Two equal positive charges Q are fixed at points (a, 0) and (-a, 0) on the x-axis. An opposite charge -q at rest is released from point (0, a) on the y-axis. The charge -q will

Question

Two equal positive charges Q are fixed at points (a, 0) and (-a, 0) on the x-axis. An opposite charge -q at rest is released from point (0, a) on the y-axis. The charge -q will (A) Move to infinity (B) Move to origin and rest there (C) Undergo SHM about the origin (D) Execute oscillatory periodic motion but not SHM

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/cb6846c0202e49fe30ecff7d8b0a9d80-.png" /> In question Net force on q is not proportional to ( x ) as F propto(-x) [For SHM] but Net force on q is F=-(K q ⋅ Q x/(x2+a2)(1)2) This is condition for periodic motion

Q. Two equal positive charges $Q$ are fixed at points $(a, 0)$ and $(- a, 0)$ on the $x$ -axis. An opposite charge $- q$ at rest is released from point $(0, a)$ on the $y$ -axis. The charge $- q$ will

Move to infinity

Move to origin and rest there

Undergo SHM about the origin

Execute oscillatory periodic motion but not SHM

In question
Net force on $q$ is not proportional to $(x)$
as $F \propto (- x)$ [For SHM] but Net force on $q$ is
$F = - \frac{K q \cdot Q x}{( x ^{2} + a ^{2} ) ^{\frac{1}{2}}}$
This is condition for periodic motion

Q. Two equal positive charges Q are fixed at points (a,0) and (−a,0) on the x-axis. An opposite charge −q at rest is released from point (0,a) on the y-axis. The charge −q will