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!