If you fix the potential as a boundary condition, your boundary condition doesn't contain permittivity. You solve laplacian(phi) = 0 - there's no permittivity here either. Therefore, you'll get the same field irrespective of the dielectric permittivity. Your surface charge will, however, change proportionally to epsilon.

Hi Radomir,

Thanks a lot for your quick reply. I also found in a COMSOL forum that the software does not solve the Effective potential, which includes the polarisation potential. Is this also true, ir if I solve my system including electrodes I will get the effective potential.

I'm somehow new to COMSOL, and I would like to know this kind of details.

Thanks

Hi

If you use only one material without space charges, the electic field does'nt change with the dielectric permittivity. You solve Laplacien equation which consider an homogeneous materiel.   Only charge in electrodes surface wil change (capacity of the system).

But if you consider two materials, the fild will change with permittivity. 

It is correct that if the space between electrodes are filled with dielectric material of single DE constant the field intensity will not change with the value of DE constant but interesting results will be observed if materials of different DE constant is used to fill the space particularly if thin layer of vacuum or air is considered as interface layer between the two dielectrics.

Hi.

The problem is not with the simulation. The electric field is the gradient of the electric potential. If you also connect a voltage source to the boundaries of your structure in practice and change the permittivity of the space, you will not see any changes in the electric field. Permittivity determines the relation between the electric field and electric flux density. If you have a constant electric charge, the permittivity can determine how the voltage and electric field attenuate by increasing the distance from the charge source. However, by connecting a voltage source or fixing the voltage in any other form, you are indeed fixing the electric field too, because the electric field is the gradient of your voltage.