How Does Gauss's Law Change in Dielectric Materials?

[AXIS]

Hey everyone, this is my first question on here, but I've seen you all be very helpful to others to I was hoping someone could clarify something for me.




1. How does Gauss law modify in dielectric materials?

 

[Claude Bile]

Gauss' law in matter is given by;

$$\nabla.\vec D = \rho_{f}$$

Where D is the electric displacement vector and rho is the free charge density. D is realted to the Electric field thus;

$$\vec D = \epsilon \vec E$$

Where epsilon is the permittivity of the medium.

Claude.

 

[AXIS]

Hmm okay not sure I follow you.

So what exactly is different about it compared to in any other material?

 

[Claude Bile]

Sorry, I misread your earlier post.

In a dielectric, the free charge density will be zero.

Claude.

 

[AXIS]

Ahhh nice one,

I think I get it now,

Cheers matey

 

[ehild]

In a dielectric, the free charge density will be zero.

Claude.

Claude, this is wrong. There can be free charges in a dielectric. You thought of metals, I think.

Gauss's law is the same for all substances: $$Div \B{D}=\rho_{free}$$.

$$\B{D}=\epsilon\B{E}$$. The difference is in the permittivity $$\epsilon$$.

ehild

 

[Claude Bile]

Yes, you are quite right.

In optical media , Gauss' Law is frequently approximated to be divD = 0.

Thanks ehild for pointing out by error.

Claude.