Conditions for applying Gauss' Law

  • #1
jv07cs
40
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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