*A*, held a fixed distance

*d*apart. To keep things simple we assume that the space between the plates is air (or vacuum, although this space could be filled by an insulating liquid or solid "dielectric").

*capacitance*of a given parallel-plate capacitor from its area

*A*and separation distance

*d*. As the constant

*k*has units of coulombs

^{2}per joule (C

^{2}/J) that remain after canceling out the

*A*and

*d*units, we redefine the C

^{2}/J units of capacitance as farads (F).

The capacitance of a capacitor is

*fixed*once it is constructed, as the only way to change the capacitance is to change its "build" parameters: the area

*A*and/or the separation distance

*d*.

*Q*and –

*Q*.

So just how much charge could a capacitor store on its plates? When does a capacitor "know" when to stop charging?

(The units for charge (in coulombs, C) and potential (in volts, V; or joules/coulomb, J/C) again work out to coulombs

^{2}per joule, or farads for the capacitance.)

The key is to realize that the capacitance of a capacitor, once built, is fixed. However, one can change the amount of potential applied to the capacitor (by connecting different batteries, etc.). With a given value of capacitance, then applying a high potential to the capacitor will allow it to store more charge, and applying a low potential to the capacitor will have it store less charge.

Thus capacitance can be said to be a measure of "charge-storing efficiency" with respect to a given potential, in that a large capacitance capacitor will store more charge than one with a small capacitance, if they are connected to the same potential source (such as a battery).

*q*travels quickly because it's "easy" to move, while the last electron

*q*moves very slowly because it's "hard" to move.

*q*has a "start-up" energy cost of effectively zero, as it moved from and to plates that are nearly or effectively uncharged (and start with zero potential, such that ∆

*EPE*=

*q*·(0) = 0 J).

*q*must be removed from a very positively charged plate, which requires a lot of work, and also must be moved onto a very negatively charged plate, again requiring a lot of work (the difficulty of which is indicated by the slowness of moving this electron). Since the plates are at or nearly at their final potential value ∆

*V*, the last electron

*q*has an energy "end cost" of ∆

*EPE*=

*q*·∆

*V*(don't worry about negative signs in this argument).

Thus the first electron moved costs nothing (or nearly nothing), while the last electron requires a much higher

*q*·∆

*V*cost to move. The average cost of moving each electron, then, can be said to be (1/2)·

*q*·∆

*V*, such that the total cost of moving all electrons is (1/2)·

*Q*·∆

*V*, where the total charge of all

*N*electrons moved is

*Q*=

*N·q*. (This argument is deliberately trying to avoid calculus to integrate the gradually increasing cost of each electron over all electrons moved, but the result is the same.)

*C*=

*Q*/∆

*V*relation, the two parameters

*Q*and ∆

*V*in the electric potential energy equation can each be substituted out, yielding two other equivalent equations for electric potential energy

*EPE*.

Endless hours of amusement await you when solving capacitor energy problems, so use caution when you use these equations, and more importantly, use only the equation you really need.

*electric potential energy*carried by these charges that makes capacitors hazardous. And more importantly, a key advantage capacitors have over batteries in storing electric potential energy for later use, is that the energy from a capacitor can be released in a very brief amount of time (as opposed to a small steady amount over a long period of time), as we will see in a subsequent presentation.

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