Charge distribution on dielectric surfaces between conducting plates

[Qyzren]

Two identical, oppositely-charged, conducting plates of area 2.50cm² are separated by a dielectric 1.80 mm thick, with a dielectric constant of 3.60. The resultant electric field in the dielectric is 1.2*10^6 V/m.

a)Compute the charge per unit area on the conducting plate.
b)Compute the charge per unit area on the surfaces of the dielectric.

The Attempt at a Solution

for part a). I used gauss's law for dielectric.
Q/A = K*epsilon_0*E = 3.60*8.85*10^-12*1.2*10^6 = 3.82*10^-5 C/m² which is correct.

However i haven't got a single clue how to do part b. But the answer is suppose to be 2.76*10^-5 C/m². Can someone please help me figure out how this answer was obtained?

 

[Meir Achuz]

For b, they must mean bound surface charge density, which equals the normal component of the polarization vector P.

 

[Qyzren]

thank you for help. i managed to figure it out now with your tip :)