question_answer

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                  AIEEE  Solved  Paper-2003

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)       \[\frac{{{q}_{2}}}{{{b}^{2}}}-\frac{{{q}_{3}}}{{{a}^{2}}}\sin \theta \]

Correct Answer: B

Solution :

In case of system of charges, for the force on an individual charge. We consider force on it due to all the remaining charges. Also, to find the net force in any direction we resolve all the individual force into their components i.e., along x-axis, y-axis and 2'-axis and add it vectorially. The situation is shown in the figure below             Force on \[-{{q}_{1}}\] \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{b}^{2}}}\hat{i}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{3}}}{{{a}^{2}}}[sin\theta \hat{i}-cos\theta \hat{j}]\] From above, x" component of force is                                 \[{{F}_{x}}=\frac{{{q}_{1}}}{4\pi {{\varepsilon }_{0}}}\left[ \frac{{{q}_{2}}}{{{b}^{2}}}+\frac{{{q}_{3}}}{{{a}^{2}}}\sin \theta  \right]\]                                     (neglecting \[\hat{j}\] component) \[\therefore \]  \[{{F}_{x}}=\propto \left[ \frac{{{q}_{2}}}{{{b}^{2}}}+\frac{{{q}_{3}}}{{{a}^{2}}}\sin \theta  \right]\]