Two equal negative charges -q are fixed at th | vedclass

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(D) Let the positive charge $Q$ be at a general point $P(x, 0)$ on the $x-$ axis. The distance between each charge $-q$ and the charge $Q$ is $r = \sq

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Execute oscillatory but not $SHM$

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(D) Let the positive charge $Q$ be at a general point $P(x, 0)$ on the $x-$ axis. The distance between each charge $-q$ and the charge $Q$ is $r = \sqrt{a^2 + x^2}$.The electrostatic force exerted by each charge $-q$ on $Q$ is $F = \frac{1}{4 \pi \varepsilon_0} \frac{Qq}{a^2 + x^2}$.The components of these forces perpendicular to the $x-$ axis cancel each other out,while the components along the $x-$ axis add up.The net force on $Q$ is $F_{net} = 2F \cos 	heta$,where $\cos 	heta = \frac{x}{r} = \frac{x}{\sqrt{a^2 + x^2}}$.Substituting the values,we get $F_{net} = 2 \left( \frac{1}{4 \pi \varepsilon_0} \frac{Qq}{a^2 + x^2} ight) \left( \frac{x}{\sqrt{a^2 + x^2}} ight) = \frac{2 Qqx}{4 \pi \varepsilon_0 (a^2 + x^2)^{3/2}}$.Since the force $F_{net}$ is directed towards the origin (restoring force) and is not directly proportional to the displacement $x$ (due to the $(a^2 + x^2)^{3/2}$ term in the denominator),the motion is oscillatory but not simple harmonic motion $(SHM)$.

Two equal negative charges $-q$ 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 will

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