The potential due to a ring of charge is given by:
\[ V(s,\phi,z)=\frac{1}{4\pi\epsilon_0}\frac{Q}{2\pi}\int_0^{2\pi}\frac{d\phi'}{\sqrt{s^2+R^2-2sR\cos(\phi-\phi')+z^2}}\]
Expand this potential in a power series to fourth order, in the plane of the ring, for \(s<R\). Make sure to integrate. Warning: Make sure you keep all of the terms up to fourth order and none of the terms of higher order. Figuring out which terms to keep is tricky and is the most important lesson from this homework problem.
In the plane \(z=0\),
\[ V(s,\phi,0)=\frac{1}{4\pi\epsilon_0}\frac{Q}{2\pi}\int_0^{2\pi}\frac{d\phi'}{\sqrt{s^2+R^2-2sR\cos(\phi-\phi')}}\]
For \(s<R\) work toward modifying to be in the form \((1+z)^p\) where \(|z|<1\).
\[ V(s,\phi,0)=\frac{1}{4\pi\epsilon_0}\frac{Q}{2\pi}\int_0^{2\pi}(s^2+R^2-2sR\cos(\phi-\phi'))^{-\frac{1}{2}}d\phi'\]
Factor out the big component \((R)\).
\[V(s,\phi, 0) = \frac{1}{4\pi\epsilon_0}\frac{Q}{2\pi}\int_0^{2\pi}\left[R^2\left( \frac{s^2}{R^2}+1-2\frac{s}{R}\cos(\phi-\phi')\right) \right]^{-\frac{1}{2}}d\phi'\]
\[V(s,\phi, 0) = \frac{1}{4\pi\epsilon_0}\frac{Q}{2\pi R}\int_0^{2\pi}\left( \frac{s^2}{R^2}+1-2\frac{s}{R}\cos(\phi-\phi')\right)^{-\frac{1}{2}}d\phi'\]
Rearrange for \(z=\frac{s^2}{R^2}-2\frac{s}{R}\cos(\phi-\phi')\)
\[V(s,\phi,0) = \frac{1}{4\pi\epsilon_0}\frac{Q}{2\pi R}\int_0^{2\pi}\left(1+\frac{s^2}{R^2}-2\frac{s}{R}\cos(\phi-\phi')\right)^{-1/2}d\phi'\]
Now using \((1+z)^p\approx 1 + zp + \frac{p(p-1)}{2!}z^2+\frac{p(p-1)(p-2)}{3!}z^3+\frac{p(p-1)(p-2)(p-3)}{4!}z^4\) With our integrand pieces being \(p=-\frac{1}{2}\) and \(z=\frac{s^2}{R^2}-2\frac{s}{R}\cos(\phi-\phi')\) Keeping only terms of fourth order or lower we have \begin{align*} z&=\frac{s^2}{R^2}-2\frac{s}{R}\cos(\phi-\phi')\\ z^2&=\frac{s^4}{R^4}-4\frac{s^3}{R^3}\cos(\phi-\phi')+4\frac{s^2}{R^2}\cos^2(\phi-\phi')\\ z^3 &= 12\frac{s^4}{R^4}\cos^2(\phi-\phi')-8\frac{s^3}{R^3}\cos^3(\phi-\phi')\\ z^4 &= 16 \frac{s^4}{R^4}\cos^4(\phi-\phi') \end{align*}
Including all of the coefficients, our approximation to fourth order becomes \begin{align*} (1+\frac{s^2}{R^2}-2\frac{s}{R}\cos(\phi-\phi'))^{-\frac{1}{2}} \approx 1&-\frac{1}{2}\left(\frac{s^2}{R^2}-2\frac{s}{R}\cos(\phi-\phi')\right)\\ &+\frac{3}{8}\left(\frac{s^4}{R^4}-4\frac{s^3}{R^3}\cos(\phi-\phi')+4\frac{s^2}{R^2}\cos^2(\phi-\phi')\right)\\ &-\frac{5}{16}\left( 12\frac{s^4}{R^4}\cos^2(\phi-\phi')-8\frac{s^3}{R^3}\cos^3(\phi-\phi')\right)\\ &+\frac{35}{128}\left(16 \frac{s^4}{R^4}\cos^4(\phi-\phi')\right) \end{align*}
Combining like terms in \(\cos(\phi-\phi')\), we get: \begin{align*} \left(1-\frac{s^2}{2R^2}+\frac{3s^4}{8R^4}\right) &+ \left(\frac{s}{R}-\frac{3s^3}{2R^3}\right)\cos(\phi-\phi') + \left(\frac{3s^2}{2R^2} - \frac{15s^4}{4R^4} \right)\cos^2(\phi-\phi')\\ &+ \left(\frac{5s^3}{2R^2} \right)\cos^3(\phi-\phi')+\left(\frac{35s^4}{8R^4}\right)\cos^4(\phi-\phi') \end{align*}
Recall we are integrating so we have \begin{align*} V(s,\phi,0)= \frac{1}{4\pi\epsilon_0}\frac{Q}{2\pi R} &\bigg[ \left( 1-\frac{s^2}{2R^2}+\frac{3s^4}{8R^4}\right)\int_0^{2\pi}d\phi'\\ &+ \left(\frac{s}{R}-\frac{3s^3}{2R^3}\right)\int_0^{2\pi}\cos(\phi-\phi')d\phi'\\ &+ \left(\frac{3s^2}{2R^2} - \frac{15s^4}{4R^4} \right)\int_0^{2\pi}\cos^2(\phi-\phi')d\phi'\\ &+\left(\frac{5s^3}{2R^2} \right)\int_0^{2\pi}\cos^3(\phi-\phi')d\phi'\\ &+\left(\frac{35s^4}{8R^4}\right)\int_0^{2\pi}\cos^4(\phi-\phi')d\phi' \bigg] \end{align*}
The odd powers of cosine integrated over a full period will be equal to zero. The remaining non-zero integrals will be: \begin{align*} \int_0^{2\pi}d\phi'&=2\pi\\ \int_0^{2\pi}\cos^2(\phi-\phi')d\phi' &= \pi\\ \int_0^{2\pi}\cos^4(\phi-\phi')d\phi' &= \frac{3\pi}{4} \end{align*}
Which will give us a final answer of
\[V(s,\phi,0) \approx \frac{1}{4\pi\epsilon_0}\frac{Q}{R}\left[ 1+\frac{s^2}{4R^2}+\frac{9s^4}{64R^4}\right]\]