Gravity/Charged Particle Equilibrium

1. Oct 7, 2009

bbhill

1) Consider negatively charged -Q ring with radius R and a particle with charge +q and mass m. Particle is confined to movement along the axis that bisects the ring which also points along the force of gravity. Taking into account the force of gravity,

a) What condition must q/m meet so that the particle is stable, and
b) Suppose the above condition holds and the system undergoes small oscillations. Show that the equilibrium oscillation freq is identical to the case where gravity is ignored and the equilibrium position is shifted from the center of the ring.

F(ring on particle)= qE = $$(-kQqz)/R^{3}$$ where z is distance of particle from the center of the ring.

F(gravity on particle) = -mg

Fnet (I think) = Fring - Fgravity

U(potential of ring on particle) = $$(-kQq)/\sqrt{z^{2}+R^{3}}$$

So what I did at first was just set the Fring equal to Fgravity and I came up that q/m = (gR^{3})$$/$$(-kQqz) where z is the distance from the ring.

Now I know there must be some way to get this in simpler terms, so I'm wondering if I am even going about this the right way. . .

Any help would be appreciated.

Thanks!

2. Oct 8, 2009

rl.bhat

q/m = (gR^{3}/(-kQqz)
this condition is wrong. And you need not put the negative sign.

3. Oct 8, 2009

bbhill

Thanks, I corrected those mistakes.

I ended up getting $$q/m=gR^{3}/kQz=gR^{2}/kQ$$ since the force of the ring is at a local maximum when R = z.

I've been staring at this for a while and I still have no clue about how even to go about the second part though. I believe that the angular frequency of this system would be $$\sqrt{kq/m}=\sqrt{gR^{2}/Q}$$. However, I am confused about how to go about finding the ang freq for a particle that has an equilibrium position away from the center of the ring. . .