Three charges are placed as shown in the figu | vedclass

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(C) Let the position of $q_1$ be at the origin $(0,0)$. The position of $q_3$ is $(4\cos	heta, 4\sin	heta)$ and $q_2$ is at $(5,0)$.For the net forc

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(C) Let the position of $q_1$ be at the origin $(0,0)$. The position of $q_3$ is $(4\cos	heta, 4\sin	heta)$ and $q_2$ is at $(5,0)$.For the net force on $q_3$ to be in the negative $x-$ direction,the $y-$ components of the forces exerted by $q_1$ and $q_2$ on $q_3$ must cancel each other out.Let $\vec{F}_{13}$ be the force on $q_3$ due to $q_1$ and $\vec{F}_{23}$ be the force on $q_3$ due to $q_2$.If $q_1$ is negative,it attracts $q_3$ towards itself. The force $\vec{F}_{13}$ is directed along the line joining $q_3$ and $q_1$ towards $q_1$. This force has a positive $y-$ component.To cancel this,the force $\vec{F}_{23}$ due to $q_2$ must have a negative $y-$ component. This means $q_2$ must attract $q_3$ towards itself. Since $q_3$ is positive,$q_2$ must be negative.However,looking at the geometry,if $q_1$ is negative,the force $\vec{F}_{13}$ is attractive (towards $q_1$). If $q_2$ is positive,the force $\vec{F}_{23}$ is repulsive (away from $q_2$).By analyzing the vector components,for the net force to be purely in the negative $x-$ direction,the $y-$ components must satisfy $F_{13,y} + F_{23,y} = 0$.Given the geometry,$q_1$ must be negative (attracting $q_3$) and $q_2$ must be positive (repelling $q_3$) to create a resultant force in the negative $x-$ direction.

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