# Fields in a capacitor before and after adding a dielectric layer

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
relevant eduation

Delta2

kuruman
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The problem statement does not specify whether the dielectric is added at constant charge (capacitor connected to a battery throughout) or at constant charge (capacitor is charged then disconnected from battery then dielectric is added). Do you know which is the case?

Also, D = ε0E in the vacuum between the plates. If you apply Gauss's law to a pillbox that has one face in the conducting plate and the other face in the vacuum between the plates, you will see why.

The problem statement does not specify whether the dielectric is added at constant charge (capacitor connected to a battery throughout) or at constant charge (capacitor is charged then disconnected from battery then dielectric is added). Do you know which is the case?

Also, D = ε0E in the vacuum between the plates. If you apply Gauss's law to a pillbox that has one face in the conducting plate and the other face in the vacuum between the plates, you will see why.
we have a constant charge in capacitor. If the D-field in vacuum is equal to sigma and in medium it becomes equal to sigma free (for free charge). since I know that the D-field should be constant so sigma = sigma free. am I right? I have calculated the E-field in vacuum and got that E = Q / epilon_0A and medium I got it to Q / epsilonA. But in both cases which is the free charge, one should not have bound charge or calculate total E in the capacitor by summing up E vacuum with E medium

Delta2
Delta2
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I think D is the same everywhere and before and after the dielectric is added and equal to Q/A where Q the charge of capacitor that remains constant as you said.
it is E that changes, according to the equation ##D=\epsilon E_{dielectric}## for inside the dielectric and ##D=\epsilon_0E_{vacuum}## in the vacuum.

I think D is the same everywhere and before and after the dielectric is added and equal to Q/A where Q the charge of capacitor that remains constant as you said.
it is E that changes, according to the equation ##D=\epsilon E_{dielectric}## for inside the dielectric and ##D=\epsilon_0E_{vacuum}## in the vacuum.
In this case, it is in a vacuum D = sigma Because we have no free charges in vacuum E = Q_tot / Aepsilon_0 In medium is D = sigma E = Q_tot / Aepsilon and the E-field for the bound charge becomes E_mediumA-E_vakuumA = Q_b / A epsilon?

Delta2
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If by ##Q_{tot}## you mean ##Q_{free}+Q_{bound}## then I don't agree. Both your equations for E should have ##Q_{free}## instead of ##Q_{tot}##.

The last equation for ##Q_{bound}## I think is correct though. Just a minor correction $$E_{medium}-E_{vacuum}=\frac{Q_{bound}}{\epsilon_0A}$$

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If by ##Q_{tot}## you mean ##Q_{free}+Q_{bound}## then I don't agree. Both your equations for E should have ##Q_{free}## instead of ##Q_{tot}##.

The last equation for ##Q_{bound}## I think is correct though. Just a minor correction $$E_{medium}-E_{vacuum}=\frac{Q_{bound}}{\epsilon_0A}$$
please can you explain to me how this is possible with E (medium) -E (vacuum) = Q_bound / epsilon_0A

Delta2
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What do you get if you apply Gauss's law for E, that is $$\iint E\cdot dS=\frac{Q_{enclosed}}{\epsilon_0}$$ and taking as gaussian surface a pillbox that has one side inside the vacuum and one side inside the dielectric. Isn't ##Q_{enclosed}=Q_{bound}## for this gaussian surface?

kuruman
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please can you explain to me how this is possible with E (medium) -E (vacuum) = Q_bound / epsilon_0A
I will restate what has already been said or implied but differently.
1. Gauss's law at the conductor-dielectric interface gives ##D_{\text{diel.}}={\sigma_{\text{free}}}.##
2. At the dielectric-vacuum interface D is continuous: ##D_{\text{vac.}} = D_{\text{diel.}}={\sigma_{\text{free}}}.##
3. You also have ##D_{\text{diel.}} = \epsilon E_{\text{diel.}}##, ##D_{\text{vac.}} = \epsilon_0 E_{\text{vac.}}## and that the discontinuity of the electric field at the interface is proportional to the bound charge: ##E_{\text{vac.}} - E_{\text{diel.}}=\dfrac{\sigma_{\text{bound}}}{\epsilon_ 0}.##

Put it together.

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Delta2
Delta2
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@kuruman did you intend to write ##Q_{free}## and ##Q_{bound}## instead of ##\sigma_{free}## and ##\sigma_{bound}##, or the denominator with the area ##A## seems redundant.

kuruman
kuruman
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Thanks. I caught my error and corrected it in time Δt but not before you posted your correction in time dt.

Delta2