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 2003]

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)            \[\frac{{{q}_{2}}}{{{b}^{2}}}+\frac{{{q}_{3}}}{{{a}^{2}}}\cos \theta \]

Correct Answer: C

Solution :

\[{{F}_{\text{2}}}\] = Force applied by \[{{q}_{2}}\] on \[-{{q}_{1}}\] \[{{F}_{\text{3}}}\] = Force applied by \[(-{{q}_{3}})\] on ?\[{{q}_{1}}\] x-component of Net force on \[-{{q}_{1}}\] is \[{{F}_{x}}={{F}_{\text{2}}}+{{F}_{\text{3}}}\text{sin}\theta \]   \[=k\frac{{{q}_{1}}{{q}_{2}}}{{{b}^{2}}}+k.\frac{{{q}_{1}}{{q}_{3}}}{{{a}^{2}}}\sin \theta \] Þ \[{{F}_{x}}=k\,\left[ \frac{{{q}_{1}}{{q}_{2}}}{{{b}^{2}}}+\frac{{{q}_{1}}{{q}_{3}}}{{{a}^{2}}}\sin \theta  \right]\] Þ \[{{F}_{x}}=k\cdot {{q}_{1}}\,\left[ \frac{{{q}_{2}}}{{{b}^{2}}}+\frac{{{q}_{3}}}{{{a}^{2}}}\sin \theta  \right]\]Þ \[{{F}_{x}}\propto \,\left( \frac{{{q}_{2}}}{{{b}^{2}}}+\frac{{{q}_{3}}}{{{a}^{2}}}\sin \theta  \right)\]