Question

Three charges $-q_1,+q_2$ and $-q_3$ are placed as shown in Fig. 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 $\frac{q_2}{b^2}+\frac{q_3}{a^2}\sin \theta$


Net force on charge $-q_1$ due to other two charges is
${\overrightarrow{F}}_1={\overrightarrow{F}}_{12}+{\overrightarrow{F}}_{13}$
${\overrightarrow{F}}_1=\frac{k{q}_1{q}_2}{b^2}\hat{i}+\frac{k{q}_1{q}_3}{a^2}(\sin \theta \hat{i}+\cos \theta \hat{j})$
Clearly, the x-component of total force is proportional to $\frac{q_2}{b^2}+\frac{q_3}{a^2}\sin \theta$