Capacitance capacitor parallel plate with dielectric (Gauss)

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The discussion revolves around calculating the capacitance of a parallel plate capacitor with a dielectric. The user attempts to apply Gauss's law and considers the electric displacement field (D) and electric field (E) in their calculations. Key points include the realization that D is not equal to E, as D should be expressed as εE, where ε is the permittivity. Additionally, the electric field outside the plates is assumed to be zero based on the infinite parallel plates assumption. The user is advised to review their equations for accuracy in the context of the problem.
Ark236
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Homework Statement


I have an elementary doubt with the calculation of the capacitance of a capacitor of parallel plates that has a dielectric in the middle.

https://ibb.co/b0W4BQ

Homework Equations



∫D⋅ds=Q
D=E+γP
C=Q/V

The Attempt at a Solution


Suppose the top plate has a positive charge and the bottom plate has a negative charge.

To calculate vector displacement in the upper plate, we occupy the gauss law, use a Gaussian box. On the outside of the top plate the polarization P is 0 and there is an electric field E. At the bottom of the plate there is an electric field E and a polarization P due to the dielectric.
If we consider only the top plate:

∫D⋅ds=[D][/out]A+[D][/down]A=Q

[D][/out]=E
[D][/down]=E+γP

If we consider the complete system (both plates)

∫D⋅ds=2*[D][/out]A+2*[D][/down]A=Q

[D][/out]=E
[D][/down]=E+γP

The field inside is 2 E.

What is the error, I do not understand. D is zero outside from capacitor?
Thank you
 
Last edited:
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At a first glance I see some equation problems since D is not equal to E D=epsilonxE. On the other side the elctric field outside the plates is considered to be 0 since the infinite parallel plates asumption is used when Gauss Law is applied. You should check your relevant equations
 
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