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(B) Let the charge $4q$ be at position $x=0$ and charge $q$ be at position $x=l$. The charge $Q$ is placed at $x=l/2$.For the net force on charge $q$

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$-q$

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(B) Let the charge $4q$ be at position $x=0$ and charge $q$ be at position $x=l$. The charge $Q$ is placed at $x=l/2$.For the net force on charge $q$ to be zero,the force exerted by $4q$ on $q$ must be equal and opposite to the force exerted by $Q$ on $q$.The force exerted by $4q$ on $q$ is $F_1 = \frac{k(4q)(q)}{l^2}$.The force exerted by $Q$ on $q$ is $F_2 = \frac{k(Q)(q)}{(l/2)^2}$.For the net force to be zero,$F_1 + F_2 = 0$,which implies $F_1 = -F_2$.$\frac{k(4q)(q)}{l^2} = -\frac{k(Q)(q)}{(l/2)^2}$.$\frac{4q}{l^2} = -\frac{Q}{l^2/4}$.$4q = -4Q$.Therefore,$Q = -q$.

Two electric charges $4q$ and $q$ are placed at a distance $l$ apart. $A$ third charge $Q$ is placed between them at the midpoint. If the resultant force on charge $q$ is zero,then the value of $Q$ is:

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