Finding the force exerted by one plate on another

[mr_coffee]

Hello everyone, I'm stuck on a mutliple choice question: A parallel-plate capacitor has a plate area of 0.3 m2 and a plate separation of 0.1 mm. If the charge on each plate has a magnitude of 5 10-6 C then the force exerted by one plate on the other has a magnitude of about:
1 E4 N
9 E5 N
5 N
9 N
0

Here is what i did:
I know the E-field right outside a conductor is E = (Q/A)/Eo <--permitivity. So i found E = (5e-6/.3)/8.85e-12 = 1883239 N/C; Then i knew F = QE, so I said, F = (5e-6)(1883239) = 9.4N, which was wrong, then i thought, hm..maybe i messed up a unit somehow, so i tried 9e5N, also wrong, then i thought, well if they are both positive, that means the net force is going to be 0, because ulike charges repel, same charge, 0 net force. also wrong. What did i do wrong?

I also ran into odd problem, I'm wondering if the homework problem is wrong...
A 20 F capacitor is charged to 200 V. Its stored energy is:
4000 J
0.1 J
2000 J
0.4 J
4 J

Easy enough, U = .5*CV^2;
U = .5(20)(200)^2 = 400000J;
The answer is .4J though?

 

[quantumdude]

Sorry for the late reply. If you're still interested, then here are my comments.

I know the E-field right outside a conductor is E = (Q/A)/Eo <--permitivity.

Check that expression, because I don't think that you can use it when dealing with parallel plate capacitors.

See the following websites:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html
http://dept.physics.upenn.edu/~uglabs/lab_manual/electric_forces.pdf