What is the tension in a charged ring?

[Keith_McClary]

There is a solution by Martin Gales based on the change in electrostatic energy when the radius is changed.

 

[haruspex]

There is a solution by Martin Gales based on the change in electrostatic energy when the radius is changed.

That is a solution to the valid case of a wire with non infinitesimal thickness. The original question in this thread took it to be infinitesimal.

 

[etotheipi]

Yeah, if the ring is toroidal and we have some way of finding the capacitance $C=\frac{\pi R}{ln (\frac{8R}{r})}$ by messy calculation, looking it up, or some other way, then it's not too bad I suppose.

However for a ring of infinitesimal thickness, I think it was concluded that the tension has to diverge, not least because itself energy also diverges. @TSny also gave a very nice analysis a little earlier on, considering the force between two charged arcs, and showing the logarithmic divergence as the gap between them shrinks.

 

[Paul Colby]

I also believe (but can't prove) that the potential is zero everywhere within the disk of which the ring forms the edge if one treats branch cuts appropriately.

Yeah, I can now prove this. We define a complex function on real three space, $\omega(x,y,z)$, given by,

$\omega(x,y,z) = (x^2+y^2+(z + i R)^2)^{-1}$.​

With this our potential (second equation in #90) becomes,

$\Phi(x,y,z) = q(\sqrt{\omega(x,y,z)} +\sqrt{\omega^\ast(x,y,z)})$. (*)​

The important thing to be gleaned from this expression is that given any path in real three space, $(x(t),y(t),z(t))$, the $\omega$ function and its conjugate, $\omega^\ast$, provides two paths in the $\omega$ plane that are conjugates of each other. Any path which terminates on a branch of the first term of (*) implies that the conjugate path taken in the second term will terminate on the opposite branch thus having the opposite sign. Therefore, the potential is zero at every point on the branch or inside the disk bounded by the ring. This implies $E_x$ and $E_y$ are zero also. The third component, $E_z$, is zero because $\Phi$ is an even function of $z$.

So, what is provided by (*) is in fact a closed form expression for the field produced by a ring-source. No integrals required.

 

[Paul Colby]

Yeah, I can now prove this.

Just to close out my suggestion it seems there is a problem with what I'm suggesting. The "solution" posted in #90 etc is indeed a ring-source of some kind however it is definitely not a simple ring of charge. The potential at the ring center is easily shown to be zero. For an actual charge ring, the work bringing a test charge in from infinity can't be zero as claimed in #94. This is corroborated by the exact expression in terms of complete elliptic integrals. To add to the confusion in the far field using the divergence theorem the net charge is indeed what one wants. To add even more obscure issues, following a path in real space around the ring flips the sign of the total charge. One might expect this to be due to a branch discontinuity in either the potential or it's derivatives which, since everything vanishes on the disk bounded by the ring, has no apparent jumps. I'm completely baffled by this.

 

[pablochocobar]

Homework Statement: The ring I'll assume to have negligible thickness, and linear charge density $\lambda$, as well as a radius $R$.
Relevant Equations: N/A

I tried considering a little piece of the ring (shaded black below) subtending angle $d\theta$, and attempted to find the electric field in the vicinity of that piece by a summation of contributions from the rest of the ring:

View attachment 265375

$$dE_x = \frac{dq}{4\pi \epsilon_0 d^2} \cos{\phi} = \frac{\lambda R d\theta}{4\pi \epsilon_0 \cdot 2R^2(1-\cos{\theta})}\cos{\phi}$$ $$E_x = \int_0^{2\pi} \frac{\lambda \sin{\frac{\theta}{2}}}{8\pi \epsilon_0 R(1-\cos{\theta})} d\theta = \frac{\lambda}{8\pi \epsilon_0 R} \int_0^{2\pi} \frac{\sin{\frac{\theta}{2}}}{1-\cos{\theta}} d\theta $$ Problem is, that thing diverges because the denominator goes to zero at the boundaries (the indefinite is easy enough to solve by changing $1-\cos{\theta} = 2\sin^2{\frac{\theta}{2}}$). If we call $$I = \int_{0}^{2\pi} \frac{\sin{\frac{\theta}{2}}}{1-\cos{\theta}} d\theta$$ then the electrostatic force on the piece in terms of this is $$F_e = (\lambda R d\theta) \frac{\lambda I}{8\pi \epsilon_0 R} = 2T\sin{\frac{d\theta}{2}} \approx T d\theta$$ and the tension would be, under the usual equilibrium constraint, $$T = \frac{\lambda^2 I}{8\pi \epsilon_0} = \frac{q^2 I}{32 \pi^3 R^2 \epsilon_0}$$but that doesn't make much sense if $I$ is infinite. I wondered if anyone could help out? Perhaps it is the case that the electric field diverges at the ring, but in that case, how could I modify my calculations to get a sensible answer? Because I am fairly sure that in reality the tension will not be infinite . Thanks!

