Dielectric boundaries and symmetry

[mmwave]

I have been reviewing the method of images for electrostatic problems and noticed the following:

Textbooks point out that to use Gauss' Law successfully there must be geometric symmetry that makes the Efield constant over some reasonable surface. I thought this was just to make the math tractable but then I noticed the following issue.

1. Point charge inside a thin spherical shell of a different dielectric constant. You can calculate the Efield inside without any knowledge of the shell using Gauss' Law.

2. Point charge near a planar dielectric boundary. You must know the distance from the charge to the boundary and both dielectric constants to calculate the Efield on the charge side of the boundary.

In one case you only need to know about the region of interest, in the second case you need to know everything. Is symmetry alone the difference?

For the spherical shell I can see that the field due to any polarization charge will 'cancel out' inside the shell and so the Efield is the same as if there were no shell. What about other cases? How about an infinite planar charge and a parallel dielectric boundary?

 

[sdeliver645]

Inquiry

Hi mmwave:

"...point charge inside a thin spherical shell of a different dielectric constant. You can calculate the Efield inside ..."

Allow me give my explanation of why the E field inside a sphere is zero. It is essentially because the E field scales as 1/r^2 whereas the "area", or "solid angle" you are exposed to scales as r^2. Let me try and clarify.

{
When inside a spherical shell, draw any cone (both nappes) centered on your position. The nappe of the cone intersecting the surface closest to your position will enclose a small area, and the nappe opposite to that will enclose a larger area. The E field generated at your position due to both these areas will be such that they will EXACTLY cancel. So, symmetry (specifically, the spherical symmetry of the shell (r^2) and the spherical symmetry of Gauss' Law (1/r^2) is responsible for the net effect.
}

"...how about an infinite planar charge and a parallel dielectric boundary"

The electric field of an infinite plane is a constant. It does not depend on the distance to the plane. If the charge is in some dielectric medium (i.e. epsilon != epsilon_0), simply replace epsilon_0 with epsilon.

"...the second case [infinite plane] you need to know everything."

Not quite. Please clarify.

---
I am sure I have misunderstood your question. Your question seems to be asking a deeper question that I am not getting. Please try and re-phrase it (P.S. make quotes from Griffiths or Jackson if that is what you are reading).

 

[Graeme Robinson]

Yes, symmetry plays a crucial role in electrostatic problems and the method of images. In the first case, the spherical shell, the symmetry of the problem allows us to use Gauss' Law to calculate the electric field without needing to know the details of the shell. This is because the electric field inside the shell is constant and does not depend on the specific dielectric properties of the shell. The presence of the shell does not change the overall symmetry of the problem.

However, in the second case, the planar dielectric boundary, the symmetry of the problem is broken. The electric field on the charge side of the boundary depends on the distance from the charge to the boundary and the dielectric constants of both sides. This is because the presence of the boundary creates a discontinuity in the electric field, breaking the symmetry of the problem.

In general, the presence of dielectric boundaries can significantly complicate electrostatic problems, as they introduce additional factors that need to be considered. In the case of an infinite planar charge and a parallel dielectric boundary, the electric field will be affected by the presence of the boundary and will not be constant. The exact behavior of the electric field will depend on the specific geometry and dielectric properties of the problem.

In summary, symmetry is a key factor in simplifying electrostatic problems and the method of images. It allows us to make assumptions and use simplified equations to solve for the electric field. However, in cases where symmetry is broken, more information and calculations are needed to accurately determine the electric field.