In this physics question I sent to u is it that electric field in air is Q/E nut A and the electric field in di electric slab is Electric field in air / K

• The electric field in the air gap (or vacuum) between the plates of a capacitor is given by:

  $$E_{air} = \frac{Q}{\epsilon_0 A}$$

  where $Q$ is the charge on the plates, $\epsilon_0$ is the permittivity of free space, and $A$ is the area of the plates. This is derived from $E = \frac{\sigma}{\epsilon_0}$ where $\sigma = \frac{Q}{A}$.

• The electric field inside a dielectric slab inserted into the capacitor is indeed reduced by the dielectric constant $K$:

  $$E_{dielectric} = \frac{E_{air}}{K}$$

You've got the right understanding of those formulas!