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

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(A) Let the charges be placed at positions $x=0$,$x=\frac{l}{2}$,and $x=l$.The charge $q$ is at $x=0$,$Q$ is at $x=\frac{l}{2}$,and $4q$ is at $x=l$.T

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$-q$

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(A) Let the charges be placed at positions $x=0$,$x=\frac{l}{2}$,and $x=l$.The charge $q$ is at $x=0$,$Q$ is at $x=\frac{l}{2}$,and $4q$ is at $x=l$.The net force on charge $q$ is the sum of the electrostatic forces exerted by $Q$ and $4q$.According to Coulomb's Law,the force $F$ between two charges $q_1$ and $q_2$ separated by distance $r$ is $F = k \frac{q_1 q_2}{r^2}$,where $k = \frac{1}{4\pi\varepsilon_0}$.For the net force on $q$ to be zero:$F_{net} = F_{Qq} + F_{4qq} = 0$$k \frac{Qq}{(\frac{l}{2})^2} + k \frac{(4q)q}{l^2} = 0$$k \frac{Qq}{\frac{l^2}{4}} + k \frac{4q^2}{l^2} = 0$$4k \frac{Qq}{l^2} + 4k \frac{q^2}{l^2} = 0$Dividing by $\frac{4k}{l^2}$ (assuming $l 
eq 0$):$Qq + q^2 = 0$$Qq = -q^2$$Q = -q$

Three charges $q$,$Q$,and $4q$ are placed in a straight line of length $l$ at points distant $0$,$\frac{l}{2}$,and $l$ respectively from one end. In order to make the net force on $q$ zero,the charge $Q$ must be equal to

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