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(A) Let the two unit negative charges be $q_1 = -1 \ C$ and $q_2 = -1 \ C$,separated by a distance $d$.For the system to be in equilibrium,the net for

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$0.25$

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(A) Let the two unit negative charges be $q_1 = -1 \ C$ and $q_2 = -1 \ C$,separated by a distance $d$.For the system to be in equilibrium,the net force on each charge must be zero.Consider the equilibrium of charge $q_1$ at point $A$. The force exerted by $q$ at $B$ and $q_2$ at $C$ on $q_1$ must sum to zero:$F_{AB} + F_{AC} = 0$$\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q}{(d/2)^2} + \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{d^2} = 0$Dividing by $\frac{q_1}{4 \pi \varepsilon_0}$,we get:$\frac{q}{(d/2)^2} + \frac{q_2}{d^2} = 0$$\frac{q}{d^2/4} + \frac{-1}{d^2} = 0$$\frac{4q}{d^2} = \frac{1}{d^2}$$4q = 1$$q = 0.25 \ C$

Two unit negative charges are placed on a straight line. $A$ positive charge $q$ is placed exactly at the midpoint between these unit charges. If the system of these three charges is in equilibrium,the value of $q$ (in $C$) is

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