Electrostatics Question 686
Question: Three charges $ q _{1},+q _{2} $ and $ q _{3} $ are place . The x-component of the force on$ -q _{1} $ is proportional to
Options:
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 $
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Answer:
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 $ ( F _{x} ) $ 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 $