Gauss law for surface charge distributions

I have read that the Maxwell's equations summarize the entire classical electrodynamics, but while studying electrostatics I encountered the gauss's law for electric fields fails when the imaginary closed surface coincides with the charged surface. For example, if we want to calculate the electric field of a charged sphere with uniform surface charge density $$\sigma$$ on the surface of the sphere, we face that there is a discontinuity in the electric field. This discontinuity cannot be observed by using Gauss's law for electric fields. Thus, I observed that I can't apply Gauss law here and have to resort to Coulomb's law (I am considering the sphere to be at rest in an inertial frame). So, am I wrong in saying that Maxwell's equations are not complete. Is there any way of summarizing classical electrodynamics without encountering such exceptions where the laws do not hold?

Here, as I am unable to solve it using Gauss's law, I solved using Coulomb's law and here is the proof.

As coulomb's law (here we can apply Coulomb's law because this is an electrostatics problem) is only valid for a point charge, I consider a tiny square element (it approaches a point as limit of its dimensions tends to 0). To find the area of this tiny area element I parameterized the spherical surface as: $$\vec r = (R\cos(\theta), R\sin(\theta)\sin(\phi), R\sin(\theta)\cos(\phi))$$. Here $$\theta$$ is the angle made by the vector and x-axis while $$\phi$$ is the angle made by the projection of the vector in y-z plane with the z-axis. So, $$\theta \in [0, \pi]$$ and $$\phi \in [0, 2\pi)$$. Then, we can say that

$$dA=\left|\frac{\partial \vec r}{\partial \theta} \times \frac{\partial \vec r}{\partial \phi}\right|d\theta d\phi=R^2 \sin(\theta) d\theta d\phi$$

After applying the limits mentioned above and performing the following integral,

$$\vec E = \iint \frac{k(\sigma\cdot dA) \vec r}{|\vec r|^3} = \frac{\sigma}{2\epsilon_0} \hat r$$

Here $$\sigma$$ refers to the surface charge density of the sphere.

• Maxwell's equations are for electro dynamic not electro static. Commented May 23 at 16:01
• But isn't electrostatic state a special case of electrodynamic state. Thus, the laws should consider the rest case to be a part of the mobile case. In other words I wanted a summary which covers both electrostatics and dynamics. Commented May 23 at 16:20
• Just because this problem can't be solved using the integral form of the Gauss law, this doesn't mean Maxwell's equations are incomplete. You made a logical jump. The problem can be solved in EM theory, either by direction integration or by using the well-known results from electrostatics about electrostatic field of charged surfaces. Maxwell's equations are incomplete for completely different reasons, mainly because they are restrictions on EM fields (and matter), but do not give boundary conditions for fields in infinity, or detailed behaviour of matter. Commented May 23 at 16:27
• It also almost goes without saying, but, of course, Maxwell's equations are a classical approximation of the actual laws of nature regarding electromagnetism which are the laws of Quantum Electrodynamics (QED) and permit some important phenomena, like quantum tunneling which is necessary for everyday electronic devices to work, which are absent from Maxwell's equations. Commented May 28 at 15:31

Before providing possible approaches:

• Maxwell equations, along with Lorentz force, are the governing equations of classical electromagnetism (treating irregularities as described below);
• Coulomb force directly comes from the combined application of Lorenz force (in steady condition) and Gauss's law for the electric field, when you deal with point charges. It adds nothing to Maxwell equations, and it's a direct consequence.

The differential/local form of Maxwell equations only hold if the fields are "regular enough", being a set of differential equations. If you want/need to use differential form of the equations, I'd try two approaches:

Approach 1. You could "make them work" using generalized functions, like Dirac's delta and steps to treat point charges, surface charges,...

Approach 2. You could 1. partition the domain in a set of regions where the fields are "regular enough"; 2. connect these regions with boundary/jump conditions across the manifolds where "irregularities" arise

Global/integral form of the equations work seamlessly even with functions not regular enough for the differential form to hold, but they usually provide you less detailed information, if you can't exploit symmetry and apply them on particular domains that give you all the information you need. These equation, for a steady domain $$V$$, read

$$\begin{cases} \oint_{\partial V} \mathbf{d} \cdot \mathbf{\hat{n}} = Q^{int}_{V} \\ \oint_{\partial S} \mathbf{e} \cdot \mathbf{\hat{t}} + \frac{d}{dt} \int_{S} \mathbf{b} \cdot \mathbf{\hat{n}} = 0 \\ \oint_{\partial V} \mathbf{b} \cdot \mathbf{\hat{n}} = 0 \\ \oint_{\partial S} \mathbf{h} \cdot \mathbf{\hat{t}} - \frac{d}{dt} \int_{S} \mathbf{d} \cdot \mathbf{\hat{n}} = I_{S} \\ \end{cases}$$

being $$Q^{int}_V$$ the total electric charge in the volume $$V$$ and $$I_S$$ the electric current through surface $$S$$.

