Three charges q, Q and +4q are placed in a st | vedclass

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(C) Let the charges be placed at positions $x=0$,$x=\frac{d}{2}$,and $x=d$ respectively.To make the net force on the charge $q$ at $x=0$ equal to zero

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$-q$

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(C) Let the charges be placed at positions $x=0$,$x=\frac{d}{2}$,and $x=d$ respectively.To make the net force on the charge $q$ at $x=0$ equal to zero,the force exerted by $Q$ and the force exerted by $+4q$ must be equal in magnitude and opposite in direction.Using Coulomb's law,the force exerted by $Q$ on $q$ is $F_Q = \frac{k q Q}{(d/2)^2}$ and the force exerted by $+4q$ on $q$ is $F_{+4q} = \frac{k q (4q)}{d^2}$.For the net force to be zero,the sum of these forces must be zero:$\frac{k q Q}{(d/2)^2} + \frac{k q (4q)}{d^2} = 0$$\frac{k q Q}{d^2/4} + \frac{4 k q^2}{d^2} = 0$$\frac{4 k q Q}{d^2} + \frac{4 k q^2}{d^2} = 0$Dividing by $\frac{4 k q}{d^2}$ (assuming $q 
eq 0$):$Q + q = 0$$Q = -q$

Three charges $q, Q$ and $+4q$ are placed in a straight line of length $d$ at points at distances $0, \frac{d}{2}$ and $d$ respectively. In order to make the net force on $q$ to be zero,the value of $Q$ should be

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