Two equal negative charges -q each are fixed | vedclass

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(D) The force on charge $Q$ at $(x, 0)$ due to the two charges $-q$ at $(0, a)$ and $(0, -a)$ is directed towards the origin.The distance of each char

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execute oscillatory but not simple harmonic motion

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(D) The force on charge $Q$ at $(x, 0)$ due to the two charges $-q$ at $(0, a)$ and $(0, -a)$ is directed towards the origin.The distance of each charge $-q$ from $Q$ is $r = \sqrt{x^2 + a^2}$.The magnitude of the force from each charge is $F = \frac{1}{4\pi\epsilon_0} \frac{qQ}{x^2 + a^2}$.The net force along the $X$-axis is $F_{net} = -2F \cos	heta = -2 \left( \frac{1}{4\pi\epsilon_0} \frac{qQ}{x^2 + a^2} ight) \frac{x}{\sqrt{x^2 + a^2}} = -\frac{2qQ}{4\pi\epsilon_0} \frac{x}{(x^2 + a^2)^{3/2}}$.For small $x$ (near the origin),$x^2 \ll a^2$,so $F_{net} \approx -\left( \frac{2qQ}{4\pi\epsilon_0 a^3} ight) x$.Since $F_{net} \propto -x$,the motion is simple harmonic for small displacements.However,for larger displacements,the restoring force is not proportional to $x$,so the motion is oscillatory but not simple harmonic.

Two equal negative charges $-q$ each are fixed at the 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 :-

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