Mutual Electrostatic Energy Derivation.

1. May 7, 2007

MathematicalPhysicist

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

but why then (1)=0 here?

thanks.

2. May 7, 2007

Meir Achuz

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.

3. May 8, 2007

MathematicalPhysicist

a correction:
$$U_{12}=\frac{1}{4\pi}\int E_1 \cdot E_2 dV=-\frac{1}{4\pi}\int E_1 \cdot \nabla\phi_2 dV= \frac{1}{4\pi}\int \phi_2 \nabla \cdot E_1 \cdot da= \int \phi_2 * \rho_1 dV$$
$$(1)=\nabla(\phi E)=E\cdot \nabla \phi+ \phi * \nabla \cdot E$$
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