# Gauss's Law in Matter

1. Jan 23, 2012

### Apteronotus

Hi,

The Poisson equation (or Gauss Law) in a vacuum is given by
$\nabla^2\phi=-\frac{\rho}{\epsilon_0}$
where $\rho \mbox{ and } \epsilon_0$ are the charge density and vacuum permittivity or (electric constant of space).

My question is what is the Gauss's Law in a dielectric material? Do we simply replace the vacuum permittivity by the permittivity of the material?
ie.
$\nabla^2\phi=\frac{\rho}{\epsilon_r}$
where above $\epsilon_r$ is the relative permittivity (or permittivity of the dielectric material).

Thanks,

2. Jan 23, 2012

### conquest

Hi,

I would check some books for follow up.

How is it Gauss's law in vacuum? What is then carrying the charge?

Beyond that I think it matters whether you're talking linear dielectric or non-linear dielectric.

For a linear dielectric if you state it using the fields use the displacement field instead of the electric field and then it accounts for the free charge with the new permittivity.

3. Jan 23, 2012

### Apteronotus

From what I've read, Guass's law is usually given in its integral form, saying that "the amount of electric flux through a surface is proportional to the amount of charge within that closed surface".
Then the differential form is derived using divergence theorem relating the divergence of the electric field to the surface charge density $\rho$.

So in short everything seems to be in a vacuum. But how do these equations compare if say we have a linear, homogeneous, isotropic material?

4. Jan 23, 2012

### pabloenigma

Poissons equation is bit different from Gauss Law(In addition to it,Poissons eqn requires that the field is derivable from a scalar potential)

Coming to the point,In matter, Gauss Law is expressed as the surface Integral of the the displacement vecotr being equal to the the total enclosed free charge.
If the matter is linear and Homogeneous,it can be written in form of Poissons equation with source Free charge divided by epsilon.
The equation you proposed is dimensionally incorrect,as you can see.

5. Jan 23, 2012

### Staff: Mentor

6. Jan 23, 2012

### Apteronotus

Thank you all for your responses. jtbell the resource you linked to is very informative, thank you.

Following equation 813, would it be correct to deduce
$\nabla^2\phi=-\frac{\rho_{f}}{\epsilon \epsilon_0}$

7. Jan 23, 2012

### pabloenigma

Yes,for Linear and Homogeneous dielectrics

8. Jan 24, 2012

### clem

and isotropic

9. Jan 24, 2012

### Dickfore

In a dielectric material, Gauss's Law holds for the electric displacement field $\mathbf{D}$:
$$\nabla \cdot \mathbf{D} = \rho_f$$
where $\rho_f$ is the free charge density.

The dielectric is modeled by a relative dielectric constant, which meaning (for an isotropic medium) is:
$$\mathbf{D} = \varepsilon_0 \, \varepsilon_r \, \mathbf{E}$$
The electric field, being irrotational, is still derivable from the electrostatic potential:
$$\mathbf{E} = -\nabla \, \phi$$