# Electric potential in a capacitor

Im a writing a school report regarding the electric energy and potential of a capacitor. In short: An experiment was carried out where we charged a capacitor, and then decharged it with an electric engine, lifting up a predefined weight. We then measured how high the weight was lifted with different voltages. i then found the effiency of the system by comparing the mechanical energy with the electrical energy.

The second half of the report is then to find the electric potential of the capacitor, but I'm not to strong at electrics, and can't find anything about this in my physics book.

My question is now; how can i find the electric potential of the capacitor?

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electric potential is voltage –  Mark Eichenlaub Feb 6 '11 at 21:53

From the context, I'm going to assume that by "electric potential" you mean the energy in the capacitor, rather than the usual meaning of "electric potential" which is simply the voltage.

The formula for electric potential energy (in Joules) is $0.5 C V^2$, where C is the capacitance in Farads, and V is the voltage in volts. Here is a reference which has the formula. (Ignore the calculus for now, you'll get there eventually.)
http://en.wikipedia.org/wiki/Capacitor#Energy_storage

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Thank you, yeah that was also what i found to be the only meaningfull thing to calculate, so ill go by that... –  user1787 Feb 7 '11 at 0:51

I think what you are asking is the energy stored by a capacitor and the amount of work it can do. Voltage is charge divided by capacitance, $V~=~q/C$. See below for the reason for this. We know that the power $P$ (wattage) in a circuit is given by $P~=~VI$. $I$, the current, is the time rate of change in the charge, or $I~=~dq/dt$. Power is the time rate of change in energy $P~=~dE/dt$ and so we have $dE~=~Vdq$ So put all of this together and you get the work $W~=~\int dE$ done by a capacitor is $$W~=~\int \frac{qdq}{C}~=~\frac{q^2}{2C}~=~\frac{qV}{2}$$

The capacitor is defined as a charge per unit area. If you have a surface with a constant charge distribution $\sigma$ so that $q~=~\int\sigma da$ $\simeq~\sigma A$ the electric field is a constant vector field normal to the surface. The inside of a capacitor is two such plates with a charge across two plates. Electric field has units of $volt/m$ and so the voltage $V~=~\int Edx$. Capacititance is $C~=~q/A$, which is the charge distribution on the flat plate. This all leads to $V~=~q/C$

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thank you, that was kinda what i was thinking it was, but since the assignment clearly says potential, it seemed odd, and i didn't know what to write about. I asked some of my fellow students, and they had all just done calculations for the energy, so i guess that is the way to go... –  user1787 Feb 7 '11 at 0:49