electric dipole moment

**joschua** · Nov6-09, 12:47 PM

Hi

I want to calculate the electric dipole moment of point charges along the z-axis with distances a and with the charge distribution

$LaTeX Code: \\varrho (\\vec{x}) = q \\delta (\\vec{x}) - 2q \\delta (\\vec{x} - \\vec{a}) + q \\delta (\\vec{x} - 2 \\vec{a})$

and of course $LaTeX Code: \\vec{a} = a \\vec{e}_{z}$

I did the following:

$LaTeX Code: \\vec{p} = \\int \\vec{x}single-quote \\varrho (\\vec{xsingle-quote}) d^{3}xsingle-quote$

$LaTeX Code: = q \\int \\vec{x}single-quote \\delta (\\vec{x}) d^{3}xsingle-quote - 2q \\int \\vec{x}single-quote \\delta (\\vec{x} - \\vec{a}) d^{3}xsingle-quote + q \\int \\vec{x}single-quote \\delta (\\vec{x} - 2 \\vec{a}) d^{3}xsingle-quote$

Now I have some questions:

1.) I guess I should write a prime in the arguments of the delta functions. Is this true? (The definition of my electric dipole moment is with prime, the given distribution without but that makes no sense? I should write a prime to all x vectors or no primes. correct?

2.) How to evaluate the integrals further? I know that the delta function is only one at the points of the charges and everywhere else zero but what to do with the x-vectors?

If this would be a normal integral I would do integration by parts, but this makes no sense here.

In general I know the relation that

$LaTeX Code: \\int f(x) \\delta (x-a) dx = f(a)$

but here I have no Function f because x is a vector and I am in 3-d space.

I am confused. Please help me.

edit:

I wanted to post it in Classical Physics and not here. Wrong forum. Sorry... maybe a nice mentor will move it? :)

**clem** · Nov6-09, 02:44 PM

1. All integration variables are x', including in the delta.
2. The f(x) here is just f(x)=x. You are integrating only along the x axis.

**joschua** · Nov9-09, 01:47 PM

thanks, I got it