Simple Parallel-Plate Capacitor Question please HeLP

[BuBbLeS01]

Simple Parallel-Plate Capacitor Question...please HeLP!

Homework Statement

A parallel-plate capacitor is formed from two 2.32 cm × 2.32 cm electrodes spaced 2.85 mm apart. The electric field strength inside the capacitor is 85600000 N/C. What is the charge (in nC) on each electrode?

Homework Equations

The Attempt at a Solution

E = Q / Eo * A
A = 0.0232 * 0.0232 = 5.38e^-4
Q = (E * Eo * A) 8 1,000,000,000 = 407.75 nC
What am I doing wrong?

 

[BuBbLeS01]

Do I divided 407.75 by 2? Since there are 2 electrodes? I don't know what else I could be doing wrong.

 

[BuBbLeS01]

Thats not right either...does anyone know where I am going wrong?

 

[olgranpappy]

you multiplied incorrectly, I guess. plugging into my calculator, I get 407.942041 nC

or, to the right number of sig figs 408 nC

 

[BuBbLeS01]

How did you get that?
K = 9 x10^9
Eo = 8.85 x10^-12
A = .0232 *.0232
right?

 

[olgranpappy]

K? That doesn't come in anywhere... I used E and A and epsilon_0. Just multiply them all together.

 

[BuBbLeS01]

I know I put that to see if you were using the same number to get Eo...I am not getting the right answer though