Three charges -q_1,+q_2,and -q_3 are placed a | vedclass

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(C) Let $F_2$ be the force applied by $+q_2$ on $-q_1$. Since they have opposite signs,the force is attractive and acts along the positive $x$-axis. T

Vedclass · Accepted Answer

$\frac{q_2}{b^2} + \frac{q_3}{a^2} \sin 	heta$

Vedclass · Answer

(C) Let $F_2$ be the force applied by $+q_2$ on $-q_1$. Since they have opposite signs,the force is attractive and acts along the positive $x$-axis. Thus,$F_2 = k \frac{q_1 q_2}{b^2}$.Let $F_3$ be the force applied by $-q_3$ on $-q_1$. Since they have the same sign,the force is repulsive. The force $F_3$ acts along the line joining them,making an angle $	heta$ with the negative $y$-axis. The magnitude is $F_3 = k \frac{q_1 q_3}{a^2}$.The $x$-component of $F_3$ is $F_{3x} = F_3 \sin 	heta = k \frac{q_1 q_3}{a^2} \sin 	heta$,acting in the positive $x$-direction.The net $x$-component of the force on $-q_1$ is $F_x = F_2 + F_{3x} = k \frac{q_1 q_2}{b^2} + k \frac{q_1 q_3}{a^2} \sin 	heta$.Therefore,$F_x = k q_1 \left( \frac{q_2}{b^2} + \frac{q_3}{a^2} \sin 	heta ight)$.Thus,$F_x \propto \left( \frac{q_2}{b^2} + \frac{q_3}{a^2} \sin 	heta ight)$.

Three charges $-q_1$,$+q_2$,and $-q_3$ are placed as shown in the figure. The $x$-component of the force on $-q_1$ is proportional to

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