Electric potential continuous at boundaries?

  • Thread starter CheMech
  • Start date
  • #1
6
0
Why is it that we assume electric potential to be continuous across boundaries in electrostatics problems (like, say we have a situation with concentric spheres with different equations for electric field across boundaries)? This is the case as far as I've seen at least. I am in introductory E&M right now, and I've searched around for answers to this question with no avail.

The only answers I've seen are that discontinuous potential across a boundary implies an infinite electric field at that point (without further elaboration) and that it has something to do with Laplace's equation, which I have not learned anything about. Is there a sort of simpler answer that can be explained at an introductory physics-for-engineers class level?

Help is appreciated! :)
 

Answers and Replies

  • #2
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,979
7,283
First begin with the electromagnetic fields since these are the really observable quantities. In your case, we have to deal with the electric field in the static case. For static fields the Maxwell equations decouple into equations for the electric and the magnetic field. So you can treat them independently from each other.

The electrostatic equations read

[tex]\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E}=\frac{1}{\epsilon} \rho,[/tex]

where [itex]\epsilon[/itex] is the dielectric constant of the material (which I assume to be (piecewise) homogeneous and isotropic for simplicity) and [itex]\rho[/itex] is the charge density.

These equations are local equations since they only contain differentials of the fields, and you have to be careful at boundaries since there the field can have singularities. However, you have to remember the definitions of the vector operations, here the curl and the divergence of a vector field, through integrals to give these operations clear meanings when you consider the neighborhood of a boundary (where, e.g., two dielectrics meet; or you have conductors around etc.).

The curl tells you to take a line integral along some closed path and the enclosed surface element. To investigate the neighborhood of a boundary surface, you chose a rectangular path with two sides tangent to the surface and two very small sides perpendicular to the surface. Let's call the enclosed rectanglular surface [itex]S[/itex] and its boundary [itex]\partial S[/itex]. Then Stoke's Law tells you

[tex]0=\int_S \mathrm{d} \vec{F} \cdot \vec{\nabla} \times \vec{E}=\int_{\partial S} \mathrm{d} \vec{x} \cdot \vec{E}.[/tex]

Now you can neglect the contributions from the two perpendicular pieces of the rectangle since we make them infinitesimally small. Then the above integral tells you that the components of [itex]\vec{E}[/itex] tangent to the boundary must be continuous.

A similar argument can be made with Gauß's Theorem and a little prism parallel to the boundary. That tells you that the normal component of [itex]\epsilon \vec{E}[/itex] has a jump equal to the surface charge along the boundary.

Now that we know the behavior of the electric field across boundaries, where the matter can change discontinuously (i.e., either the dielectric constant changes or you have a conductor, and their may be some surface charges sitting on the boundary etc.), we can see what this means for the electric potential. Since electrostatic fields are curl free according to the first equation of electrostatics this means that (at least locally) there exists a scalar potential, [itex]\Phi[/itex], such that

[tex]\vec{E}=-\vec{\nabla} \Phi.[/tex]

Plugging this into the second equation of electrostatics gives

[tex]\Delta \Phi:=\vec{\nabla} \cdot \vec{\nabla} \Phi=-\frac{\rho}{\epsilon}.[/tex]

The latter is known as Poisson's Equation (for [itex]\rho=0[/itex] the Laplace equation). Now since the components of [itex]\vec{E}[/itex] tangent to the boundary are continuous and the normal components have at most finite jumps across the boundary, the potential itself must be continuous across the boundary. That is so because, suppose you know the electric field, the potential is given (up to an uninteresting constant) by the line integrals connecting a fixed point, [itex]\vec{x}_0[/itex], with the point under consideration, [itex]\vec{x}[/itex] (let's call this lines [itex]C(\vec{x}_0 \rightarrow \vec{x})[/itex]):

[tex]V(\vec{x})=-\int_{ C(\vec{x}_0 \rightarrow \vec{x})} \mathrm{d} \vec{x} \cdot \vec{E}.[/tex]

The first equation of electrostatics guaranties that the value of the potential is independent of the particular line chosen (as long as the considered region in space is simply connected). Since [itex]\vec{E}[/itex] has at most finite jumps in the normal component across the boundary, [itex]V[/itex] thus must be continuous.
 
