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## Homework Statement

Find the potential at off-axis points due to a thin circular ring of radius a and mass M. Let R be the distance from the center of the ring to the field point and [tex]\theta[/tex] be the angle between the line connection the center of the ring with the field point and the axis of the ring. Assume R>>a so that terms of order (a/R)

^{3}and higher may be neglected.

## Homework Equations

[tex]\Phi[/tex]=[tex]\int[/tex][tex]\frac{-Gdm}{r}[/tex]

## The Attempt at a Solution

The main problem is to express r as a function of a, R, and [tex]\theta[/tex]. I approached it by first looking at the potential due to an arbitrary point on the ring, and letting the field point lie in the yz plane (the ring is in the xy plane). r can then be decomposed via the pythagorean theorem into two parts: the distance in the xy plane from the y intercept of the ring to the point on the ring, and the distance from the y-intercept up to the field point. I'll call the former l and the latter k. k can be found via the law of cosines as k

^{2}=a

^{2}+R

^{2}-2aRcos(90-[tex]\theta[/tex]). lf I then let [tex]\phi[/tex] be the angle from the y-axis to the radial line connecting the point on the ring with the center of the ring, l is the third side of an equilateral triangle and so sin([tex]\frac{\phi}{2}[/tex])=[tex]\frac{l}{2a}[/tex]. At this point, dm can be set up. dm=[tex]\rho[/tex]*dx where dx is the portion of the ring that we are looking at. Since the segment of the ring we are looking at is marked off by a d[tex]\phi[/tex], and for a small change like that, l makes a good estimate, dx=dl, and so dm=[tex]\rho[/tex]dl. dl of course can be put in terms of d[tex]\phi[/tex]=a*cos([tex]\frac{\phi}{2}[/tex])*d[tex]\phi[/tex]. Putting all of this together yields the following integral:

[tex]\frac{-GM}{4*\pi*a}[/tex][tex]\int[/tex][tex]\frac{2*a*cos(\frac{\phi}{2})d\phi}{\sqrt{4*a^{2}sin^{2}(\frac{\phi}{2})+k^{2}}}[/tex]

This really complex looking integral simplies (way too nicely) under a change of variables to:

[tex]\frac{-GM}{4*\pi*a}[/tex]*[tex]\int[/tex]sec([tex]\varphi[/tex])d[tex]\varphi[/tex] from 0 to 2[tex]\pi[/tex].

This yields 0. If you can spot what I did wrong, that'd be great. If you can think of an easier method, I'd appreciate that as well. Thanks for any insight you guys can provide!

Arjun