- Homework Statement:
We have a capacitor that with two conductive plates with distance r, We fill to half of them a dielectric material (r / 2). Our task is to calculate D and E fields before and after we lay the material.
I have calculated the D and E fields before and got to:
E = Q / ε0A and D = Q / A = σ, but My question is that Gauss' law says that ∫Dda = Qfdå has no free charges in vacuum so it will be zero and then the D-field will be equal to zero which is right?
After we add the medium I get that (since I assumed here that D = σ before we add the medium):
D = D = -Q / A = -σ_f because it is the free charges that D is affected by in the negative plate therefore we have minus signs.
E (vacuum) gap = Q / ε0A and E (medium) = Qf / εA then (D = εEmedium)
Am I doing the right thing?
Is not it that you have Q_b in the medium just so that you get that E (medium) = Q_b / epsilon A
- Relevant Equations:
- D=epsilon E