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Homework Help: Variations of electrostatic potential energy

  1. Feb 21, 2013 #1

    (this isn't homework but I thought it wasn't worthy of the main forums)

    I'm looking to clarify some thoughts about the various forms of electrostatic potential energy.

    As I understand, the expression

    [itex]\displaystyle \frac{\epsilon_{o}}{2} \int \textbf{E} \cdot \textbf{E} \; dV[/itex]

    over all space gives the total work required to assemble the charge distribution that produces the field [itex]\textbf{E}[/itex].

    Two questions:

    1. Suppose I have two charge distributions. Is the total potential energy of the system equal to the self energy of each distribution alone plus the interaction energy between the two systems? (are there other components of potential energy?)

    2. How do you calculate the interaction energy of two charge distributions? I know how to do it for point charges ([itex]q_1 \phi_2[/itex] or [itex]q_2 \phi_1[/itex]).


    If distribution 1 has field [itex]\textbf{E}_1[/itex] and distribution 2 has field [itex]\textbf{E}_2[/itex], the TOTAL energy should be

    [itex]\displaystyle \frac{\epsilon_{o}}{2} \int \textbf{E}_1 \cdot \textbf{E}_1 + 2\textbf{E}_1 \cdot \textbf{E}_2 + \textbf{E}_2 \cdot \textbf{E}_2 \; dV[/itex].

    Speculation: Is the first term the self energy of distribution 1, the second term the interaction energy, and the last term the self energy of distribution 2?

    Is there any way to find expressions for the energies of charge distributions using potential?

    Thank you very much!
  2. jcsd
  3. Feb 22, 2013 #2


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    Yes. However, for a system of point charges the self energy is usually ignored because it is a fixed quantity that doesn't depend on the relative positions of the charges and, moreover, it is infinite!
    Yes, that is correct. The "self energy of distribution 1" (represented by your first integral) would include the self energy of the particles making up distribution 1 as well as the interaction energy of the particles within distribution 1.
    You could substitute ##\vec{E} = -\vec{\nabla} \phi## into your integral [itex]\displaystyle \frac{\epsilon_{o}}{2} \int \textbf{E} \cdot \textbf{E} \; dV[/itex]
    Last edited: Feb 22, 2013
  4. Feb 23, 2013 #3
    Thanks, TSny! Very helpful! I think I understand it now.
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