question_answer

Two charges, each equal to q, are kept at \[x=-a\] and\[x=a\] on the x−axis. A particle of mass m and charge\[{{q}_{0}}=\frac{q}{2}\] is placed at the origin. If charge\[{{q}_{0}}\]is given a small displacement\[(y<<a)\] along the y−axis, the net force acting on the particle is proportional to: [JEE MAIN 2013]

A) y

B) − y

C) \[\frac{1}{y}\]

D) \[-\frac{1}{y}\]

Correct Answer: A

Solution :

[a]

\[{{F}_{net}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{2q{{q}_{0}}}{{{y}^{2}}+{{a}^{2}}}\cos \theta \]

\[=\frac{2q{{q}_{0}}y}{4\pi {{\varepsilon }_{0}}{{\left( {{y}^{2}}+{{a}^{2}} \right)}^{3/2}}}=\frac{2q{{q}_{0}}}{4\pi {{\varepsilon }_{0}}}\times \frac{y}{{{a}^{3}}}\]as

\[(y<<a)\]

\[\therefore \] \[F\propto y\]