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 :
[b] Force on charge \[{{q}_{1}}\], due to \[{{q}_{2}}\] is \[{{F}_{12}}=k\frac{{{q}_{1}}{{q}_{2}}}{{{b}^{2}}}\] Force on charge \[{{q}_{1}}\]due to \[{{q}_{3}}\] is \[{{F}_{13}}=k\frac{{{q}_{1}}{{q}_{3}}}{{{a}^{2}}}\] The X- component of the Force \[\left( {{F}_{x}} \right)\] on \[{{q}_{1}}\]is \[{{F}_{12}}+{{F}_{13}}\sin \theta \] \[\therefore \,\,\,\,\,{{F}_{x}}=k\frac{{{q}_{1}}{{q}_{2}}}{{{b}^{2}}}+k\frac{{{q}_{1}}{{q}_{2}}}{{{a}^{2}}}\sin \theta \] \[\therefore {{F}_{x}}\propto \frac{{{q}_{2}}}{{{b}^{2}}}+\frac{{{q}_{3}}}{{{a}^{2}}}\sin \theta \]
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