Gujarati
Hindi
1. Electric Charges and Fields
normal

Two equal $-ve$ 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

A

Execute $SHM$ about the origin

B

Move to the origin and remain at rest

C

Move to infinity

D

Execute oscillatory but not $SHM$

Solution

Component of force on charge $+Q$ at $P$ along $x$ – axis,

${\mathrm{F}_{\mathrm{x}}=\frac{2 \mathrm{Qq}}{4 \pi \varepsilon_{0}\left(\mathrm{a}^{2}+\mathrm{x}^{2}\right)} \cos \theta} $

${=\frac{2 \mathrm{Qq}}{4 \pi \varepsilon_{0}\left(\mathrm{a}^{2}+\mathrm{x}^{2}\right)} \times \frac{\mathrm{x}}{\sqrt{\mathrm{a}^{2}+\mathrm{x}^{2}}}} $

${=\frac{2 \mathrm{Qqx}}{4 \pi \varepsilon_{0}\left(\mathrm{a}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}}$

Which is not directly proportional to $\mathrm{x}$. So, motion is oscillatory but not $\mathrm{SHM}.$

Standard 12
Physics

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