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Mutual Electrostatic Energy Derivation.

  1. May 7, 2007 #1


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    i don't quite understand the derivation of mutual electrostatic energy of two charged system:
    [tex]U_12=\frac{1}{4\pi}\intE_1(dot)E_2dV=-\frac{1}{4\pi}\intE_1(dot)\nebla\phi_2 dV= \frac{1}{4\pi}\int \phi_2(dot)\nebla(dot)E_1=\int \phi_2*\rho_1dV[/tex]
    i undersantd that we are using here: [tex] (1)=\nebla(\phiE)=E(dot)\nebla\phi+\phi*\nebla(dot)E[/tex]

    but why then (1)=0 here?

  2. jcsd
  3. May 7, 2007 #2

    Meir Achuz

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    Your Latex is garbled. Use U_{12} for subsrcipts and \cdot instead of \dot.
    There are too m any other misprints to make sense of it.
  4. May 8, 2007 #3


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    a correction:
    [tex]U_{12}=\frac{1}{4\pi}\int E_1 \cdot E_2 dV=-\frac{1}{4\pi}\int E_1 \cdot \nabla\phi_2 dV= \frac{1}{4\pi}\int \phi_2 \nabla \cdot E_1 \cdot da= \int \phi_2 * \rho_1 dV[/tex]
    [tex] (1)=\nabla(\phi E)=E\cdot \nabla \phi+ \phi * \nabla \cdot E[/tex]
    where E is a vector field, and phi is a potential field.
    i want to understand why in the first equations (the first line) why we get that (1)=0, i hope now the latex is better.
    Last edited: May 8, 2007
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