§ {Activity: Power Series for Dipoles} combines both the math (power series) and physics (electrostatic potential) we've been discussing.

All of the cases can be approximated by exploiting one of the common power series that it was suggested that you memorize — there's no need to compute the coefficients using differentiation. But it's important to understand the approximation process: Make sure you are expanding in terms of something small! This requires rewriting each potential, typically to make it look like \begin{eqnarray*} (1 + \hbox{something small})^{\hbox{power}} \end{eqnarray*} In the series, you are adding terms of different powers. Therefore, the something small must be dimensionless, otherwise the power series is dimensionally nonsense.

You may be tempted to start by adding up these fractions, and putting them over a common denominator. In most cases, however, the resulting expression is more difficult to expand in a power series than the one you started with.

After expanding these potentials near the origin or far away, look at the leading term or two and ask yourself if they are what you expect. Think especially about the symmetry or anti-symmetry of the charge distribution and the symmetry or anti-symmetry of your answer. It is very easy to ignore the sign of variables and inadvertently to drop absolute value signs. Also think about whether the leading terms give you the behavior you expect? Where is each series valid?

For two positive charges $+Q$ positioned at $z=\pm D$: \begin{eqnarray*} V(0,0,z) &=& {Q\over 4\pi\epsilon_0}\, {2\over D} \left( 1 + {z^2\over D^2} + {z^4\over D^4} + …\right) \quad\qquad\qquad \hbox{for $|z|<D$} \\ V(0,0,z) &=& {Q\over 4\pi\epsilon_0}\, {2\over |z|} \left( 1 + {D^2\over z^2} + {D^4\over z^4} + …\right) \quad\qquad\qquad \hbox{for $|z|>D$} \\ V(x,0,0) &=& {2Q\over 4\pi\epsilon_0} {1\over D} {\left(1 - {1\over 2}{{x^2}\over {D^2}} + {3\over 8}{{x^4}\over {D^4}} + …\right)} \qquad\qquad \hbox{for $|x|<D$} \\ V(x,0,0) &=& {2Q\over 4\pi\epsilon_0} {1\over |x|} {\left(1 - {1\over 2}{{D^2}\over {x^2}} + {3\over 8}{{D^4}\over {x^4}} + …\right)} \qquad\qquad \hbox{for $|x|>D$} \end{eqnarray*}

For a positive charge $+Q$ positioned at $z=+D$ and a negative charge $-Q$ positioned at $z=-D$: \begin{eqnarray*} V(0,0,z) &=& {Q\over 4\pi\epsilon_0}\, {2\over D} \left({z\over D} + {z^3\over D^3} + {z^5\over D^5} + …\right) \qquad\qquad\hbox{for $|z|<D$} \\ V(0,0,z) &=& {Q\over 4\pi\epsilon_0}\, {2\over |z|} \left({D\over z} + {D^3\over z^3} + {D^5\over z^5} + …\right) \qquad\qquad\hbox{for $|z|>D$} \\ V(x,y,0) &=& 0 \end{eqnarray*}