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What does the energy of an electrostatic field mean?

  1. Aug 14, 2011 #1
    Hello everyone. So that is my question: What does the energy of an electrostatic field mean?

    A more detailed description of my background or understanding will help you to provide an answer:
    The process to deduce such a thing is usually calculate the work done in order to bring charge configuration to its position. However I've asked my self, suppose a charge configuration has an energy of 1 MeV. What does that mean/imply?

    Morever, what does it mean that in such deduction the energy of "creating charges" was not considered?

    Try to answer first my first question, the others are just guides for your answers.

    Thanks everyone very much in advance.
     
  2. jcsd
  3. Aug 14, 2011 #2
    Hello Bruno,

    Good question. As I see it, it's simply the energy it took for you, given the particles, to put them into that specific configuration (if you brought them in from infinity -- you take infinity, because "there" the particles don't interact and you can move any particle freely about without it costing you anything).

    You know the force that works between charged particles (Lorentz' force law) and you know the expression for the energy it takes you to move a particle (definition of work: integrating force over distance). These two are enough to tell you how much energy it takes to put every charged particle in its specific place (you simply fill in the Lorentz' force in the work integral and then integrate the position of the particles from infinity to their respective location).

    How you concretely calculate this: first you move over particle 1 from infinity to its location. This costs you nothing because the other particles are at infinity and thus not reacting with your first particle. Then you bring in particle 2. This is simply the coulomb force between to particles, integrated from infinity to particle 2's location; a short calculation gives [itex]W = k \frac{q_1 q_2}{r_{12}}[/itex] (and you will recognize this as "the potential energy", you see it's by definition simply the energy needed to construct the situation). Now we bring in particle 3, the total force working on it is the vector sum of the coulomb force of 1 on 3 and that of 2 on 3. Because the total force is this sum, you can also write the work integral as a sum (because the integral is linear in force); if you look at the expressions of each of the two integrals, you see they're equivalent to the situation you just had where you were adding particle 2! (write it down and think about it!) So analogously, without computation you get [itex]W = k \frac{q_1 q_2}{r_{12}} + k \frac{q_1 q_3}{r_{13}} + k \frac{q_2 q_3}{r_{23}} [/itex]. The form for the construction of N particles should now become obvious (again, think about it).

    The sum you get in the end is what you call "the energy of the charge configuration" or its "potential energy" ("potential" because you can get it out again).

    Let's assume we started from extended charges (by this I mean "charge 1" etc weren't point charges, but extended objects such that each infinitesimal piece had an infinitesimal charge on it). In the above situation the charges themselves were given, and we called the above W-expression "the energy of the charge configuration", but if you're really ambitious, you can start from even less: you can even decide to make charge 1, charge 2, etc (and afterwards put them into place, again starting from infinity). It should be obvious that this will cost even more energy. But how much more?

    The key question is then: how much does it take to make an extended charge? Well, imagine what you need to do for it: you need to bring an infinite amount of infinitesimal building blocks together, each with an infinitesimal amount of charge. The amount of times I've just said "infinite(simal)" should make it clear we're gonna need a (volume) integral. But you should also notice another thing: what we actually have to do is analogous what we did just now: bringing charges together from infinity. The only difference now is that we have an infinite number of charge, and that each charge in itself is infinitesimally small (hence the integral instead of a sum as above). You can get the integral that does the job by rewriting the above expression W; it's not directly obvious, but not hard, so I can also explain that, but I'll leave the current post as it is, as it already contains quite some information and I want to see if you understand all the previous stuff first.
     
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