Voltage drop across a capacitor

[zooboodoo]

Homework Statement

The left plate of a parallel plate capacitor carries a positive charge Q, and the right plate carries a negative charge -Q. The magnitude of the electric field between the plates is 100 kV/m. The plates each have an area of 2 × 10-3 m2, and the spacing between the plates is 6 × 10-3 m. There is no dielectric between the plates.

Calculate the energy stored in the capacitor.

Homework Equations

q=vc , PE=c*v^2/2 , PE=Q^2 / 2C , PE=QV / 2

The Attempt at a Solution

I calculated the Capacitance by C=kEoA/Separation = (2.95133333333333e-012)
I also calculated Q by E * Eo *A= Q = 1.7708e-9

I wasn't sure if "Voltage drop" entailed a different equation? I've gone through the powerpoints from my class and have not seen any other equations. My book hopefully comes in the mail soon :-\ any help would be appreciated

 

[Mk]

I wasn't sure if "Voltage drop" entailed a different equation

I believe that would entail Q=∆V*C which you've already got up there

 

[zooboodoo]

Sigh, power of 10 error. It was the right equation, thanks though.