Edit. Here you can find a possible approach, using two different models of the thin distribution of electric charge (with no detail known, except for radial symmetry and total charge), one with Dirac's delta and one with uniform charge in a finite-thickness layer close to the surface of the sphere. As you can see, if you use the Dirac's delta model, the electric field is discontinuous at the surface, so it's meaningless to look for a value there (if you don't further regularize the integral of the Dirac delta, finding the average of the values on the sides of the jump); otherwise, if you use the finite approximation, the electric field is continuous and you can find a value of the electric field for each value of $$r$$.

Edit 2. - Evaluation of the electric field using Gauss' law and generalized function. In this paragraph, I'll show you how to use a differential equation with generalized functions, as the one required to represent a surface distribution in a 3-dimenisonal domain.

Using spherical coordinates, and the assumption of spherical symmetry, here the charge density is represented by the generalized function

$$\rho(r) = \sigma \delta(r-R) \ .$$

Gauss' law $$\nabla \cdot \mathbf{e} = \frac{\rho}{\varepsilon}$$ in spherical coordinates reads

$$\frac{1}{r^2} \frac{d}{d r} \left( r^2 e_r \right) = \frac{\rho}{\varepsilon} \ ,$$

and it can be recast as

$$\frac{d}{d r} \left( r^2 e_r \right) = r^2 \frac{\rho}{\varepsilon} \ .$$

Direct integration of the last equation reads

\begin{aligned} r^2 e(r) - r_0^2 e(r_0) & = \frac{\sigma}{\varepsilon} \int_{r_0}^{r} x^2 \delta(x - R) d \, x = \\ & = \frac{\sigma}{\varepsilon} R^2 u(r-R) \ , \end{aligned}

being $$u(r)$$ the step function

$$u(r) = \begin{cases} 0 & , \quad r < 0 \\ 1 & , \quad r > 0\end{cases}$$

Remark. This is the discontinuous function that appears in the diagrams above, and that you can regularize. But if you use this "Dirac's delta"-approach this function is discontinuous and, as I've already mentioned, it's meaningless to look for a value in $$r=R$$; if you really want, you can set $$\frac{1}{2}$$ keeping in mind that it's just the average value of the jump.

Setting $$r_0 = 0$$, the analytic expression (piece-wise continuous) of the radial component of the electric field reads

$$e(r) = \frac{\sigma}{\varepsilon} \frac{R^2}{r^2} u(r-R) = \begin{cases} 0 & , \quad r < R \\ \frac{\sigma}{\varepsilon} \frac{R^2}{r^2} &, \quad r > R \ .\end{cases}$$

• As I have mentioned in the question itself. I used an imaginary surface as a sphere of radius R (for the first law) and then used that to calculate the electric field at the surface of a sphere which is also of radius R and overlaps the imaginary surface, but it doesn't work because I can't decide what to do with the charges on the surface (as we can only do something about the charges which are inside or outside the surface). Commented May 23 at 16:22
• You're right, if you model charges as a infinitely thin surface. Usually we use this model when we're not interested in the actual distribution in the thin layer close to the surface. But, as we neglect the detail in the small thickness and lump all the charges in the infinitesimal-thin layer, you should give up in looking for details in the thin layer and rely on your model. I'll edit my answer with an example of this approach Commented May 23 at 16:29
• Yeah. You're right to some extent. What I mean to say is that the sphere doesn't have thickness. If there is even an infinitesimal thickness, then you would have to visualize two spheres of radii R and R + dR where dR represents an infinitesimal length. Commented May 23 at 16:47
• Edited. Please, have a look if it's more clear now Commented May 23 at 17:09
• Yeah. The graph is extremely clear and matches with my predictions however this is the not the purpose. The purpose was to find out how we could derive these graphs using Gauss law for electric fields. Especially, the graph with discontinuity. This is because the first order partial differential equations known as the Maxwell's equations can only be applied in the case of volume charge distributions and cannot be used in case of surface or linear charge distributions (as we can't find the electric field on the distributions). Commented May 29 at 6:20

Assuming $$\sigma$$ as a mathematically surface charge density, and integrating for a point very close to the surface (outside or inside), the limit is $$E = \frac{\sigma}{2\epsilon_0}$$.

