Question

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 :

• A. $\frac{q_2}{b^2}-\frac{q_3}{a^2}cos\theta$
• B. $\frac{q_2}{b^2}+\frac{q_3}{a^2}sin\theta$
• C. $\frac{q_2}{b^2}+\frac{q_3}{a^2}cos\theta$
• D. $\frac{q_2}{b^2}-\frac{q_3}{a^2}sin\theta$

Answer 1

The correct option is B $\frac{q_2}{b^2}+\frac{q_3}{a^2}sin\theta$

Force applied by $+q_2$ on $-q_1$ is attractive and in $+X$ direction and equals :$F_{-q1q2}=k\frac{q_1{q}_2}{b^2}$ (using Coulomb,s law)

Now force on $-q_2$ due to $-q_3$ is repulsive and has $X$ and $Y$ components as :

$\left(F_{-q1-q3}\right)=k\frac{q_1{q}_3}{a^2}\sin \theta$

$(F-q1-q3)=-k\frac{q_1{q}_3}{a^2}\cos \theta$

So net force on $-q_1$ in $+X$ direction is :

$F_x=k{q}_1(\frac{q_2}{b^2}+\frac{q_3}{a^2}\sin \theta )$