What is the tension in a charged ring?

[TSny]

What if you consider a small element on the right, and calculate the tension on that and take the limit as the size of the element goes to zero?

I don't know if that would work out. It's just an idea.

Actually that's a pretty good idea, have some (no longer infinitesimal) control region on the right that subtends an angle $\alpha$ or something, and try to come up for an expression for the electrostatic force on that.

It's slightly more complicated maths-wise and I'm not so sure how best to do it yet, but I think that would give something useful! Let me think about it for a while

Thanks!

This a good approach. For an infinitesimal segment subtending angle $\alpha$, the force on the segment is due to the net electric field of all of the charge of the ring excluding the charge in the segment itself. (The field due to the segment does not exert a net force on the segment itself.) So you just need to modify the limits of the integral $I$ in the OP. Everything else in that post looks good. When I carry out this procedure, I get an expression for $T$ that agrees with $T = F/2$ in post #39 in the limit of infinitesimally small $\alpha$. In particular, you get the logarithmic divergence as $\alpha$ approaches zero.

 

[haruspex]

The problem says to neglect the thickness, so I am working with a zero thickness line

But it does not say space has only two dimensions, whch is what is needed for your field line counting. Your argument results in an inverse law, not an inverse square law.
In 3D, the field lines from the ring diverge into the third dimension, effectively becoming sparser.

Indeed, if you consider the infinite cylinder in 3D, axis along Z, you will see that the field lines in the XY plane do exactly what you are saying.

 

[Delta2]

We all are chickens, hehe, no one dared to deal with the case where we take the ring to be a torus. We would have to do a "fearsome" triple integral over the region of the torus to find the electric field. Something tells me that working in cylindrical coordinate system might be the best for working with torus, but i am not sure. The good thing when we take the ring to be a torus , is that the volume charge density becomes a nice finite continuous function over the region of the torus, so we get rid of the infinity of the dirac delta function for the charge density which is apparent when we take the ring to be a circle.

 

[Adesh]

We all are chickens, hehe, no one dared to deal with the case where we take the ring to be a torus. We would have to do a "fearsome" triple integral over the region of the torus to find the electric field. Something tells me that working in cylindrical coordinate system might be the best for working with torus, but i am not sure. The good thing when we take the ring to be a torus , is that the volume charge density becomes a nice finite continuous function over the region of the torus, so we get rid of the infinity of the dirac delta function for the charge density which is apparent when we take the ring to be a circle.

I really appreciate the idea, and yes I worked for the magnetic field of a torus and cylindrical coordinates fits best (the phi component vanishes). I got this honor to give you the 1000th like, sir.

 

[Delta2]

I really appreciate the idea, and yes I worked for the magnetic field of a torus and cylindrical coordinates fits best (the phi component vanishes). I got this honor to give you the 1000th like, sir.

Thanks @Adesh :D, I was wondering the same thing not too long ago, that is who would be the person that will give me the 1000th like !

 

[etotheipi]

In 3D, the field lines from the ring diverge into the third dimension, effectively becoming sparser.

Yes, I think you pinned the tail on the donkey there; when we perform the "shrink" operation on the ring, not all of the field lines in the plane will stay in the plane, but they'll migrate upward and downward to make the field approach a radial one.

And congratulations @Delta2 on 1000 upvotes! I think it's time for a physics party

 

[archaic]

https://www.quora.com/A-ring-has-un...of-tension-in-the-ring/answer/Raushan-Raj-225

 

[etotheipi]

@archaic it is a similar problem, but as far as I can tell they ignore the electric forces on that charge element from the rest of the ring, and only consider the electric force due to the central added charge. That seems a bit problematic, because the field from the rest of the ring becomes infinite at the charge element

 

[archaic]

@archaic it is a similar problem, but as far as I can tell they ignore the electric forces on that charge element from the rest of the ring, and only consider the electric force due to the central added charge. That seems a bit problematic, because the field from the rest of the ring becomes infinite at the charge element

hm i think that the way it is being thought about is that since at the beginning, before having the charge in the middle, the ring is at rest, or, in a way, the "internal" tension at one ring element is equal to the electric force on this element by the others. now we imagine a charge popping into the middle of the ring, this will cause it to expand isotropically until it comes to rest again. at this point, i guess that we could think of the same concept of internal tension (though with lower value this time, since, in some sense, the distance between the elements is larger) cancelling the electric force (which would also be lower in value) of the ring elements on the one we're looking at. though, this time, we have another force; that of the test charge. i guess, then, that in the new "internal tension" there's a component going against the charge's force, hence, in essence, we're calculating this tension.

