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

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Question

The charge moves towards $x$ - axis. Let at certain instant its position from point $\mathrm{O}$ is $\mathrm{x}$.

Instantaneous resultant force act

Vedclass · Accepted Answer

execute oscillatory not $SHM$

Vedclass · Answer

The charge moves towards $x$ - axis. Let at certain instant its position from point $\mathrm{O}$ is $\mathrm{x}$.

Instantaneous resultant force acting over it is given as

${\mathrm{F}_{\mathrm{R}}=2 \mathrm{F} \cos 	heta} $

${=2 \frac{\mathrm{qQ}}{4 \pi \varepsilon_{0}\left(\mathrm{a}^{2}+\mathrm{x}^{2}ight)} \frac{\mathrm{x}}{\sqrt{\mathrm{a}^{2}+\mathrm{x}^{2}}}} $

${\mathrm{F}_{\mathrm{R}}=\frac{2 \mathrm{q} \mathrm{Q}}{4 \pi \varepsilon_{0}} \frac{\mathrm{x}}{\left(\mathrm{a}^{2}+\mathrm{x}^{2}ight)^{3 / 2}}}$

This force is not of the form $\mathrm{F}=\mathrm{kx}$

Therefore, the motion of the charge $\mathrm{Q}$ is oscillatory but not simple harmonic.

Two equal negative charges $-q$ are fixed at points $(0, -a)$ and $(0, a)$ in the $x-y$ plane. A positive charge $Q$ is released from rest at a point $(2a, 0)$. The charge $Q$ will

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