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. $\frac{q_2}{b^2}-\frac{q_3}{a^2}sin\theta$
• 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 $(-q_3)$ on $-q_1$
x-component of Net force on $-q_1$ is
$F_x=F_2+F_{3}sin\theta =k\frac{q_1{q}_2}{b^2}+k.\frac{q_1{q}_3}{a^2}sin\theta$
$\Rightarrow F_x=k[\frac{q_1{q}_2}{b^2}+\frac{q_1{q}_3}{a^2}sin\theta ]$
$\Rightarrow F_x=k.q_1[\frac{q_2}{b^2}+\frac{q_3}{a^2}sin\theta ]$