Proving Electric Flux Density: eE Determination

[mym786]

How to prove that D = eE.

Use a general proof. I know how to prove it assuming a point charge.

 

[Hootenanny]

How to prove that D = eE.

Use a general proof. I know how to prove it assuming a point charge.

I'm not sure what you mean by "prove". That is simply a definition: D=eE because we say so.

 

[mym786]

How did we get this formula ?

 

[Hootenanny]

How did we get this formula ?

As I said in my previous post, by definition. There really is nothing to prove. The electric displacement field is simply defined that way.

 

[yungman]

Check out "Introduction to Electrodynamics" by Griffiths 3rd edition, page 175.

$$\epsilon_0 \nabla \cdot \vec E =\rho_v + \rho_{f} = \rho_v + \nabla \cdot \vec P$$

$$\epsilon_0 \nabla \cdot \vec E + \nabla \cdot \vec P = \rho_{f} \;\Rightarrow \; \nabla \cdot (\epsilon_0 \vec E + \vec P) = \rho_{f}$$

So we define:

$$\vec D = \epsilon_0 \vec E + \vec P \;\Rightarrow \; \nabla \cdot \vec D = \rho_{f}$$

For linear and isotropic material:

$$\vec P= \epsilon_0 \chi_e \vec E \;\Rightarrow \; \vec D = \epsilon_0(1+ \chi_e) \vec E$$

We define:

$$\epsilon_r = 1+ \chi_e \;\hbox { and } \epsilon = \epsilon_0 \epsilon_r \;\Rightarrow \vec D = \epsilon \vec E$$

 

[dgOnPhys]

Hi yungman

I think you have a typo; your first line should go

$$\epsilon_0 \nabla \cdot \vec E =\rho_v + \rho_{f} = - \nabla \cdot \vec P + \rho_{f}$$

 

[yungman]

Hi yungman

I think you have a typo; your first line should go

$$\epsilon_0 \nabla \cdot \vec E =\rho_v + \rho_{f} = - \nabla \cdot \vec P + \rho_{f}$$

Yes, sorry!