# Homework Help: Wondering why my method of finding charge doesn't work?

1. Oct 21, 2011

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
Equations my professor used:
Vmax=Emax*d
Qmax=CVmax

Equations I used:
E=σ/ε

3. 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!

2. Oct 21, 2011

### Staff: Mentor

σ = Q/Area (not Q/L)

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

3. Oct 21, 2011
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