Mutual Electrostatic Energy Derivation.

[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(\phi<b>E</b>)=<b>E</b>(dot)\nebla\phi+\phi*\nebla(dot)<b>E</b>

but why then (1)=0 here?

thanks.

 

[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.

 

[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.