Wondering why my method of finding charge doesn't work?

[pradeepk]

Homework Statement

A parallel plate capacitor has a capacitance of C=5.00pF when there is air between the plates. The separation between the plates is 1.50mm. What is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 x 10^4 V/m?

Homework Equations

Equations my professor used:
Vmax=Emax*d
Qmax=CVmax

Equations I used:
E=σ/ε

The Attempt at a Solution

So the way my professor solved which is probably the easiest way and the right way is first he solved for the maximum potential:
Vmax=Emaxd
= (3.00 x 10^4 V/m)(1.50 x 10^-3m)= 45V
Then you can solve for the charge:
Qmax=CVmax
= (5pF)(45.0V)= 225pC

So the way I did it was that since we have parallel plates, we know the electric field between them is E=σ/ε. So:
3.00 x 10^4 V/m= σ/(8.85 x 10^-12 C^/Nm^2)
σ=2.655 x 10^-7

And since σ=Q/L
Q=σL
=(2.655 x 1-^-7)(1.50 x 10^-3 M)
=3.98 x 10^-10 C A different answer than my professor.

I know that my units are all messed up. But I just get confused because there are so many different ways to solve for the electric field..so shouldn't they all equal the same thing? Or is this different because it is a capacitor? Any help would be appreciated. Thanks!

 

[Doc Al]

And since σ=Q/L
Q=σL
=(2.655 x 1-^-7)(1.50 x 10^-3 M)
=3.98 x 10^-10 C A different answer than my professor.

σ = Q/Area (not Q/L)

You need to make use of the given capacitance to find the charge.

 

[pradeepk]

ohhh ok I feel stupid now. So if we know the are then we could use this way. My professor however on a different problem used A=2∏r^2 for the area of the parallel capacitors. That doesn't really make sense to me since they are rectangles..