# Electric flux density

1. Mar 2, 2013

### Nikitin

Hey. So if I have a condenser, where the electric field between the plates is equivalent to 50kV/m, the electric flux density D= ε0*50kV/m.

If I insert a dielectic between the plates, the electric field will decrease to 10kV/m, and kappa (the relative permittivity) equals 5. So, will the flux density D = ε0κ*10kV/m, or D = ε0κ*50kV/m?

The first one makes more sense to me, but I am unsure.

Last edited: Mar 2, 2013
2. Mar 2, 2013

### tiny-tim

Hey Nikitin!
(is this a marxist dialectic? )

D (the electric displacement field) is related only to free charge

(as opposed to E, which is related to total charge, and P, which is only related to bound charge)

so the presence of the dielectric (and the bound charge in it) makes no difference …

if there is a volume density of free charge, divD is always equal to it

if there is a surface density of free charge, |D| is always equal to it

(and in particular, D for a parallel plate capacitor depends only on the (free) charge on the plates, not on the dielectric between the plates:

D = σn, and E is then calculated from D and from the dielectric)​

3. Mar 2, 2013

### Nikitin

So D should be equal for the condenser regardless if there is a dielectric present or not?

4. Mar 2, 2013

### tiny-tim

Yes (assuming the surface charge density on the plates is the same).

5. Mar 2, 2013

### Nikitin

Uhm, you can have condensers where the surface-charge density isn't equal among the plates?

Well, at least the charge Q on both the plates is the same. Right?

6. Mar 2, 2013

### tiny-tim

Yes, the charge on both plates is the same.

But on eg a spherical capacitor, although the charge is the same, obviously the charge density is different.

7. Mar 2, 2013