Question

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

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

Answer 1

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

$F_2=$ Force applied by $q_2$ on ${-q}_1$

$F_3=$ Force applied by $\left({-q}_3\right)$ on $-q_1$

$X-$ component of net force on $-q_1$ is

$F_x=F_2+F_3\sin \theta$

$=K\frac{q_1{q}_2}{b^2}+K\frac{q_1{q}_3}{a^2}sin\theta$

$\Rightarrow F_x=K[\frac{q_1{q}_2}{b^2}+\frac{-q_1{q}_3}{a^2}\sin \theta ]$

$\Rightarrow F_x=K{q}_1[\frac{q_2}{b^2}+\frac{q_3}{a^2}\sin \theta ]$

or, $(\frac{q_2}{b^2}+\frac{q_3}{a^2}\sin \theta )$