Question

Two equal negative charges $-q$ are fixed at 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

• A. execute simple harmonic motion about the origin
• B. move to the origin and remains at rest
• C. move to infinity
• D. execute oscillatory but not simple harmonic motion

Answer 1

Answer: (D)

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 $O$

$F=2\frac{1}{4\pi {\epsilon }_0}\frac{-qQ}{\left(a^2+x^2\right)}\times \frac{x}{{\left(a^2+x^2\right)}^{1/2}}$
i.e., $F=-\frac{1}{4\pi {\epsilon }_0}\cdot \frac{2qQx}{{\left(a^2+x^2\right)}^{3/2}}$
As the restoring force $F$ is not linear, motion will be oscillatory (with amplitude $2a$ ) but not simple harmonic.