Evaluating Potential of A Uniformly Charged Circular Ring - kQ/a

[Phymath]

A uniformly charged infinitely thin circular ring of radius a has total charge Q. Place the ring in the xy-plane with its center at the origin. Use cylindrical coordinates...

Find the potential at a point in the xy-plane a distance 2a from the orgin give your answer as a number times kQ/a
...
I figured the potential to be for some point in the xy-plane,$$(2a, \Phi , 0)$$
$$k_e j a \int_0 ^{2 \pi} \frac{d \theta}{\sqrt{(2a)^2+a^2-2(2a)a cos(\Phi - \theta)}}$$
$$= k_e j a \int_0 ^{2 \pi} \frac{d \theta}{\sqrt{4a^2+a^2-4a^2 cos(\Phi - \theta)}}$$
where, $$j = \frac{Q}{2 \pi a}$$
$$= k_e j \int_0 ^{2 \pi} \frac{d \theta}{\sqrt{5-4cos(\Phi - \theta)}}$$
did I set up this intergral right? how do I evaluate it? hints answers it all helps

 

[Phymath]

Did I at least set it up correctly? I don't KNOW what to do! i know because the ring is symetric you can drop the phi...

$$= k_e j \int_0 ^{2 \pi} \frac{d \theta}{\sqrt{5-4cos(\theta)}}$$

 

[Justin Lazear]

You want the potential in the same plane as the ring? Btw, you're forgetting a negative.

And Mathematica likes that integral about as much as I do.

--J

 

[Phymath]

ok check this out let me know if this a valid move...

$$r = \sqrt{5-4cos(\theta)}$$ which is polar so...
polar ints have the form
$$\frac{1}{2} \int r^2 d \theta$$

thus...

$$k_e j \frac{1}{2} \int_0 ^{2 \pi} \frac{1}{5-4cos(\theta)} d \theta$$
thus making $$= \frac{k_e Q}{a} \frac{1}{6}$$
let me know what you think...

 

[Phymath]

anyone? come on this can't be that hard

 

[Justin Lazear]

You can't use r² because it's not an area integral. You've already accounted for curvature when you say $ds = r d\theta$.

I don't see a problem with your setup, but your integral isn't easily computed. Mathematica says that it's

$$\frac{2Q \mbox{EllipticK} [-8]}{a \pi}$$

Then again, my brain isn't in such great shape at the moment, so I could very well be wrong.

--J

 

[Phymath]

can i get an actual value for EllipticK(-8)
$$\int_0^{2\pi}\sqrt{1+8sin^2(\theta)}d\theta$$

 

[Justin Lazear]

$$\frac{2Q \mbox{EllipticK} [-8]}{a \pi} = \frac{0.536591 Q}{a}$$

--J