What is the tension in a charged ring?

[Kyathallous]

 

[TSny]

Welcome to PF!

If you let $\phi = \pi$, you can see that the expression above yields 0 for $|\vec r|$. But, the correct value is clearly $2R$.

From the right triangle $abo$ in the figure below, $\dfrac{r}{2} = R \sin\left(\dfrac{\phi}{2}\right)$.

So, $r = 2R \sin\left(\dfrac{\phi}{2}\right)$.

 

[Kyathallous]

Oh.. Right. Can't believe I messed that up. I tried to work further on my method using your correct but, I ran head first into some logarithms of 0. At the moment I don't see how I might be able to fix this, but if I do I will surely let you know.


edit: I don't think I'll be able to make much of it as i am stuck on a logarithmic function. The smaller the element size in my solution will become, the higher the functions value gets.

Welcome to PF!

If you let $\phi = \pi$, you can see that the expression above yields 0 for $|\vec r|$. But, the correct value is clearly $2R$.

From the right triangle $abo$ in the figure below, $\dfrac{r}{2} = R \sin\left(\dfrac{\phi}{2}\right)$.

So, $r = 2R \sin\left(\dfrac{\phi}{2}\right)$.

View attachment 354369

 

[TSny]

I tried to work further on my method using your correct but, I ran head first into some logarithms of 0. At the moment I don't see how I might be able to fix this, but if I do I will surely let you know.

edit: I don't think I'll be able to make much of it as i am stuck on a logarithmic function. The smaller the element size in my solution will become, the higher the functions value gets.

You are getting the expected divergent behavior for the tension for the idealized circular ring with zero thickness. The tension is infinite in this case. See some of the earlier discussion in this thread. Post #39 has some numerical calculations.