Electric Field on the surface of charged conducting spherical shell

[hutchphd]

How is that included in the limit? Clearly they are not independent. Why should this work?

 

[haruspex]

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 $kQ/2R^2$.

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.

So I am convinced now, and must find a counter to my argument in post #15.

In reality, the charge is not uniform. There is a finite number of charges with gaps between. Moreover, there is no such thing as a point charge. At the quantum level they are more smeary.
Consider a negative overall charge, and one electron of it that ventures very slightly beyond radius R. Because of the smeared nature of the other charges it does not immediately feel the full force of $kQq/R^2$. Rather, the force will increase smoothly, reaching that over an atomic or subatomic distance.
Likewise, venturing to a shorter radius will not immediately reduce the force to zero.
Thus, the flaw in my reasoning in post #15 is assuming that the usual expressions for the internal and external fields are exact for even the slightest deviation from radius R.

 

[kuruman]

I think I have managed to explain to myself what has been said here so far to my satisfaction. I present it below to summarize my ideas for anyone who might be interested.

Start with $d\vec E=\dfrac{k dq(\vec r-\vec r')}{|\vec r-\vec r')|^3}$. Consider an element on the sphere at $\theta$ and $\phi$. The point of observation is on the $z$-axis. Then
$\vec r = z~\hat z$
$\vec r' = R \sin\theta \cos\phi~\hat x+R \sin\theta \sin\phi~\hat y+ R \cos\theta ~\hat z$
$dq=\sigma R^2\sin\theta~ d\theta d\phi$
$(\vec r-\vec r')=(-R \sin\theta \cos\phi~\hat x-R \sin\theta \sin\phi~\hat y+(z- R \cos\theta) ~\hat z)$
$|\vec r-\vec r'|=(R^2+z^2-2zR\cos\theta)^{1/2}$
$d\vec E=\dfrac{k \sigma R^2\sin\theta (-R \sin\theta \cos\phi~\hat x-R \sin\theta \sin\phi~\hat y+(z- R \cos\theta) ~\hat z)}{(R^2+z^2-2zR\cos\theta)^{3/2}}d\theta~ d\phi$
Integration over $\phi$ results in an axial contribution only,
$dE_z=\dfrac{2 \pi k \sigma R^2\sin\theta (z- R \cos\theta)}{(R^2+z^2-2zR\cos\theta)^{3/2}}d\theta.$
The theta integral is done using the standard substitution $\int_0^{\pi}f(\cos\theta)\sin\theta~d \theta \rightarrow \int_{-1}^{+1}f(u)~du.$ Then
$$E_z=2 \pi k \sigma R^2\int_{-1}^{+1}\frac{(z- R u)}{(R^2+z^2-2zRu)^{3/2}}du.$$The integral can be easily done by a second substitution, $y=R^2+z^2-2zRu$.$$\begin{align}\int_{-1}^{+1}\frac{(z- R u)}{(R^2+z^2-2zRu)^{3/2}}du & =\left. \frac{u z-R}{z^2 \sqrt{R^2-2 R u z+z^2}}\right |_{-1}^{+1} \nonumber \\ &=\frac{z-R}{z^2 \sqrt{R^2-2 R z+z^2}}-\frac{-R-z}{z^2 \sqrt{R^2+2 R z+z^2}}\nonumber \\ & =\frac{z-R}{z^2 \sqrt{(R-z)^2}}+\frac{R+z}{z^2 \sqrt{(R+z)^2}}.\nonumber\end{align}$$ Thus, $$\begin{align}E_z=\frac{2 \pi k \sigma R^2}{z^2}\left[\frac{z-R}{ \sqrt{(R-z)^2}}+\frac{R+z}{ \sqrt{(R+z)^2}}\right]=\frac{2 \pi k \sigma R^2}{z^2}\left[\frac{z-R}{ \sqrt{(R-z)^2}}+1\right]\end{align}$$ We distinguish three cases.
Case I: $R >z$, inside the shell.
Then $\sqrt{(R-z)^2}=R-z$ in which case $$E_z=\frac{2 \pi k \sigma R^2}{z^2}\left[\frac{z-R}{ R-z}+1\right]=\frac{2 \pi k \sigma R^2}{z^2}\left[-\frac{R-z}{ R-z}+1\right]=0$$ as expected.

