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hutchphd
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How is that included in the limit? Clearly they are not independent. Why should this work?
That is very interesting since it implies the matter is open to practical demonstration. Even as a thought experiment, there must be a well defined force on the patch, whatever it is, implying a well-defined field.vela said:Griffiths considers the force per unit area ##\vec f## on a patch of charge and shows that it's equal to
$$\vec f = \frac 12 (\vec E_{\rm{above}} + \vec E_{\rm{below}}) \sigma,$$ where ##\sigma## is the surface charge density and ##\vec E_{\rm{above}}## and ##\vec E_{\rm{below}}## are respectively the electric field just above and below a small patch of charge. In this case, this result would again say that the electric field at ##r=R## has magnitude ##kQ/2R^2##.
As noted in post #33 (which you Liked) it does tell you the pressure exerted on the sphere. That could have practical relevance.alan123hk said:apart from theoretical research value, there does not seem to be much practical application demand and value.
It is not clear to me that Gauss's law gives a different result from Gauss's law. After all, Coulomb's law, an experimental result, is summarized by Gauss's law. The integral form of Maxwell's equation known as Gauss's law is $$\int_S\vec E\cdot \hat n~dS =\frac{1}{\epsilon_0}\int_V \rho(r)~dV$$ whereDelta2 said:It is worth noting that the excellent analysis of post #38 by @kuruman , as well as mine of post #31, is based on Coulomb's law which is not one of Maxwell's equations. The relevant Maxwell equation is Gauss's law in integral form (plus symmetry argument) which agrees with Coulomb's law in the cases r<R and r>R.
But what Gauss's law gives for r=R? If we take a gaussian surface that is the same as the charged sphere then we have a problem: The charges lies exactly on this surface so is it considered enclosed or not?
If we consider it enclosed then Gauss's law gives ##\frac{kq}{R^2}##, if we consider it not enclosed then Gauss's law gives 0. In any case the result is different than what we get by Coulomb's law.
This mathematical formula ## \vec f = \frac 12 (\vec E_{\rm{above}} + \vec E_{\rm{below}}) \sigma,~## is that we can calculate the pressure exerted on the boundary based on the average value of the electric field strength immediately above/below the boundary, but we do not need to know the electric field strength on the boundary.haruspex said:As noted in post #33 (which you Liked) it does tell you the pressure exerted on the sphere. That could have practical relevance.
You seem to have missed the point.alan123hk said:This mathematical formula ## \vec f = \frac 12 (\vec E_{\rm{above}} + \vec E_{\rm{below}}) \sigma,~## is that we can calculate the pressure exerted on the boundary based on the average value of the electric field strength immediately above/below the boundary, but we do not need to know the electric field strength on the boundary.
According to my understanding, here we assume that all charges are distributed on this zero-thickness boundary, and try to find the electric field strength at this zero thickness boundary accurately.
Of course, there may be also a theoretical electric field strength value on this boundary, but I think it seems unnecessary in most practical engineering applications to find out what this value is.
I found a reference (http://cloud.crm2.univ-lorraine.fr/pdf/uberlandia/Estevez_Delta_Dirac.pdf) which addresses this question. It defines the Dirac Delta function askuruman said:Let's apply Gauss's law to the situation of this thread. We expand the radius of the concentric charged shell from ##a## to ##R##. Then $$E 4\pi R^2 =\frac{1}{\epsilon_0}\int_0^R \frac{Q}{4\pi R^2}\delta (r-R)4\pi r^2~dr$$How to handle the upper limit in view of the delta function in the integrand? It cannot be changed because it has to match the left-hand side. One can expand ##a## in the charge distribution to ##(R+\epsilon)## in which case the integral is zero or to ##(R-\epsilon)## in which chase the result is ##Q##. This is consistent with the results obtained from doing the Coulomb integrals in cases I and II, post #38.
That's nice but it seems like "semi-nonsense" when we apply Gauss's law like this. We decide that half of the charge is enclosed, and the other half is not enclosed when the gaussian spherical surface is right at ##r=R##. While in my opinion the logical thing to do is to either say the whole charge is enclosed, or the whole charge is not enclosed.vela said:I found a reference (http://cloud.crm2.univ-lorraine.fr/pdf/uberlandia/Estevez_Delta_Dirac.pdf) which addresses this question. It defines the Dirac Delta function as
$$\int_a^b f(x)\delta(x-x_0)\,dx = \begin{cases}
\frac 12[f(x_0^+)+f(x_0^-)] & x_0 \in (a,b) \\
\frac 12 f(x_0^+) & x_0 = a \\
\frac 12 f(x_0^-) & x_0 = b \\
0 & x_0 \not\in [a,b]
\end{cases}$$ where ##f## is piecewise continuous on ##[a,b]##. The other references I checked didn't really discuss the case where ##x_0## is one of the endpoints of the integration.
Consider this case. A charge ##Q## is located at the origin, and you use a hemispherical Gaussian surface centered on the origin. There's no flux across the flat face, and only half the flux from the charge passes through the rest.Delta2 said:That's nice but it seems like "semi-nonsense" when we apply Gauss's law like this. We decide that half of the charge is enclosed, and the other half is not enclosed when the gaussian spherical surface is right at ##r=R##. While in my opinion the logical thing to do is to either say the whole charge is enclosed, or the whole charge is not enclosed.
You present another example where we should consider half the charge. Both of the cases involve charge densities that are expressed as Dirac delta functions. Fine. I just can't digest the fact that we seem to use some sort of "Solomonian" solutions in physics, that is we can't decide if the charge is enclosed or not enclosed so we consider half the charge enclosed and half not enclosed.vela said:Consider this case. A charge ##Q## is located at the origin, and you use a hemispherical Gaussian surface centered on the origin. There's no flux across the flat face, and only half the flux from the charge passes through the rest.
Well this notion works well for all charge densities except the ones that involve Dirac delta functions.vela said:Perhaps you need to abandon the notion that a charge has to either be enclosed or not enclosed
The delta function can be defined more rigorously bykuruman said:What bothers me now is that, if we accept the results of case III or the Griffiths argument presented by @vela in post #33, the implication is that one can do a delta function integral when the upper limit is the value where the argument of the delta function vanishes: $$E 4\pi R^2 =\frac{1}{\epsilon_0}\int_0^R \frac{Q}{4\pi R^2}\delta (r-R)4\pi r^2~dr=\frac{Q}{2}.$$Can one assert that this is true for the specific case of a charged conductor but not in general? Although Griffiths's argument is based on physical grounds, the Coulomb integral in case III is a mathematical argument.
So if I understand you correctly, by choosing an appropriate sequence of functions ##\delta_n(x)##, one may get an arbitrary factor ##k##, not just ##\frac{1}{2}##, multiplying ##f(0)## after ##k## has been determined independently by some other method. This I didn't know. Thank you for pointing it out to me.vela said:When the argument of the delta function vanishes at one of the endpoints of integration, the value of the integral depends on how you define the delta function. In this problem, the factor of 1/2 makes physical sense.