Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is E = 3.10×105 V/m. When the space is filled with dielectric, the electric field is E = 2.20×105 V/m.

2) dielectric constant K = Eo / Ek = 3.1×10^5 / 2.2×10^5 = 1.41

1) Eo = sigma / eo, where ‘sigma’ is the charge density

so that charge density = Eo eo = 3.1×10^5 x 8.854 x10^-12 = 2.744 x10^-6 C/m^2.

If this is a Mastering Physics problem, then take what @knr got for 1 (2.744e-5) and halve it to get charge density on each surface of the dielectric.

Increasing dielectric constant has he effect of seperating the plates,

Hence

E1x e1 = E2 x e2

Hence e2/e1 = E1/E2 = (3.10 x 10^5 )/(2.2X10^5)

So e2 = 1.41 e1 = 1.41 relative to a vacuum