# A positively charged ring is placed in the y-z plane with its center at origin and a negatively charged particle is placed at some distance on x axis. Will it execute SHM?

Susmita
425 Points
4 years ago
Let the particle of charge q be placed at a height x from the centre of the circular ring.
Take any point on the circumference of the ring.Let R be the radius of the ring.
So distance of charge q from this point will be
$r=\sqrt{R^2 +x^2}$
If Q is charge on the ring,then potential at r is
$\phi=\frac{Q}{4\pi \epsilon r}$
And electric field
$E=-\frac{d}{dx}\phi =\frac{Qx}{4\pi \epsilon (R^2+x^2)^\frac{3}{2}}$
If R>>x,ie,if the charge q is placed too close to the centre of the loop then E will approximate to
$E=-\frac{d}{dx}\phi =\frac{Qx}{4\pi \epsilon R^3}$
Force on any charged particle is qE.
So the equation of force will become

$Or,m\frac{d^2x}{dt^2}=-\frac{qQx}{4\pi \epsilon R^3}$
We are putting the negative sign as q is negative.
$Or,\frac{d^2x}{dt^2}+\frac{qQ}{4\pi \epsilon m R^3}x=0$
$Or,\frac{d^2x}{dt^2}+(\omega)^2 x=0$
Where
$\omega = \frac{qQ}{4\pi \epsilon m R^3}$
So it is equation of SHM.
Yes the charge will execute SHM with angular frequency w.