2
$\begingroup$

Assume that a solid conducting torus (toroidal ring), with a cross-section of a circle of (minor) radius $r$, is negatively charged. Solving Poisson's equation, we can find the charge distribution of this torus, thereby calculating the electric field of the torus everywhere in space. Consider the electric field at an arbitrary point ($x_0, y_0, z_0$) to be $E_1$ when $r$ tends to zero. Assume the major radius $R$ of the torus remains unchanged.

Now, repeat the experiment using a conducting torus (toroidal ring), with a cross-section of an equilateral triangle of side $a$. Solving Poisson's equation, we may find the charge distribution of this torus to have great values on the triangle's vertices. Consider the electric field at the same point ($x_0, y_0, z_0$) to be $E_2$ when $a$ tends to zero. Assume the major radius $R$ of the torus remains unchanged.

Recall that we used the same surplus charge of $Q$ (electrons) to charge the tori in both experiments.

Question #1:
Can we say that since both $r$ and $a$ are approaching zero, the torus in both experiments can behave as a uniformly charged infinitesimally thin ring/thread and we have $E_1=E_2$?

Question #2:
Can we say that since both $r$ and $a$ are approaching zero, the torus in both experiments can behave as a charged nonconducting infinitesimally thin ring/thread with a uniform linear charge density of $\lambda$?

Question #3:
Can we say that, by infinitely decreasing the size of the cross-section of a charged conducting closed wire, its conductivity no longer plays a decisive role in the strength and pattern of its electric field so that we can mimic the field by replacing the conducting wire with a uniformly charged nonconducting thread of the same size?

Improve this question
$\endgroup$
5
  • $\begingroup$ My guess would be yes, yes and yes - but a proper mathematical treatment is needed to check, of course $\endgroup$
    – CompassBearer
    Commented Jul 3 at 22:27
  • $\begingroup$ We can certainly model even infinitesimally small sources with multipole moments, so that answer is... yes, but. One always has to understand these "infinitesimally small quantities and objects" as practically useful approximations of finite size fields and objects. $\endgroup$
    – FlatterMann
    Commented Jul 3 at 22:34
  • $\begingroup$ The field at the vertices of an ideal triangle is infinitely large, and I think taking the limit is not "allowed" in any sense I know of. But I bet that you will get a better answer in the math SE than here if you ask it there. $\endgroup$
    – hyportnex
    Commented Jul 3 at 23:08
  • $\begingroup$ The field at the center of the torus is zero by symmetry. Probably you mean something else, can you elaborate? $\endgroup$
    – Jos Bergervoet
    Commented Jul 4 at 8:33
  • 1
    $\begingroup$ @JosBergervoet Yes, you are right. Indeed, I meant the electric field at an arbitrary point in space. I would update the question. $\endgroup$
    – Mohammad Javanshiry
    Commented Jul 4 at 8:45

1 Answer 1

Reset to default
2
$\begingroup$

I assume the charge is uniformly distributed around the major radius for all these tori.

You are right. Shape and conductivity make smaller and smaller differences as a torus approaches a thin ring.

The electric field is produced by the distribution of charge. The charge distributions all approach the same limit. So do the electric fields.

As the size gets small, it doesn't matter what part of the cross-section contains charge. All parts approach the same limiting position.

A conductor allows the charge to redistribute itself, while a non-conductor keeps it where it was. Again, it doesn't matter where it is.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.