- #1

zooboodoo

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## 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