Gauss's Law with Dielectric

1. Feb 1, 2008

breez

My textbook (high school level) derives an instance of Gauss's Law with Dielectric for the case in which the dielectric material fills the gap between a parallel-plate capacitor entirely.

So you get surface int (D dot Area-vector) = q, where D = (dielectric constant)(epsilon_0)(Electric field), and q is the free charge (on capacitor)

In the following example, the book shows how to compute the electric field inside a dielectric material between a parallel-plate capacitor, but the material in this case does not fill the entire gap. However, they utilize the version of Gauss's Law above. I guess the above version of Gauss's law holds as along as a portion of the dielectric material is inside the Gaussian surface used? After the derivation, the text did state that "this holds true generally and is the most general form of Gauss's Law."

btw, this is a high school text, so all e-fields, dielectric constants, etc are assumed to be uniform.

2. Feb 2, 2008

Hootenanny

Staff Emeritus
This is one of the reasons why Gauss' law is so useful because we can chose the surface which we integrate over arbitrarily. The only thing that matters is that we enclose all the charge inside the surface.

3. Feb 2, 2008

Staff: Mentor

That form of Gauss's law holds true in general. A polarization charge is created on the surfaces of the dielectric that creates a field opposing the field from the free charges, thus reducing the electric field within the dielectric. For your capacitor example: As long as uniform fields are maintained, it doesn't matter if the dielectric fills only a portion of the gap or the whole gap: The induced polarization charge (and resultant field within the dielectric) is the same.