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
\[V(s,\phi,0) \approx \frac{1}{4\pi\epsilon_0}\frac{Q}{R}\left[ 1+\frac{s^2}{4R^2}+\frac{9s^4}{64R^4}\right]\]
A conical surface (an empty ice-cream cone) carries a uniform charge density \(\sigma\). The height of the cone is \(H\), and the radius of the top is \(R\).
Sense-making: It is interesting to note that the result increases with \(H\), so as I make the cone larger, the potential at this point goes up. This may be initially unintuitive, as the point of interest is getting farther away from where the charge is, but because the cone has charge all the way around it, and the charge density is constant, you actually are accumulating more charge than you need to balance out the increasing distance.
The cone that I'll be considering has its vertex at the origin and opens up on the positive z-axis. I'll use cylindrical coordinates in my solution (but I could also use spherical if I wanted). Starting with the general expression for calculating potential due to a surface charge distribution: \begin{equation} V(\vec{r})=\int \int \frac{k \,\sigma \,dA'}{|\vec{r}-\vec{r}\,'|} \end{equation} then I'll use what I know. For a cone with height \(H\) and base radius \(R\), the equation of the cone is \(z'=\frac{H}{R}\,s'\).
Draw two infinitesimal displacement vectors on the surface of the cone: \begin{equation} d\vec{r}_1' = s'\,d\phi'\,\hat{\phi}' \qquad d\vec{r}_2 '= ds'\,\hat{s}' + dz'\,\hat{z}' = \left(\hat{s}' + \frac{H}{R}\,\hat{z}'\right)\,ds' \end{equation} which in turn yields \begin{equation} dA' = |d\vec{r}_1'\times d\vec{r}_2'| = \sqrt{1+\frac{H^2}{R^2}}\; s'\, ds'\,d\phi' \end{equation}
Using this differential area element, and inserting the expression discussed in class (and in previous HW) for the distance \(|\vec{r}-\vec{r}\,'|\) in cylindrical coordinates, the potential becomes: \begin{equation} V(\vec{r}) = \int_0^{2\pi} \int_0^R \frac{k \,\sigma\sqrt{1+\frac{H^2}{R^2}} \,s'\,ds'\,d\phi'} {\sqrt{s^2+s'^2-2ss'\cos(\phi-\phi') +(z-z')^2}} \end{equation} Since we know where the potential is being evaluated, we can simplify the denominator; we have \(z=H\) and \(s = 0\), so that: \begin{align} V(s=0,z=H) &= \int_0^{2\pi} \int_0^R \frac{k \,\sigma\sqrt{1+\frac{H^2}{R^2}} \,s'\,ds'\,d\phi'} {\sqrt{{s'}^2 + (H-z')^2}} \\ &= \int_0^{2\pi} \int_0^R \frac{k \,\sigma\sqrt{1+\frac{H^2}{R^2}} \,s'\,ds'\,d\phi'} {\sqrt{{s'}^2 + (H-\frac{H}{R}s')^2}} \\ &= k \,\sigma\sqrt{1+\frac{H^2}{R^2}} \,\int_0^{2\pi} \int_0^R \frac{s'\,ds'\,d\phi'} {\sqrt{{s'}^2 + \left(\frac{H}{R}\right)^2(R-s')^2}} \end{align}
- Using technology or an integral table to evaluate the integral, we finally obtain: \begin{equation} V(s=0,z=H) = \pi k\,\sigma H \,\ln{\left[\frac{\sqrt{2}+1}{\sqrt{2}-1}\right]} \end{equation}