What Are the Electric Field Components at a Charge Density Boundary?

In summary, the conversation discusses the boundary conditions for electrostatics, specifically the equations for the electric field components above and below a boundary and the question of whether the electric field in these equations is the sum of the external field and the field due to the charge at the boundary. The conversation also addresses the misconception of multiple electric fields and clarifies that there is only one electric field, which includes all contributions. The answer to the question is that both the external field and the contribution from the surface charge exhibit a discontinuity in the electric field across the boundary.
  • #1
Sebas4
13
2
Homework Statement
Boundary conditions electrostatics.
Relevant Equations
Meaning of the electric field variables in the boundary condition equations.
Hey, I have a really short question about electrostatics.
The boundary conditions are :
[tex] \mathbf{E}^{\perp }_{above} - \mathbf{E}^{\perp}_{below} = -\frac{\sigma}{\varepsilon_{0}}\mathbf{\hat{n}} [/tex],
[tex] \mathbf{E}^{\parallel }_{above} = \mathbf{E}^{\parallel}_{below}[/tex].

My question is what is [itex] \mathbf{E}^{\perp }_{above} [/itex], [itex] \mathbf{E}^{\perp }_{below} [/itex],
[itex] \mathbf{E}^{\parallel }_{above}[/itex] and [itex]\mathbf{E}^{\parallel}_{below}[/itex], is it the total electric field component near the boundary?
So is the electric field in this equation the sum of the external field and the electric field due to the charge at the boundary?

I will try to explain my question with an example.
Let's say we have an infinite plane with homogeneous charge density [itex]\sigma[/itex].
The electric field above the plane
[tex] \mathbf{E} = \frac{\sigma}{2\varepsilon_{0}}\mathbf{\hat{z}} [/tex].
The electric field below the plane is
[tex] \mathbf{E} = - \frac{\sigma}{2\varepsilon_{0}}\mathbf{\hat{z}} [/tex].
We have a homogeneous external field pointing in the z-direction, [itex]\mathbf{E}_{external} = \mathbf{E}_{0} \mathbf{\hat{z}}[/itex].
The electric field just below the surface of the plane is
[tex] \mathbf{E}_{total below} = \left(\mathbf{E}_{0} - \frac{\sigma}{2\varepsilon_{0}}\right) \mathbf{\hat{z}} [/tex].
The electric field just above the surface of the plane is
[tex] \mathbf{E}_{total above} = \left(\mathbf{E}_{0} + \frac{\sigma}{2\varepsilon_{0}}\right) \mathbf{\hat{z}} [/tex].
If we plug this in, in the boundary condition we get
[tex] \mathbf{E}_{total above} - \mathbf{E}_{total below} = \left(\mathbf{E}_{0} + \frac{\sigma}{2\varepsilon_{0}}\right) \mathbf{\hat{z}} - \left(\mathbf{E}_{0} - \frac{\sigma}{2\varepsilon_{0}}\right) \mathbf{\hat{z}} = \frac{\sigma}{\varepsilon_{0}} \mathbf{\hat{z}} [/tex].
This is true, according to the boundary condition.

I have also another question, this also works for non-homogeneous charge density boundaries? (I think so).
 
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  • #2
Sebas4 said:
Homework Statement:: Boundary conditions electrostatics.
Relevant Equations:: Meaning of the electric field variables in the boundary condition equations.

My question is what is Eabove⊥, Ebelow⊥,
Eabove∥ and Ebelow∥, is it the total electric field component near the boundary?
So is the electric field in this equation the sum of the external field and the electric field due to the charge at the boundary?
Yes. Total vector field
Notice the external field here does not really enter into the boundary condition because of the geometry. But the E field is always the total E field (or its components).
 
  • #3
It is a common misconception that there are several electric fields. This most likely stems from the fact that the equations governing electromagnetism (Maxwell’s equations) are linear, which means that solutions to them can be superpositioned and it is therefore easy to colloquially say things like ”the electric field of charge A” when what would be more precise would be ”the contribution to the electric field from charge A” (which is more of a mouthfull). However, there is no way to independently measure such a contribution.

The above also means that the answer to your question is ”both”. Both the electric field and the contribution from the surface charge will exhibit this discontinuity because all other contributions will be continuous across the surface and so any discontinuity must arise from the surface charge.
 
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