Chapter 2 Griffiths EM Problem: E-Field from a charged ring

In summary: Cartesian basis vectors. In fact, it is not even clear what ##\hat r## would mean outside the integral as the point where it would be considered would be ambiguous.Edit: What you can do is to express the curvilinear basis in a Cartesian basis and then take the Cartesian basis vectors out of the integral as they are constant.
  • #36
For this problem, the symmetry argument is a mathematical one, in that for every component of ## \vec{E} ## that points radially outward, there is a corresponding point on the opposite side of the ring where the radially outward component will cancel it. The other way of looking at the symmetry is to say that if the resultant ## \vec{E} ## has a radially outward component, there is no favored direction, and thereby the radially outward component must be zero. I like the first approach better, but in any case, it pays to be able to use all the tools that are available in solving problems, and symmetry is one of those tools.
 
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  • #37
warhammer said:
So what I infer from your response here is that computation of field is a much better option than Potential since the former delivers more information as compared to the latter or did you mean something else? 😅
For my money both involve invoking symmetry. If symmetry is invoked at the appropriate time in each I don't see that finding the potential vs. finding the field directly makes much difference.
Usually the potential is easier.
.
 
  • #38
Charles Link said:
For this problem, the symmetry argument is a mathematical one
Symmetry arguments are always mathematical, even if not always presented as such.

Charles Link said:
in that for every component of E→ that points radially outward, there is a corresponding point on the opposite side of the ring where the radially outward component will cancel it. The other way of looking at the symmetry is to say that if the resultant E→ has a radially outward component, there is no favored direction, and thereby the radially outward component must be zero. I like the first approach better, but in any case, it pays to be able to use all the tools that are available in solving problems, and symmetry is one of those tools.
The formal symmetry argument is that ##\vec E## is a vector under rotations and that rotations about the z-axis are a symmetry of the problem. Any component perpendicular to the z-axis would therefore rotate into a different direction under such a rotation. As the solution is unique, the perpendicular component must therefore be equal to zero because it is the only way that a vector can be equal to its rotation about an axis perpendicular to it.

This argument works not only for a circle, but also for any number of equal point charges evenly distributed around the circle.
 
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  • #39
kuruman said:
Yes, I don't particularly like Griffiths's "curly r" notation so I tend to ignore it when it raises its head. I stand corrected.

kuruman said:
It is appropriate to ask here but the answer you will get from me is that Griffiths's "curly r" is the main reason I did not adopt the book when I taught intermediate E&M.
That's really interesting. Having taught intermediate EM from Griffiths many moons ago I immediately picked up on the difference in notation. Not sure I agree with your concern but I very much like the book otherwise.

/
 
  • #40
Orodruin said:
Symmetry arguments are always mathematical, even if not always presented as such.
Perhaps I should have referred to it as arithmetic or computational. The one symmetry method essentially is computing the integral, and summing it by adding opposites of the radial component together, and then sweeping half the ring, getting zero for the integral of the radial component. When the opposites are present in such a fashion, (writing this for the OP @warhammer ), it is zero by inspection.
 
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  • #41
warhammer said:
Edit- I think due to confusions that arise on curly r usage, the curly r is merely a separation vector no? Whereas the \hat r that you have described above is a basis vector thing. Griffiths goes like curly \hat r=frac{\vec r} {r}
Right. Curly r is the displacement vector from the infinitesimal charge to the point of interest. If you put a hat on the vector, that denotes the unit vector in the same direction.

warhammer said:
Edit 2- @vela I think I have another confusion to request guidance on. You have used cylindrical system in your response and then expressed it in cartesian basis vectors through the transformations. What I have based my response on is the assertion that the circle is contained within xy plane with center as origin and z axis perpendicular through it; the source element "on" circle represented as ##x* \hat x + y* \hat y## (because the circle too is on the xy plane and it must have some locative abscissa and ordinate). Then the destination is a point on the z axis which is represented by ##z* \hat z##. Using this destination and source terminology I based my "curly r" vector above. Is this approach incorrect, and if yes then if you could please also guide how and why...
I wouldn't say it's incorrect as long as it's understood that your ##x## and ##y## refers to a point on the circle (as opposed to some other point in the ##xy##-plane). If you write ##r \cos\phi \,\mathbf{\hat x} + r \sin\phi\, \mathbf{\hat x}##, it's clear that the point in question lies on a circle in the ##xy##-plane centered at the origin on a circle of radius ##r##. It's just two different ways to represent the same idea.

That said, because of the symmetry in the problem, cylindrical or spherical coordinates are the natural choice. You can, of course, express everything in Cartesian coordinates initially, but I don't see any advantage to doing so.
 
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  • #42
warhammer said:
This I believe satisfies what I set out to achieve, a brutely mathematical way to show the cancellation rather than symmetry usage blindly.
It's brute force integration that is the blind approach. Using symmetry shows insight.
 
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  • #43
warhammer said:
I think this habit is extremely unhealthy for someone who wishes to keep continuing with Physics, because Physics is not Mathematics
Mathematicians use symmetry too!
 
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  • #44
vela said:
That said, because of the symmetry in the problem, cylindrical or spherical coordinates are the natural choice. You can, of course, express everything in Cartesian coordinates initially, but I don't see any advantage to doing so.
That is exactly how I was taught to set up electrostatic and magnetostatic problems. It eliminates mistakes early on.
 
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  • #45
vela said:
That said, because of the symmetry in the problem, cylindrical or spherical coordinates are the natural choice. You can, of course, express everything in Cartesian coordinates initially, but I don't see any advantage to doing so.
Absolutely. It makes no sense to use Cartesian Coordinates when we have cylindrical or spherical symmetry involved. However I think by your response I have put to bed a few conceptual doubts; extremely thankful to you!
 
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  • #46
Well, if it comes to the subtleties of singularities it's better to translate back to Cartesian coordinates, because they don't introduce additional coordinate singularities as cylindrical and spherical coordinates do (in the standard setup along the ##z## axis). The problem at hand can be simply solved by calculating the electrostatic potential along the ##z##-axis and then take the derivative to get the field or directly the field. For the potential you get for ##\vec{r}=(0,0,z)##
$$\Phi(\vec{r}) = \frac{\lambda}{4 \pi \epsilon_0} \int_0^{2 \pi} \mathrm{d} \varphi a \frac{1}{\sqrt{z^2+a^2}}=\frac{\lambda a}{2 \epsilon_0 \sqrt{z^2+a^2}},$$
and thus
$$E_z=-\partial_z \Phi=\frac{\lambda a z}{2 \epsilon_0 (z^2+a^2)^{3/2}}.$$
 
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