Three charges are placed as shown in the figu | vedclass

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(B) The net force on $q_3$ is in the negative $x-$ direction,which means the net electric field $\vec{E}$ at the position of $q_3$ must also be in the

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$\frac{27}{32}\, \mu C$

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(B) The net force on $q_3$ is in the negative $x-$ direction,which means the net electric field $\vec{E}$ at the position of $q_3$ must also be in the negative $x-$ direction. This implies that the vertical component of the electric field at $q_3$ must be zero.Let $	heta$ be the angle at $q_1$. From the geometry,$\sin 	heta = \frac{3}{5}$ and $\cos 	heta = \frac{4}{5}$.The electric field due to $q_1$ at $q_3$ is $E_1 = \frac{k|q_1|}{r_1^2}$ and due to $q_2$ is $E_2 = \frac{k|q_2|}{r_2^2}$.For the vertical component to be zero,the vertical components of $\vec{E}_1$ and $\vec{E}_2$ must cancel each other out.From the geometry,the angle between the line $q_1q_3$ and the horizontal is $	heta$. The angle between $q_2q_3$ and the horizontal is $90^\circ - 	heta$.Thus,$E_1 \sin 	heta = E_2 \cos(90^\circ - 	heta) = E_2 \sin 	heta$ is incorrect based on the geometry; rather,the vertical components are $E_1 \sin 	heta$ and $E_2 \cos 	heta$ (relative to the horizontal).Setting vertical components equal: $\frac{k|q_1|}{(4 	imes 10^{-2})^2} \sin 	heta = \frac{k|q_2|}{(3 	imes 10^{-2})^2} \cos 	heta$.Using $\sin 	heta = 3/5$ and $\cos 	heta = 4/5$,we get $\frac{|q_1|}{16} 	imes \frac{3}{5} = \frac{|q_2|}{9} 	imes \frac{4}{5}$.$|q_2| = |q_1| 	imes \frac{9}{16} 	imes \frac{3}{4} = 2.00\, \mu C 	imes \frac{27}{64} = \frac{27}{32}\, \mu C$.

Three charges are placed as shown in the figure. The magnitude of $q_1$ is $2.00\, \mu C$,but its sign and the value of the charge $q_2$ are not known. Charge $q_3$ is $+4.00\, \mu C$,and the net force on $q_3$ is entirely in the negative $x-$ direction. The magnitude of $q_2$ is

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