A capacitor is two conductors of arbitrary shape separated by some insulate material (i.e. air). Capacitors are used to store electrical charge and electrical potential energy. In this lesson, it's be our goal to prove that and to derive equations which will allow us to calculate the total electrical potential energy stored in any kind of capacitor.
Consider the capacitor in Figure 1. The two conductors are initially electrically neutral with an equal number of positive and negative chaged particles. Suppose than an electric field is turned on and an electric force \(\vec{F}_E\) is applied to a charged particle \(q\) in Conductor B. Suppose that \(\vec{F}_E\) moves \(q\) over to Conductor A. What is the work done by this
force? Using the definition of work (\(W≡\int_c\vec{F}·d\vec{r}\)), we find that the work done by \(\vec{F}_E\) moving \(q\) a distance \(d\) to Conductor A is given by
$$W=\int_C\vec{F}_E·d\vec{r}.\tag{1}$$
Figure 1
Equation (1) represents the work done by any electrical force \(\vec{F}_E\) moving some charge \(q\) along any arbitrary path over to Conductor B. So, Equation (1) is very general and, initially, it might not look all that practically useful. After all, how do we solve the integral \(\int_c\vec{F}_E ·d\vec{r}\) if \(\vec{F}_E\) and \(c\) can be anything? Without any simplifying assumption, this integral does indeed seem pretty hard to solve. But, using the definition of the electric field (\(\vec{E}=\frac{\vec{F}_E}{q}\)), let's substitute \(\vec{F}_E=\vec{E}q\) into Equation (1) to see how that might simplify this integral:
$$W=q\int_{r_A}^{r_B}\vec{E} ·d\vec{r}.\tag{2}$$
The question is, why was this substitution useful? The answer is because the electric potential difference \(ΔV_{AB}\) (or voltage) between Conductors B and A can now be substituted into Equation (2); unlike the integral \(\int_{r_A}^{r_B}\vec{E} ·d\vec{r}\), the voltage \(ΔV_{BA}\) is something that we can actually measure and, thus, determine the value of. the voltage \(ΔV_{AB}\) is defined as \(ΔV_{AB}≡\int_{r_A}^{r_B}\vec{E} ·d\vec{r}\); if we substitute \(ΔV_{AB}\) into Equation (2) then we have
$$W=q ΔV_{AB}.\tag{3}$$
Equations (1), (2) and (3) all represent the work done by any arbitrary electric force \(\vec{F}_E\) moving a charged particle along a path \(c\) from one location to another (in our case, from Conductor A to Conductor B); but the point is that Equation (3) is the most useful since voltage is what engineers typically measure.
Let's now think about how to calculate the work done moving infinitely many point charges (each having a charge of \(dq\)) from Conductor A to Conductor B. If we move an ininitely many point charges of charge \(dq\) from Conductor A to Conductor B, then how much additional charge does Conductor B get? To answer this question, we can just add up every charge \(dq\) added to conductor B. But we have to add up an infinite number of these charges and we must take the infinite sum, \(∫\). Doing so, we find that the total charge added to Conductor B is \(Q=\int{dq}\). How much charge did Conductor A lose? WE could invoke the law of conservation charge if we wanted to be technical, but on a more intuitive level, here's an easy way to think about it: since we removed an amount of charge \(Q\) from Conductor A (in order to move it to Conductor B), the total change in charge of Conductor A is simply just \(-Q\).
Let's now ask a slightly more difficult question: what was the total work done moving all of the charges? We could alreadyy be able to figure out the amount of work done moving a single ppoint charge \(dq\) using Equation (3). If the point charge being moved becomes infinitesimally small, then what's on the opposite sidfe of the equal sign (the work) must also become infinitesimally small and Equation (3) simplifies to
$$dW=ΔV_{AB}dq.\tag{4}$$
Out of all the infinite number of charges (each with charge \(dq\)) that we want to move from Conductor A to Conductor B, Equation (4) represents the work done in moving just one of those charges from Conductor A to Conductor B. But in order to find the total amount of work associated with moving the charge \(Q\) to Conductor B, we must add up the work done in moving every charged particle to get
$$\int{dW}=\int{ΔV_{AB}dq}$$
$$W=\int{ΔV_{AB}dq}.\tag{5}$$
Put simply, Equation (3) is used to find the work done moving a single point charge (but we'd use Equation (4) instead if that point charge is infinitesimally small) and Equation (5) is used to find the work done moving many charged particles. Since the way that a capacitor is charged in real life is by having an electric field (which applies an electric force) moving many charged paarticles (electrons) from one conductor to another, Equation (5) will be mucj more useful to use in our analysis of capacitors.
What is work, though? Work, of course, measures the amount of energy transfered into or out of a system. Equation (5) is used to calculate how much energy is transfered into or out of the capacitor when you strip one of the conductors of that capacitor of its electrons and move them to the other conductor comprising the capacitor.
$$ΔKE+ΔPE=W+Q+T_{other}$$
I have written the law of conservation of energy above. In our reference frame, the capacitor obviously isn't moving away from us since capacitors are held in place in electrical circuits. There is no loss of energy in the capacitor associated with \(T_{other}\); in other words, we'll assume that the capacitor doesn't lose energy by radiating it away, nor that energy is gained by adding electrons to the system, nor any of the other energy transfer mechanisms besides work and heat. The only non-trivial idealization we'll make is that the capacitors doesn't lose any energy due to heat. In some applications, it is necessary to consider energy lost in a capacitor due to heat. But for out purposes, we'll assume that the heat loss is negligible. Making all of these assumptions, we can write \(ΔPE=W\) and simplify Equation (5) to
$$ΔPE=\int{ΔV_{AB}dq}.\tag{6}$$
In other words, as one of the conductors in a capacitor gets stripped of its electrons which get moved to the other conductor, the transfer of energy \(W\) into the capacitor is additional electrical potential energy.
Well, that concludes this lesson. Hopefully this gives you a sense that by charging a capacitor (moving charges from one conductor to another), this adds electrical energy to the system.