According to that "exactly zero thickness surface" approach, the field is $$\frac{\sigma}{2 \epsilon_0}$$ exactly at the surface. It falls sharply asymptotically to zero inside. And grows sharply outside until gets $$\frac{\sigma}{\epsilon_0}$$. Only then, it starts to decrease with the square of the distance.

I don't think it is right. For example, it is strange that the field outside (and very close to the surface) grows with the distance for a while before decay.

The correct procedure is to use volumetric density of charges $$\rho$$. In this case, the field is zero inside, grows sharply (for a thin shell) from zero to $$\frac{Q}{4\pi R_e^2\epsilon_0}$$ at the outside surface, and is $$\frac{Q}{4\pi r^2\epsilon_0}$$ outside.

As for a shell, $$\frac{Q}{4\pi R_e^2} = \rho \frac{4}{3}\frac{\pi(R_e^3 - R_i^3)}{4\pi R_e^2}$$ the field becomes at the external surface of a thin shell: $$E \approx \frac{\rho \Delta R}{\epsilon_0}$$ If we rename $$\rho \Delta R = \sigma$$ (surface density of charges) then the field at the external surface is:$$E \approx \frac{\sigma}{\epsilon_0}$$

All this can be obtained by using the Coulomb law, and integrating the field along the shell. The Coulomb law is the general solution for the Gauss (differential) law.

However, for a spherical symmetric case like that, it is possible to use the divergence theorem: $$\int_V \nabla\cdot\vec E = \int_S \vec E\cdot\vec n dS$$ because due to the symmetry, $$\vec E$$ is always radial and has the same magnitude for a given radius. For a point outside the sphere: $$\int_V \nabla\cdot\vec E = \frac{\rho 4\pi R^2 \Delta R}{\epsilon_0} = \frac{\sigma 4\pi R^2}{\epsilon_0} = \int_S \vec E\cdot\vec n dS = 4\pi r^2 E_r$$ $$E_r = \frac{\sigma R^2}{\epsilon_0 r^2}$$ For a point just at the surface: $$E_r = \frac{\sigma}{\epsilon_0}$$

A similar approach shows that the field is zero in the interior and grows from zero to $$E_r = \frac{\sigma}{\epsilon_0}$$ in the shell.

$$\vec E(r)$$ is a continuous function, but the gradient of the field inside the shell is sharper for thinner shells.

• Yeah. But doesn't this clearly show that as maxwell's equations are differential equations, thus they won't work for such charge distributions (surface charge distributions). Thus, I am saying that how is it that Maxwell's laws can't be applied directly? Or is there a way to apply them directly? Commented May 25 at 7:21
• We can use a similar procedure of integration, but to a shell of finite thickness and volumetric density of charges. The Coulomb law is a solution of the (differential) Gauss law. It is not an additional equation. Commented May 25 at 11:53
• Yeah. I know that these laws work for volume charge distributions. I also know that Coulomb's law is apparently a direct consequence of these laws. Then, how is it that I am able to solve the problem above by using Coulomb's law but can't solve it using Gauss's law at all. Commented May 25 at 17:23
• You can use Gauss law directly in a situation of spherical symmetry, by using the divergence theorem. The surface integral of the field equals the volume integral of the divergence, that is equal to the total charge inside. As the field is the same on the sphere surface (due to the symmetry), its value can be found easily. It is the case of your example, because the correct solution is zero inside, and $\frac{\sigma}{\epsilon_0}$ outside (and infinitely close to the surface). Your solution using integration is unfortunately wrong. Commented May 26 at 1:59
• Firstly, I am talking about the electric field on the surface. Consider a sphere of radius R. Then consider that this (hollow) sphere has a surface charge density $\sigma$. Then, I want to calculate the electric field at a distance R from the centre of the sphere. Now that I have clarified what my example suggests, I would like to understand how Gauss law is applicable if there is a discontinuity in the electric field (as mentioned in the above solution). Commented May 27 at 14:56