Voltage drop across a capacitor

AI Thread Summary
The discussion focuses on calculating the energy stored in a parallel plate capacitor with specific parameters, including charge, electric field strength, plate area, and spacing. The capacitance was calculated using the formula C=kEoA/Separation, resulting in a value of approximately 2.95 pF. The charge was determined using the electric field and area, yielding about 1.77 nC. There was uncertainty regarding the relevance of the "voltage drop" concept, but it was clarified that the equation Q=∆V*C is applicable. The conversation highlights the importance of correctly applying formulas and addressing potential errors in calculations.
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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
 
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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 :smile:
 
Sigh, power of 10 error. It was the right equation, thanks though.
 

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