Electrostatics Question 227
Question: Two equal negative charge q are fixed at the fixed points $ (0,a) $ and $ (0,-a) $ on the Y-axis. A positive charge Q is released from rest at the point $ (2a,0) $ on the X-axis. The charge Q will [IIT 1984; Bihar MEE 1995; MP PMT 1996]
Options:
A) Execute simple harmonic motion about the origin
B) Move to the origin and remain at rest
C) Move to infinity
D) Execute oscillatory but not simple harmonic motion
Show Answer
Answer:
Correct Answer: D
Solution:
By symmetry of problem the components of force on Q due to charges at A and B along y-axis will cancel each other while along x-axis will add up and will be along CO.
Under the action of this force charge Q will move towards O.
If at any time charge Q is at a distance x from O. Net force on charge Q $ F _{net}\Rightarrow 2F\cos \theta =2\frac{1}{4\pi {\varepsilon _{0}}}\frac{-qQ}{(a^{2}+x^{2})}\times \frac{x}{{{(a^{2}+x^{2})}^{{1}/{{}};2}}} $
i.e., $ F _{net}=-\frac{1}{4\pi {\varepsilon _{0}}}.\frac{2qQx}{{{( a^{2}+x^{2} )}^{{3}/{{}};2}}} $
As the restoring force Fnet is not linear, motion will be oscillatory (with amplitude 2a) but not simple harmonic.