Case II: $z >R$, outside the shell.
Then $\sqrt{(R-z)^2}=z-R$ in which case $$E_z=\frac{2 \pi k \sigma R^2}{z^2}\left[\frac{z-R}{ z-R}+1\right]=\frac{4 \pi k \sigma R^2}{z^2}=\frac{\sigma R^2}{\epsilon_0 z^2}.$$When we set $z=R$, $E_z=\dfrac{\sigma}{\epsilon_0}.$

Case III: $z =R$, on the shell.
Then $$E_z=\frac{2 \pi k \sigma R^2}{R^2}\left[\frac{R-R}{ \sqrt{(R-R)^2}}+1\right]=2 \pi k \sigma\left[\frac{0}{ \sqrt{(0)^2}}+1\right]=\text{undefined}$$
The article by F.M.S. Lima (reference in post #26) sets $z=R$ before doing the integral. In that case, $$\int_{-1}^{+1}\frac{(z- R u)}{(R^2+z^2-2zRu)^{3/2}}du~\rightarrow~\int_{-1}^{+1}\frac{(R-Ru)}{(R^2+R^2-2R^2u)^{3/2}}du=\frac{1}{2^{3/2}R^2}\int_{-1}^{+1}\frac{(1-u)}{(1-u)^{3/2}}du.$$This last integral is problematic if we allow the upper limit, corresponding to $\cos0$, be equal to 1. This is where the "hole" comes in. We set the upper limit equal to $1-\epsilon~~(\epsilon > 0)$ and proceed. $$\frac{1}{2^{3/2}R^2}\int_{-1}^{1-\epsilon}\frac{(1-u)}{(1-u)^{3/2}}du=\frac{1}{2^{3/2}R^2}\int_{-1}^{1-\epsilon}\frac{du}{(1-u)^{1/2}}=\left .-\frac{2\sqrt{1-u)}}{2^{3/2}R^2}\right |_{-1}^{1-\epsilon}=\frac{1}{R^2}\left(1-\sqrt{\frac{\epsilon}{2}}\right).$$We multiply by the constant and close the "hole" by setting $\epsilon=0$ to get the electric field at $R$, $$E_z=2 \pi k \sigma R^2 \frac{1}{R^2}\left(1-\sqrt{\frac{\epsilon}{2}}\right)= \frac{ \sigma}{2\epsilon_0}.$$Notably, Equation (1) gives the electric field everywhere. If we strip the constants and consider a unit sphere, the functional dependence on $z$ can be rewritten as $$f(z)=\frac{1}{2z^2}(\text{Sgn}[z-1]+1)$$ producing the plot shown below.

We also note that without the $z^{-2}$ dependence we have a shifted Heaviside step function, $h(z)=\frac{1}{2}(\text{Sgn}[z-1]+1)$ that is equal to zero for $z<1$ and equal to 1 for $z>1.$ Thus, the value of $\frac{1}{2}$ that multiplies the value $\frac{\sigma}{\epsilon_0}$ of the field arbitrarily close to the outer surface of the shell is the Heaviside half-maximum.

I have explained this to myself in a non-mathematical way as follows: Let's say that, for some reason e.g. quantum effects or discreteness of charges, the transition of the electric field from zero (just inside) to $\frac{\sigma}{\epsilon_0}$ (just outside) is not infinitely sharp. Then it can be approximated by some function, that goes smoothly from zero to $\frac{\sigma}{\epsilon_0}$. It is reasonable to say that, by symmetry, the point that is neither inside nor outside the "surface" is the inflection point of the smoothing function, i.e. the Heaviside half-maximum (see Wikipedia article).

 

[hutchphd]

This is the Cauchy principal value argument as I recall...works for me!

 

[Delta2]

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.

 

[alan123hk]

This discussion is very interesting, there are many excellent mathematical arguments
But if my understanding is correct, this is to find the electric field strength in a zero-thickness and zero-volume space, then another feeling of mine is that, apart from theoretical research value, there does not seem to be much practical application demand and value.

 

[haruspex]

apart from theoretical research value, there does not seem to be much practical application demand and value.

As noted in post #33 (which you Liked) it does tell you the pressure exerted on the sphere. That could have practical relevance.

 

[kuruman]

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.

