Electrostatics Question 257
Question: A point charge of 40 stat coulomb is placed 2 cm in front of an earthed metallic plane plate of large size. Then the force of attraction on the point charge is
Options:
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 $
Show Answer
Answer:
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 $
therefore $ F _{x}=k[ \frac{q _{1}q _{2}}{b^{2}}+\frac{q _{1}q _{3}}{a^{2}}\sin \theta ] $
therefore $ F _{x}=k\cdot q _{1}[ \frac{q _{2}}{b^{2}}+\frac{q _{3}}{a^{2}}\sin \theta ] $
therefore $ F _{x}\propto ( \frac{q _{2}}{b^{2}}+\frac{q _{3}}{a^{2}}\sin \theta ) $
