Fields in a capacitor before and after adding a dielectric layer

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  • #1
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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
 

Answers and Replies

  • #2
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.
 
  • #3
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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.
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
 
  • #4
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.
 
  • #5
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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.
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?
 
  • #6
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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  • #7
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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
 
  • #8
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?
 
  • #9
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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  • #10
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.
 
  • #11
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:oldsmile:.
 

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