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 $$\int_S\vec E\cdot \hat n~dS =\frac{1}{\epsilon_0}\int_V \rho(r)~dV$$ where
$S$= a closed surface and $\hat n$ an outward normal to it;
$V$ = the volume enclosed by the surface.
Specifically, if $S$ is a sphere of radius $R$, the radial part of the volume integral on the right has an upper limit $r=R$. Now consider a Gaussian spherical surface of radius $R$ with a concentric, conducting thin shell of radius $a$ inside it ($a<R$) having total charge $Q$ evenly distributed on its surface. The volume charge distribution is $\rho(r) = \frac{Q}{4\pi a^2}\delta (r-a)$. When we apply Gauss's law to this, we use the standard symmetry arguments to write the left-hand side of the equation as $E 4\pi R^2$ regardless of what's on the right-hand side. Here $E$ is the radial component of the field at $r=R$. Thus, $$E 4\pi R^2 =\frac{1}{\epsilon_0}\int_0^R \frac{Q}{4\pi a^2}\delta (r-a)4\pi r^2~dr=Q$$Note that if we shrink $a$ to zero, we have a point charge at the origin and an equation that gives the electric field at distance $R$ from it according to Coulomb's law. Thus, doing the Coulomb integrals to find the electric field at the surface is a superposition of a result obtained from Gauss's law and cannot give a different result from Gauss's law.

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.

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.

 

[alan123hk]

As noted in post #33 (which you Liked) it does tell you the pressure exerted on the sphere. That could have practical relevance.

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.

 

[TSny]

Using energy concepts you can argue for $E =\large \frac{kQ}{2R^2}$ at the surface.

The electrostatic energy $U$ in the system can be calculated as $$U =\frac {\varepsilon_0}{2} \int E^2 d\tau$$ where the integration is over all of space. Evaluating this for the uniformly charged spherical surface gives $$U = \frac{kQ^2}{2R}$$ If we allow the sphere to expand an infinitesimal amount $dR$, the energy change $dU$ must equal the negative of the work $dW$ done by the electric force acting on the surface charge. Let $f$ be the force per unit area acting on the surface charge. Then, $-dU = dW$ gives us $$\frac{kQ^2}{2R^2}dR = 4 \pi R^2 f dR$$ $$f = \frac{kQ^2}{(2R^2)(4\pi R^2)} = \frac{kQ}{2R^2}\sigma$$ Since $f$ should equal $\sigma E_{\rm at \, surf}$, we identify $E_{\rm at \, surf} = \large \frac{kQ}{2R^2}$.

 

[haruspex]

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.

You seem to have missed the point.
Consider e.g. a charged bubble. The repulsion results in outward pressure on the bubble, increasing its radius beyond the uncharged version. This is in principle measurable.

 

[vela]

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.

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.

 

[Delta2]

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.

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]

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.

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]

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.

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.

What if the gaussian surface is 3/4 of a sphere, again centered at the point charge, now we should consider 3/4 of the charge in order for Gauss's law to hold?

 

[vela]

Perhaps you need to abandon the notion that a charge has to either be enclosed or not enclosed. It can be neither, just like 0 is neither positive nor negative.

 

[Delta2]

Perhaps you need to abandon the notion that a charge has to either be enclosed or not enclosed

Well this notion works well for all charge densities except the ones that involve Dirac delta functions.

 

[vela]

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.

The delta function can be defined more rigorously by
$$\int_{-\infty}^\infty f(x)\delta(x-a)\,dx = \lim_{n\to\infty} \int_{-\infty}^\infty f(x)\delta_n(x-a)\,dx = f(a)$$ for a suitably chosen sequence of functions $\delta_n(x)$. (See for example https://dlmf.nist.gov/1.17 and https://mathworld.wolfram.com/DeltaSequence.html.) Most of the examples on the MathWorld page are symmetric about $x=0$, but symmetry isn't a requirement. For example,
$\delta_n(x) = n[u(x)-u(x-1/n)]$ is a perfectly valid delta sequence.

If we were to use the first example on the MathWorld page, we could say
\begin{align*}
\int_0^\infty f(x)\delta(x)\,dx &= \lim_{n\to\infty} \int_0^\infty f(x)\delta_n(x)\,dx \\
&= \lim_{n\to\infty} n \int_0^{1/2n} f(x)\,dx \\
&= \frac 12 f(0)
\end{align*} But if we used $\delta_n(x) = n[u(x)-u(x-1/n)]$ to evaluate the same integral, we'd get
\begin{align*}
\int_0^\infty f(x)\delta(x)\,dx &= \lim_{n\to\infty} \int_0^\infty f(x)\delta_n(x)\,dx \\
&= \lim_{n\to\infty} n \int_0^{1/n} f(x)\,dx \\
&= f(0)
\end{align*} 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.

 

[alan123hk]

Dose it mean that we can understand this way, even if the electron is so small, the thickness of the layer accumulated by the electrons on the surface of the spherical conductor will not be zero, so the position of the boundary of the conductor should be defined according to the distribution of charge. If we assume the gaussian surface contains half of the total charge, then the electric field will become $\frac{kQ}{2R^2}$ ?

 

[kuruman]

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.

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.