# Math Help - Uniformly charged ring?

1. ## Uniformly charged ring?

A ring of radius $r$ having the charge $Q$ distributed uniformly over its surface is fixed in $Y-Z$ plane with centre at origin. A particle of mass $m$ and charge $(-q)$ is placed at a point $P$ which is at a distance $d$ from the centre of the ring. Particle is attached to one end of a spring of spring constant $k$. The other end of spring is fixed at origin. Find the minimum velocity to be imparted to $P$ (Consider motion in $X-Y$ Plane) so that the spring is not compressed.

2. Originally Posted by fardeen_gen
A ring of radius $r$ having the charge $Q$ distributed uniformly over its surface is fixed in $Y-Z$ plane with centre at origin. A particle of mass $m$ and charge $(-q)$ is placed at a point $P$ which is at a distance $d$ from the centre of the ring. Particle is attached to one end of a spring of spring constant $k$. The other end of spring is fixed at origin. Find the minimum velocity to be imparted to $P$ (Consider motion in $X-Y$ Plane) so that the spring is not compressed.
I am not quite sure what to make of this one. For starters we will need to know what the equilibrium length of the spring is.

And I'm a bit confused by the problem statement. We can easily have an initial applied speed given to the particle but that will only hold for one instant. The added speed will simply stretch or compress the spring from this position and then we have simply harmonic motion. Is this a correct interpretation of the problem statement?

-Dan

3. Does the answer throw up any clues? -
Spoiler:
$v_{min} = \sqrt{\frac{qQd^2}{4\pi \epsilon_0m(r^2 + d^2)^{\frac{3}{2}}}}$

The only thing I could do in this problem was to find the electrostatic force exerted by the ring on the particle P which gave me (by some simple integration) :

$\bold{F_e} = \frac{1}{4\pi \epsilon_0}\cdot \frac{qQd}{(r^2 + d^2)^{\frac{3}{2}}}$ along positive x-axis.

4. Originally Posted by topsquark
I am not quite sure what to make of this one. For starters we will need to know what the equilibrium length of the spring is.

And I'm a bit confused by the problem statement. We can easily have an initial applied speed given to the particle but that will only hold for one instant. The added speed will simply stretch or compress the spring from this position and then we have simply harmonic motion. Is this a correct interpretation of the problem statement?

-Dan
I believe you are meant to assume that the spring is at its natural length at the start.

Basically, the speed needs to be sufficient that the centripetal acceleration required to keep it on a radius equal to the natural length of the spring is greater than the electrostatic force of attraction.

I'll put up a solution either tomorrow or the next day.

Actually, just noticed your post Fardeen - you've done the hard bit which was the integration of the electrostic forces so just equate the force you've computed to the centripetal acceleration for radius of $d$ so

$F_e = m \frac{v^2}{r}$

which gives you the solution you've put.

Strictly speaking I would have computed what the field of the force is generally as you also need to demonstrate that the force of attraction doesn't increase from side to side although this is obvious in this case as the force is maximal on the line of symmetry.