Sunday, January 20, 2013

Electrical Displacement Field Media Boundary Conditions

Today's post is a little plain Jane, (as far as I know), in that it's just a review of our classes' derivation of the boundary conditions for the electrical displacement field at an interface between two materials.  Here's a question.  Does anyone know of anything, cool or clever to take away from the following derivation?  At the moment, it seems necessary, but not tantalizing.

First, we'll need the Divergence Theorem, (picture 1),

stating that the volume integral of the divergence of a vector field is equal to the surface integral of the same field with respect to the normal of the surface that bounds the volume.

Given the above tools, and starting with a boundary between two materials, and the typical pillbox, (picture 2),

where delta A is the area of the top and bottom of the box, n is the unit vector normal to the top of the pillbox and the pillbox straddles the boundary between the two materials.  In the end we'll let the sides of the pillbox shrink to zero, pulling the two faces of the pillbox onto the interface, so the volume is zero.

We're using Gauss's Law, (picture 3),

where rho is the surface charge density and D is the displacement field.  Taking the volume of both sides we wind up at, (picture 4),:

This is where the divergence theorem comes in to get rid of the volume integral on the left, (picture 5),

while at the same time letting the box walls shrink to nothing, (arbitrarily squat), the charge enclosed in the pillbox becomes just a surface charge density on the pillbox and the integral of the displacement field normal to the bounding surface is just the difference of the normal components of the displacement fields on the two sides of the boundary as seen in the above formula.   This finally gives a condition that must be met between the two displacement fields on opposite sides of the boundary, (picture 6):

Stokes' Theorem:

Picture of the Day:

From 1/20/13

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