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

Question

Dr. Abubacker · Accepted 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 $O.$ Net force on charge $Q$
$F_{net} \Rightarrow 2F\cos	heta = 2\frac{1}{4\pi\varepsilon_{0}}\frac{- qQ}{(a^{2} + x^{2})} 	imes \frac{x}{\left( a^{2} + x^{2} ight)^{1/2}}$
$i.e.,\ F_{net} = - \frac{1}{4\pi\varepsilon_{0}}.\frac{2qQx}{\left( a^{2} + x^{2} ight)^{3/2}}$
As the restoring force $F_{net}$ is not linear, motion will be oscillatory (with amplitude $2a$) but not simple harmonic

Q. 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

Explanation

Reviewed by Dr. Abubacker

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