# Capacitance capacitor parallel plate with dielectric (Gauss)

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1. May 5, 2017

### Ark236

1. The problem statement, all variables and given/known data
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

2. Relevant equations

∫D⋅ds=Q
D=E+γP
C=Q/V

3. 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: May 5, 2017
2. May 5, 2017

### Diegor

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