About the dielectric, The E field and charge in the plate

[guo-hai]

Hey guys,
I know that adding a dielectric in a two plate can add the charge in the plate, so it can add the capacitance.
But I have a problem, When we place the dielectric in the E field, there is a dipole moment which produce internal E ( E polarization ) that opposed the E field, so the E field is reduced.
My question is How can E field is reduced while the charge in the plate is increased? Isn't it a contradiction?

I found this answer from this web **broken link removed**
"A decreasing E-field would cause the potential difference to decrease--if it could. But it can't here, so the battery does what it can to make the E-field correspond with the amount of potential difference. How do you adjust the charge to increase the E-field? You have to increase it. So the battery pumps more charge to the plates. The charge increases and the energy increases. "
I understand it, but I'm not still satisfied with the answer, cant anyone tell me?

Thank YOu

 

[albbg]

We know that the voltage across a capacitor is given by V=Q/C where Q is the displaced charge and C the capacitance. Furthermore in a parallel plate capacitor (simpler to analyze) the electric field is given by E=V/d where d is the distance between the two plates.

Let's suppose to charge a capacitor, with no dielectric, having capacitance Co with a voltage source Vo.
A charge Qo=Co*Vo will be then stored and an electric field Eo=Vo/d will appear.
Removing now the voltage source and inserting a dielectric medium, the charge will not change, but the polarization effect in the dielectric will generate an electric field opposed to the original one. It can be proven that this effect is proportional to the magnitude of the electric field by a constant depending from the material and called εr.
This means we will have now a "new" electric field En=Eo/εr. The voltage across the capacitor will be then Vn=En*d ==> Vn=Eo*d/εr that is:
Vn=Vo/εr. The new capacitance will be then (remember Qo is still the same): Cn=Qo/Vn ==> Cn=Qo/(Vo/εr)=Qo*εr/Vo, but since Co=Qo/Vo then Cn=εr*Co

Let's suppose now that after charging the capacitor with no dielectric, as we did before, we leave the source connected.
The initial charge and the electric filed will be the same as before, that is: Qo=Co*Vo and Eo=Vo/d
Now inserting the dielectric the polarization effect will act as an electric field, proportional to the original one, but with opposite direction. As before: En=Eo/εr
We have seen that in absence of the voltage source, the voltage across the capacitor decreases. But in this case the voltage is fixed by the voltage source itself and is higher with respect to that we would have if the voltage source is disconnected. Then to increase the voltage to a level equal to that of the voltage source the only way is to increase the charge.
Since En=Eo/εr then Vn=Vo/εr; in order to keep Vn constant to Vo means that we have to multiply by εr that means the charge has to be increased by a factor εr, that is Qn=Qo*εr.
In this case the new capacitance will be: Cn=Qn/Vo ==> Cn=Qo*εr/Vo then Cn=εr*Co as in the previous case.