- #1

Vapor88

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**1. Problem Statement:**

There are charges of q placed at distance +a and -a from the origin on the z-axis. There is a charge at the origin of -2q. Express the potential of this point-like linear quadrupole in Legendre polynomials

There are charges of q placed at distance +a and -a from the origin on the z-axis. There is a charge at the origin of -2q. Express the potential of this point-like linear quadrupole in Legendre polynomials

The distance between origin and point is r, the distance between (0,0,a) and r is [tex]l[/tex] and the distance between (0,0,-a) is [tex]l'[/tex]

## The Attempt at a Solution

The potential can be expressed as

[tex](q/(4\pi r)*(1/l + 1/l' + 2/r)[/tex]

Using law of cosines

[tex]l = \sqrt{a^2 + r^2 - 2ar cos(\theta)}[/tex]

[tex]l' = \sqrt{a^2 + r^2 + 2ar cos(\theta)}[/tex]

Solving for 1/l and 1/l' gives

[tex]1/l = \frac{1}{r} (1 + a^2/r^2 - a/r cos (\theta))^{-1/2}[/tex]

[tex]1/l' = \frac{1}{r} (1 + a^2/r^2 + a/r cos (\theta))^{-1/2}[/tex]

that's supposed to be raised to the -0.5, but i don't know how to make it look right in latex

Now

[tex] V = ( \frac{q}{4 \pi \epsilon_{o}} \frac{1}{r} [(1+( \frac{a}{r})^2 - \frac{a}{r} cos (\theta))^{-1/2} + (1+ (\frac{a}{r})^2 + \frac{a}{r} cos (\theta))^{-1/2} - 2] [/tex]

We had an example in class of a dipole that used the expansion

(1+x)^n = 1 + nx + ...

And only used the first two terms of it. Do I need to do the same thing to this, or can I go straight to the Legendre polynomials. I'm not sure where to go from here.

Also... If you take a^2/r^2 to go to zero, then the potential goes to zero as well. So I can't do that.