Electric field inside a material I was thinking about the polarisation, and how the electric field behaves inside the material of permittivity greater than one.
I think to have understood what happens to D and P, but is not clear what happens to E.
Is the electric field inside the material bigger, remains constant, or is weaker?
 A: I assume you are talking about linear materials with dielectric constant greater than $1$.
Say you have a free charge distribution $\rho$ in vacuum, which produces an E field. Now you introduce linear material. Then the free charges will be "weakened" because they will by partially screened by charges of the dipoles sticking on them. However, the opposite amount of these bound charges must appear somewhere else, because the total bound charge is zero. Usually you will find the opposite amount of bound charges on the surface of the dielectric, which also produce an E field. So it's hard to say in general whether the E field will be smaller or larger.
Let's consider some special examples then.
(1) If it's known that there are no other bound charges except those sticking to the free charges, then the field will be weakened, like in the case of a dielectric-filled capacitor.
(2) If the surface bound charges are very far away, then they can be ignored and we can say the field will be weakened. Like in the case of extended dielectric media.
(3) If the bound charge distribution has symmetry, then we may conclude the answer easily. For example, in the case of a dielectric sphere, since the surface bound charges are spherically symmetric, their field inside is zero and their field outside cancels exactly of those of the volume bound charges. So one can conclude that when you introduce the dielectric sphere, the field inside is weakened, while the field outside remains the same.
A: (For completely filled capacitors)
Q = CV
So, C = Q/V
So, C is charge stored per unit Potential Difference applied.
Now,
V = Ed
,where d is distance between plates.
$E = \dfrac{V}{d}$
Case 1) When you apply a constant V of 1V to capacitor
E across capacitor is $ \dfrac{1V}{d}$
which is constant independent of capacitance of capacitor or dielectric b/w plates.
So, E in capacitor is constant.
Case 2) You disconnect battery after applying a PD of 1V.
And then insert a capacitor.
So, C becomes C'.
Clearly Q = C' V'
So, since Q is constant and C' > C ** , **V' < V.
Since,  $E = \dfrac{V'}{d}$ ,
E decrease.
A: For the most part it is the same as for the magnetic field. For ferromagnets and paramagnets, the magnetization goes in the same direction as H, giving a larger overall magnetic field. Whereas diamagnets have the magnetization in the opposite direction of the applied field. 
Likewise for dielectrics,  the polarization will be in the opposite direction of the applied field, meaning the electric field overall is lowered. For paraelectrics and ferroelectrics, the polarization is in the same direction as the applied field,  so the electric field is increased. 
