Electric Flux Density

[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!!!