Capacitors With and Without Dielectric

[jchoca]

Homework Statement

Two parallel-plate capacitors A and B are connected in parallel across a 620.0 V battery. Each plate has area 100 cm² and the plate separations are 4.5 mm. Capacitor A is filled with air; capacitor B is filled with a dielectric of dielectric constant 2.2.

A)Find the magnitude of the electric field within the dielectric of capacitor B.
B)Find the magnitude of the electric field within the air of capacitor A.
C)What is the free charge density on the higher-potential plate of capacitor A (with proper sign)?
D)What is the free charge density on the higher-potential plate of capacitor B (with proper sign)?
E)What is the induced charge density on the surface of the dielectric which is nearest to the higher-potential plate of capacitor B (with proper sign)?

Homework Equations

E=\frac{E_{0}}{E}\newline

E_{0}=\frac{\sigma}{\epsilon_{0}}\newline

V = Ed\newline

C_{eq}=C_{1}+C_{2}+\cdots+C_{n}

The Attempt at a Solution

The part that I am mainly having issues with (at the moment) is part A. I was able to obtain part B by using V = Ed and solving for E. Then I found part C by doing E₀=\sigma/\epsilon₀ and solving for \sigma. I tried to find part A by using K=E₀/E but that is incorrect. I'm sure it has something to do with the fact that there are two parallel capacitors rather than just one isolated one with a dielectric. I am pretty confused on how to approach getting part A.

 

[jchoca]

I made another attempt at the solution for A):

\frac{C}{C_{0}} = \frac{Q}{Q_{0}}=K

I know Q_{0} since I know E_{0}.

Q=KQ_{0} = 2.684 x 10^{-8}C

\frac{Q}{A} = \epsilon_{0} = 2.68 x 10^{-6} C/m²

E = \frac{\sigma}{\epsilon_{0}} = 3.03 x 10^{5}N/C

Still wrong though...