# Electric Field Problem II

1. Sep 14, 2008

### bodensee9

Hello:

I have the following. A stationary ring of radius a lies in the yz plane and has a uniform positive charge Q. A small particle that has mass m and a negative charge -q is located at the center of the ring. (a) show that is x << a the electric field along the axis of the ring is porportional to x. (b) find the force on teh particle as a function of x. (c) show that if the particle is given a small displacement in the +x direction, it will perform SHM.

So for (a), do I do, since the E for a ring is k*Q*x/(a^2+x^2)^(3/2), where x is the displacement on the z axis, and a is the radius. So that's a because if x is very small then the equation is basically k*Q*x/a^3. So this is porportional to x.

(b) Then wouldn't the force just be q*E, and since there's a negative charge -q here, wouldn't the F = q*k*Q*x/a^3?

(c) So to show SHM, I need to show that acceleration = some constant w^2*displacement. So couldn't I just set q*k*Q/a^3 as w?

Thanks!!

2. Sep 14, 2008

### Defennder

Your answer for (a) is correct provided you got the expression for E correct (I didn't check it). (b) is odd since x=0 and it is evident by symmetry that that E-field at the centre of the ring where the particle is located is clearly 0. Unless they mean (if the particle was instead placed at some x=b), then you would have to perform the integration to get the value.

(c) is nearly correct. Just note you have to take the mass of particle into consideration. Be careful of the missing square.

3. Sep 15, 2008

THanks!