Last edited:
  • Like
Likes poseidon721
  • #3
6
0
Thank you for the reply! I don't think I understand all of it though. I feel that my background in math isn't strong enough to fully understand your explanation, though I did understand bits and pieces of it. In particular...

The curl tells you to take a line integral along some closed path and the enclosed surface element. To investigate the neighborhood of a boundary surface, you chose a rectangular path with two sides tangent to the surface and two very small sides perpendicular to the surface. Let's call the enclosed rectanglular surface [itex]S[/itex] and its boundary [itex]\partial S[/itex]. Then Stoke's Law tells you

[tex]0=\int_S \mathrm{d} \vec{F} \cdot \vec{\nabla} \times \vec{E}=\int_{\partial S} \mathrm{d} \vec{x} \cdot \vec{E}.[/tex]

Now you can neglect the contributions from the two perpendicular pieces of the rectangle since we make them infinitesimally small. Then the above integral tells you that the components of [itex]\vec{E}[/itex] tangent to the boundary must be continuous.
We have just learned Stoke's Theorem in my math class, and for now what I understand is that it relates the line integral of a vector field over the boundary of a surface to the surface integral of the curl of F over the boundary of the surface. In symbols:

[tex]\int_{\partial S}\vec{F} \cdot d\vec{r} = \int \int_{S} curl \vec{F} \cdot d\vec{S}[/tex]

I don't quite understand how from your usage of Stoke's Theorem you reach the conclusion that the components of [itex] \vec{E}[/itex] tangent to the boundary must be continuous.


ince electrostatic fields are curl free according to the first equation of electrostatics this means that (at least locally) there exists a scalar potential, [itex]\Phi[/itex], such that

[tex]\vec{E}=-\vec{\nabla} \Phi.[/tex]
This is due to the fact that conservative/gradient vector fields are continuous, correct?


Now since the components of [itex]\vec{E}[/itex] tangent to the boundary are continuous and the normal components have at most finite jumps across the boundary, the potential itself must be continuous across the boundary. That is so because, suppose you know the electric field, the potential is given (up to an uninteresting constant) by the line integrals connecting a fixed point, [itex]\vec{x}_0[/itex], with the point under consideration, [itex]\vec{x}[/itex] (let's call this lines [itex]C(\vec{x}_0 \rightarrow \vec{x})[/itex]):

[tex]V(\vec{x})=-\int_{ C(\vec{x}_0 \rightarrow \vec{x})} \mathrm{d} \vec{x} \cdot \vec{E}.[/tex]
I understand the second statement that the line integral of the electric field over a path gives the electric potential. However, I don't understand where you reached the conclusion that the normal components of the electric field over a boundary have at most finite jumps.


...the potential itself must be continuous across the boundary. That is so because, suppose you know the electric field, the potential is given (up to an uninteresting constant) by the line integrals connecting a fixed point, [itex]\vec{x}_0[/itex], with the point under consideration, [itex]\vec{x}[/itex] (let's call this lines [itex]C(\vec{x}_0 \rightarrow \vec{x})[/itex]):

[tex]V(\vec{x})=-\int_{ C(\vec{x}_0 \rightarrow \vec{x})} \mathrm{d} \vec{x} \cdot \vec{E}.[/tex]

The first equation of electrostatics guaranties that the value of the potential is independent of the particular line chosen (as long as the considered region in space is simply connected). Since [itex]\vec{E}[/itex] has at most finite jumps in the normal component across the boundary, [itex]V[/itex] thus must be continuous.
I'm also confused with this bit here. I understand that the value of the potential is path independent due to the fact that the electric field is a gradient vector field. However, once again, I don't understand how the mathematical argument presented results in the conclusion you have reached. How does the electric field having at most finite jumps in the normal component across the boundary result in V being continuous?

Thanks!
 

Related Threads on Electric potential continuous at boundaries?

  • Last Post
Replies
6
Views
3K
Replies
16
Views
17K
Replies
6
Views
2K
Replies
3
Views
726
Replies
6
Views
2K
Replies
28
Views
40K
Replies
2
Views
3K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
10
Views
2K
Top