First I am trying to calculate the field due to any circular arc which subtends angle $\alpha$.

For any sector which subtends same angle the Electric Field will be same. Ie.$| \vec E_1 | = | \vec E_2 |$ Because charge is distributed symmetrically.​

DERIVATION :

$GC\perp{AO}$
$OG = R$, $OC = R\cos\theta$
$OA = 2OC = 2R\cos\theta$The charge on the small arc subtend by angle ${d\theta}$ is, $$dq = \lambda2R\cos\theta d\theta$$
where $\lambda$ is charge per unit length = $\frac{Q}{2\pi{R}}$.
Electric Field due to ${dq}$ is, $$dE = \frac{kdq}{(2R\cos\theta)^2}$$

For a symmetric part OB which subtends same angle $d{\theta}$ the Y-component of electric field $dE_Y$ will cancel.

∴ Net Electric Field due to a arc of angle $\alpha$ at O, $$E_{net} = \int_{-\frac{\alpha}{2}}^\frac{\alpha}{2} dE\cos\theta$$ $$= \int_{-\frac{\alpha}{2}}^\frac{\alpha}{2} \frac{kdq\cos\theta}{(2R\cos\theta)^2}$$ $$= \int_{-\frac{\alpha}{2}}^\frac{\alpha}{2} \frac{k\lambda2R\cos\theta{d\theta}\cos\theta}{(2R\cos\theta)^2}$$ $$= \int_{-\frac{\alpha}{2}}^\frac{\alpha}{2} \frac{k\lambda}{2R}d\theta$$ $$= \frac{k\lambda}{2R}\left[ \theta \right]_{-\frac{\alpha}{2}}^{\frac{\alpha}{2}}$$ $$ = \frac{k\lambda\alpha}{2R}$$

Now for $\alpha=2\pi$ the Electric field will be due to all the charges at any point on the ring.
$$E_{at~any~point~on~ring} = \frac{k\lambda\pi}{R}$$

charge on the marked section = $\lambda2R2d\theta$ and E-field on marked section = $\frac{k\lambda\pi}{R}$

$$2T\sin{d\theta} = \frac{k\lambda\pi}{R}\lambda2R2d\theta$$

∵$d\theta$ is too small, ∴$\sin{d\theta}\approx{d\theta}$

$$T=\frac{Q^2}{8\pi^2\varepsilon{R}^2}$$

 

[Paul Colby]

I’d suggest fattening the ring up ever so slightly, view it as a wire of cross sectional radius $d$ where $d<<R$. Then take the limit $d\rightarrow 0$.

first rate solution. Well done.

 

[pablochocobar]

I’d suggest fattening the ring up ever so slightly, view it as a wire of cross sectional radius $d$ where $d<<R$. Then take the limit $d\rightarrow 0$.

first rate solution. Well done.

You talking about my solution?

 

[Paul Colby]

Well, the first comment was directed at the quoted text in #96. Then I looked through your attachment where a very reasonable solution is arrived at.

 

[TSny]

The charge on the small arc subtend by angle ${d\theta}$ is, $$dq = \lambda2R\cos\theta d\theta$$

This is not the correct expression for $dq$. If you integrate your expression for $dq$ over the ring, you will see that you don't get the correct total charge.

$2R\cos\theta d\theta$ is not the infinitesimal arc length of the ring corresponding to $d \theta$ in your figure. Swinging $2R\cos \theta$ through $d\theta$ gives an infinitesimal arc length that is "perpendicular" to the line segment $2R\cos \theta$. But this arc length is not along the ring. It is not hard to show that the infinitesimal arc length along the ring corresponding to $d\theta$ is $2R\cos \theta d \theta \cdot \large \frac 1 {\cos \theta}$ $= 2R d \theta$.

If you correct this and then go on to find $E_{net}$, you'll find that $E_{net}$ diverges. This was shown in the first post of the thread.

 

[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.