Mandl and Shaw, page 16, eqn (1.56)

[Jimmy Snyder]

Homework Statement

\int{E}_L^2d^3x = \int\frac{\rho(x)\rho(x')}{4\pi|x - x'|}d^3xd^3x'

Homework Equations

{E}_L = -\nabla\phi
{\nabla}^2\phi = -\rho

The Attempt at a Solution

\int{E}_L^2d^3x = \int(\nabla\phi)^2d^3x = -\int\phi\nabla^2\phi d^3x = \int\rho(x)\phi(x)d^3x
I suppose to finish up, I need to see why
\phi(x) = \int\frac{\rho(x')d^3x'}{4\pi|x - x'|}
But I don't see it. Or am I on the wrong track.

By the way, I have the 1993 revised edition.

 

[malawi_glenn]

isn't this just plain electromagnetism?

The potential at point x:

<br /> \phi(x) = \int\frac{\rho (x &#039;) d^3x&#039;}{4\pi|x - x&#039;|}<br />

where \rho is the charge distribution?

http://upload.wikimedia.org/math/0/d/c/0dcc14fba817f280f68b9a59247614e2.png

then use your "relevant equations"

 

[Jimmy Snyder]

Thanks malawi_glenn, that's what I needed.