 

[etotheipi]

I don't know if there's any expansion (because then we'd also have to re-evaluate the force from the rest of the ring), but what they could be doing is looking at the equilibrium case, putting $q$ in the centre and looking at the increment of tension (so they're defining $T = T_{real} - T_0$). That might work, but it still seems a bit strange because you go from one infinite tension to another infinite tension

 

[Keith_McClary]

To compress a continuous charge distribution into a point or a line (segment) requires infinite energy. (A 2D surface can be done with finite energy.) The tension is proportional to the difference in energy between the ring and an infinitesimally stretched ring. I think it must be infinite.

 

[Adesh]

A 2D surface can be done with finite energy.)

Can you please explain that? I’m having trouble with in imaging that.

 

[etotheipi]

Can you please explain that? I’m having trouble with in imaging that.

I might well be wrong, but I think the self energy of a configuration of charges is the integral of the energy density $\frac{1}{2} \epsilon_0 E^2$ through all of space. If the electric field diverges at some points, e.g. at a line of charge, then this will become infinite. But AFAIK the electric field is still finite at a plane of charge, so the energy density doesn't diverge anywhere and we can calculate a finite value for the self energy.

 

[Adesh]

But AFAIK the electric field is still finite at a plane of charge

Is electric field finite at some point on a continuous sheet of charge?

 

[haruspex]

Is electric field finite at some point on a continuous sheet of charge?

The infinities do not arise with charged surfaces, which is just as well as these occur in reality.

 

[Adesh]

The infinities do not arise with charged surfaces, which is just as well as these occur in reality.

Okay for charged surfaces infinities (doing a sin to mathematics) cancel each other out?

 

[haruspex]

Okay for charges surfaces infinities (doing a sin to mathematics) cancel each other out?

Sorry, can't parse that.

 

[Adesh]

Sorry, can't parse that.

If I have a continuous sheet of a charge in $xy$ plane, will electric field at a point on the sheet be finite? If yes then it means that infinities cancel each other out, infinities are arising because charges are so close to each other therefore the $r$ in the denominator is becoming very close to zero.

 

[haruspex]

If I have a continuous sheet of a charge in $xy$ plane, will electric field at a point on the sheet be finite? If yes then it means that infinities cancel each other out, infinities are arising because charges are so close to each other therefore the $r$ in the denominator is becoming very close to zero.

But these are not point charges. You have a finite charge per unit area. If you follow field lines back to the sheet from some point outside they they maintain their spacing, whereas doing that for a line or point charge they become crammed together.
The field from an infinite flat sheet of charge is constant throughout each half space. There is a discontinuity in passing through the sheet, but it remains bounded.

 

[Paul Colby]

The following may be fun to consider / discuss. I'll give it a try.

For what it is worth, the potential of a point charge is an analytic function of the charge location. For example one may write the potential of a charge, $2q$, located at $(x_c,y_c,z_c)$ as

$\Phi(x,y,z) = \frac{-2q}{\sqrt{(x-x_c)^2+(y-y_c)^2+(z-z_c)^2}}$.​

Now, $x_c$, $y_c$ and $z_c$ are simply real parameters. Nothing in principle prevents us from locating our charge at complex locations. Well, except the potential becomes complex as does the resulting electric field. We can fix this by splitting $2q$ into two and placing $q$ at, $z_c = iR$, and the other half at, $z_c = -iR$, where $R$ is a real parameter equal to the charge ring radius discussed previously. The new potential is,

$\Phi(x,y,z) = \frac{q}{\sqrt{x^2+y^2+(z-iR)^2}}+\frac{q}{\sqrt{x^2+y^2+(z+iR)^2}}$.​

Okay, so our new potential is everywhere real for real $(x,y,z)$ real as are the components of the electric field. It also obeys the required equation,

$\nabla^2\Phi = 0$,​

at every point it's defined at which I believe is everywhere except on the ring, $x^2+y^2=R^2$ in the $z=0$ plane. 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.

 

[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.