$,so $\frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{x^2} + \frac{1}{4 \pi \varepsilon_0} \frac{qQ}{x^2/4} = 0$.$\frac{Q^2}{x^2} + \frac{4qQ}{x^2} = 0 \Rightarrow Q + 4q = 0 \Rightarrow q = -\frac{Q}{4}$.

$A$ charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the three charges will be in equilibrium if $q$ is equal to:

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