# 3.9: Potential at a Large Distance from a Charged Body

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We wish to find the potential at a point P at a large distance \(R\) from a charged body, in terms of its total charge and its dipole, quadrupole, and possibly higher-order moments. There will be no loss of generality if we choose a set of axes such that P is on the \(z\)-axis.

\(\text{FIGURE III.14}\)

We refer to Figure \(III\).14, and we consider a volume element \(δτ\) at a distance *r* from some origin. The point P is at a distance *r* from the origin and a distance \(∆ \text{ from }δτ\). The potential at P from the charge in the element \(δτ\) is given by

\[4\pi\epsilon_0 \delta V = \dfrac{\rho \delta \tau }{\Delta} = \dfrac{\rho}{R}\left ( 1+\dfrac{r^2}{R^2}-\dfrac{2r}{R}\cos \theta \right )^{-1/2}\delta \tau ,\]

and so the potential from the charge on the whole body is given by

\[4\pi\epsilon_0 V =\dfrac{1}{R} \int \rho \left ( 1+\dfrac{r^2}{R^2} -\dfrac{2r}{R}\cos \theta \right )^{-1/2}\delta \tau .\]

On expanding the parentheses by the binomial theorem, we find, after a little trouble, that this becomes

\[4\pi\epsilon_0 V = \dfrac{1}{R}\int \rho \,d\tau + \dfrac{1}{R^2}\int \rho r P_1 (\cos \theta)\,d\tau + \dfrac{1}{2!R^3}\int \rho r^2 P_2 (\cos \theta)]\, d\tau + \dfrac{1}{3!R^4}\int \rho r^3 P_3 (\cos \theta )\,d\tau + ... , \]

where the polynomials *P* are the Legendre polynomials given by

\[\begin{align}P_1 (\cos \theta ) &= \cos \theta \\ P_2(\cos \theta) &= \dfrac{1}{2}(3\cos^2 \theta -1 ), \\ P_3(\cos \theta)&=\dfrac{1}{2}(5\cos^3 \theta - 3\cos \theta ). \\ \end{align}\]

We see from the forms of these integrals and the definitions of the components of the dipole and quadrupole moments that this can now be written:

\[4\pi\epsilon_0 V = \dfrac{Q}{R} + \dfrac{p}{R^2}+\dfrac{1}{2R^3}(3q_{zz}-Tr\textbf{q})+...,\label{3.9.7}\]

Here Tr **q** is the trace of the quadrupole moment matrix, or the (invariant) sum of its diagonal elements. Equation \ref{3.9.7} can also be written

\[4\pi\epsilon_0V=\dfrac{Q}{R}+\dfrac{p}{R^2}+\dfrac{1}{2R^3}[2q_{zz}-(q_{xx}+q_{yy})]+... . \]

The quantity \(2q_{zz}-(q_{xx}+q_{yy})\) of the diagonalized matrix is often referred to as “the” quadrupole moment. It is zero if all three diagonal components are zero or if \(q_{zz}=\dfrac{1}{2}(q_{xx}+q_{yy})\). If the body has cylindrical symmetry about the \(z\)-axis, this becomes \(2(q_{zz}-q_{xx})\) .

.)

\(\text{FIGURE III.15}\)

The solution to this exercise is easy if you know about *Legendre polynomials*. See Section 1.14 of my notes on Celestial Mechanics. What you need to know is that the expansion of \((1-2ax+x^2)^{-1/2}\) can be written as a series of Legendre polynomials, namely \(P_0(x)+xP_1(x)+x^2P_2(x)+...\). You also need a (very small) table of Legendre polynamials, namely \(P_0(x)=1,\,P_1(x)=x,\,P_2(x)=\dfrac{1}{2}(3x^2-1)\). Given that, you should find the exercise very easy.