What is the Energy Stored in a Parallel Plate Capacitor?

AI Thread Summary
The discussion revolves around calculating the energy stored in a parallel plate capacitor with given parameters, including charge, electric field, area, and spacing. The user initially calculated the capacitance as 2.95 x 10^-12 F but struggled with finding the potential energy due to confusion over the formulas. Clarifications were provided regarding the distinction between voltage and energy, emphasizing that voltage is energy per charge. Ultimately, the user successfully determined the voltage to be 600 V and applied the formula PE=1/2CV^2 to find the energy stored. The conversation highlights the importance of understanding the relationships between electric field, voltage, and energy in capacitors.
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Given: There are two parallel plates. 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.

I found C is 2.95 x10^-12 by taking ((8.85x10^-12)(2 x10^-3))/(6x10^-3).

I know:
the electric field is uniform and that V=E*d
The potential difference is just the magnitude of the electric field times the distance you move in the electric field

When I try to solve for V or PE, I keep getting the Power of ten error.

Can someone please help me?

Thank you:)
 
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The value you have for C looks good to me.
What formula are you using for the energy of the capacitor?
Note that in V = E*d, the E is not energy. Also, the potential V is not the energy; it is the energy per charge.
 
I'm not exactly sure what formula to use. I am kind of stuck.
 
Thank you so much for your help:)

I figured it out. I had to find the Voltage which is 600. Then plug it into the equation: PE=1/2CV^2
 
Most welcome!
 
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