Charged ring, integrate for electric potential

[Loopas]

Homework Statement

A flat ring of inner radius R1 and outer radius R2 carries a uniform surface charge density σ. Determine the electric potential at points along the axis (the x axis). [Hint: Try substituting variables.]

Homework Equations

V = (kQ)/r

The Attempt at a Solution

As you can see from my screenshot, I think I've figured it out mostly I'm just stuck on finding the value for r in the above equation. Shouldn't it be something like sqrt(x^2+y^2)? But that can't be the answer because there's no labelled y-axis...

 

[jackarms]

Well, you can't calculate the potential directly from the equation since not every point on the disk will be the same distance from a given point along the x axis. Try splitting the disk into thin disks, and then doing an integration.

 

[Loopas]

What exactly do you mean by "thin disks"? Don't I still need to find the distance between the x point and any given slice of charge on the disk?

 

[jackarms]

By thin I mean infinitely thin. With the disk you're using, the distance to the x point is going to vary as you go farther out from the center. You can find an expression for the distance in terms of the x point and the point you go radially outward, but it's going to be variable, so the solution is going to involve integration -- if you just look at an infinitesimally thin disk and find an expression for the potential along its axis, then you can consider the original disk as a collection of these disks, and integrate them all together.

 

[Loopas]

I'm having trouble finding an expression for the distance between any given point on the ring and the x-point, since there doesn't seem to be any variable that defines the distance between the center of the ring and any given piece of charge on the ring.

 

[jackarms]

Not explicitly given in the diagram, but that doesn't mean you can't create your own. You can indentify a ring of charge with a radius, or if you really want to identify any point, a radius and an angle.

 

[Loopas]

Thanks, I was finally able to figure it out, integrating from R1 to R2:

[itex]\frac{σ}{2ε_{0}}[/itex]*[itex]\int[/itex][itex]\frac{rdr}{\sqrt{x^2+r^2}}[/itex]