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 \[{{F}_{x}}\propto \,\left( \frac{{{q}_{2}}}{{{b}^{2}}}+\frac{{{q}_{3}}}{{{a}^{2}}}\sin \theta \right)\]You need to login to perform this action.
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