Conditions for applying Gauss' Law

In summary, Gauss' Law can be applied under specific conditions, including the presence of a symmetric charge distribution (spherical, cylindrical, or planar symmetry), where the electric field can be easily calculated. The law is most effective in regions where the electric field is uniform or can be approximated as such. Additionally, it is essential that the surface used for the application of Gauss' Law is closed and encloses the charge, allowing for the calculation of the total electric flux through the surface.
  • #1
jv07cs
44
2
To apply the Divergence Theorem (DT), at least as it is stated and proved in undergrad calculus, it is required for the vector field ##\vec{F}## to be defined both on the surface ∂V, so that we can evaluate the flux through this surface, and on the volume V enclosed by ∂V, so that we can evaluate the integral of ##\nabla\cdot\vec{F}## on V. In electrostatics, we use Gauss' Law (GL), which comes directly from the DT.

From my understanding, to apply Gauss' Law, we would then have to require for ##\vec{E}## to be defined both on the surface ∂V and on the volume V. If ##\vec{E}## is not defined on V, we can get around this by using the dirac delta function. But what if ##\vec{E}## is not defined on the boundary ∂V?

More specifically, I would like to ask about two examples. When we consider a uniformly charged sphere of radius ##R##, for example, and we use Gauss' Law to calculate ##\vec{E}## inside the sphere, drawing a Gaussian Sphere ##V_0## with boundary ##\partial V_0## and radius ##r < R##. To apply the theorem in the first place, we needed to know that ##\vec{E}## is indeed defined inside the sphere, which is not necessarily obvious from Coulomb's Law
Screenshot from 2024-07-25 07-30-33.png

as there are singularities since cursive r is on the denominator. So we would need to know that this integral converges. How can I be sure of this before applying Gauss' Law? Does it come from the fact that ##\nabla\cdot\vec{E} (\vec{r}) = \rho(\vec{r})/\epsilon_0##? Since the divergence of ##\vec{E}## inside the distribution is defined, as ##\rho(\vec{r})## is defined, then ##\vec{E}(\vec{r})## must also be defined.

The other question is regarding the uniformly charged infinite plane, for example. In this case, we use a pillbox, with volume ##V_0## and boundary ##\partial V_0##, as the gaussian surface and, applying Gauss' Law to calculate the field at a height ##z## and then taking the limit ##z\to 0##, we find out that the field is not defined on the plane, there is a discontinuity. If the field is not defined on the plane and the pillbox pierces the plane, then the field is not defined everywhere on the surface of the pillbox, that is ##\vec{E}## is not defined on ##\partial V_0##. From the requirement that the field is defined on ##\partial V_0## for the DT to be applied, we could not apply the DT here. Mathematically speaking, what allows us to apply Gauss' Law here?
 
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  • #2
jv07cs said:
Does it come from the fact that ∇⋅E→(r→)=ρ(r→)/ϵ0?
I think that finite ##\rho## assures it.
jv07cs said:
The other question is regarding the uniformly charged infinite plane, for example.
[tex]E_z(z)=\frac{\sigma}{2\epsilon_0}sgn(z)[/tex]
Ambiguity of
[tex]sgn(z):=2(H_a(z)-1/2)[/tex]
, where usually a =1/2 but another value in general, at z=0 would not show difficulty in so much for physics.

[EDIT] Distribution of charge area density
[tex]\sigma(z)=\sigma \delta(z)[/tex]
With a tricky relation
[tex]\int_0^{+\epsilon} \delta(z) =\frac{1}{2}[/tex]
Applying Gauss's law for [0,+\epsilon] pillbox, we might be able to say
[tex]E_z(0)=0[/tex]
which means a=1/2 as usual.
 
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