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\section*{Activity 3: Solution for electric field}
\subsection*{Find the electric field in all space due to a ring with total charge $Q$ and
radius $R$}

\begin{equation}
\vf E = {1\over 4\pi\epsilon_0} \sum_{i=1}^N
{q_i\,\rr-\rr_i\over|\rr-\rr_i|^3}
\end{equation}
For a ring of charge this becomes
\begin{eqnarray}
\vf E = \int\limits_{\hbox{ring}}
        {1\over 4\pi\epsilon_0} {\lambda(\rrp)\,|d\rrp|\,\rr-\rrp\over|\rr-\rrp|^3}
\end{eqnarray}
where $\rr$ denotes the position in space at which the electric
field is measured and $\rrp$ denotes the position of the charge.

In cylindrical coordinates, $|d\rrp|=R\,d\phi'$, where $R$ is the
radius of the ring. Thus,
\begin{eqnarray}
\vf E = {1\over 4\pi\epsilon_0} \int\limits_0^{2\pi}
       {\lambda(\rrp)\,R\,d\phi'\,\rr-\rrp\over|\rr-\rrp|^3}
\end{eqnarray}
Assuming constant linear charge density for a ring with charge Q and
radius R, $\lambda(\rrp) = {Q\over{2\pi R}}$ Thus,
\begin{eqnarray}
\vf E = {1\over 4\pi\epsilon_0} {Q\over 2\pi} \int\limits_0^{2\pi}
{d\phi'\,\rr-\rrp\over|\rr-\rrp|^3}
\end{eqnarray}
Since $\rr$ and $\rrp$ are not necessarily in the same direction, we
cannot simply leave $|\rr-\rrp|$ in curvilinear coordinates and
integrate directly. One solution to this problem is to go back and
forth between cylindrical and cartesian coordinates to represent
$\rr-\rrp$  $\medskip$
\begin{eqnarray}
\rr-\rrp &=& (x - x\Prime)\ii +(y - y\Prime)\jj + (z -
z\Prime)\kk\\\noalign{\smallskip} &=& (r\cos\phi -
R\cos\phi\Prime)\ii + (r\sin\phi - R\sin\phi\Prime)\jj + (z -
z\Prime)\kk
\end{eqnarray}
And
\begin{equation}
|\rr-\rrp| = \sqrt{r^2 - 2rR\cos(\phi - \phi') + R^2 + z^2}
\end{equation}
The electric field can now be represented by the elliptic integral
\begin{eqnarray}
\vf E = {1\over 4\pi\epsilon_0} {Q\over 2\pi} \int\limits_0^{2\pi}
{{[(r\cos\phi - R\cos\phi')\ii + (r\sin\phi - R\sin\phi')\jj +
z\kk]\,d\phi'}\over{(r^2 - 2rR\cos(\phi - \phi') + R^2 +
z^2)^{3/2}}}
\end{eqnarray}
\subsection{The $z$ axis}
For points on the $z$ axis, $r = 0$ and the integral simplifies to
\begin{eqnarray}
\vf E = {1\over 4\pi\epsilon_0} {Q\over 2\pi} \int\limits_0^{2\pi}
{{[-R\cos\phi'\,\ii + -R\sin\phi'\,\jj + z\kk]\,d\phi'}\over{(R^2 +
z^2)^{3/2}}}
\end{eqnarray}
Doing the integral results in
\begin{eqnarray}
\vf E = {Q\over 4\pi\epsilon_0} {{z\kk}\over{(R^2 + z^2)^{3/2}}}
\end{eqnarray}
\subsection{The $x$ axis}
For points on the $x$ axis, $z = 0$ and $\phi = 0$, so the integral
simplifies to
\begin{eqnarray}
\vf E = {1\over 4\pi\epsilon_0} {Q\over 2\pi} \int\limits_0^{2\pi}
{{[(r - R\cos\phi')\,\ii + -R\sin\phi'\,\jj]\,d\phi'}\over{(r^2 -
2rR\cos\phi' + R^2)^{3/2}}}
\end{eqnarray}
let $u =r^2 - 2rR\cos\phi' + R^2$, then $du = 2rR\sin\phi'd\phi'$,
and for the $\jj$ component the integral becomes
\begin{eqnarray}
\vf E_j = {1\over 4\pi\epsilon_0} {Q\over 2\pi}{1\over {2r}}
\int\limits_0^{2\pi} {{du\jj}\over{u^{3/2}}}
\end{eqnarray}
Doing the integral results in
\begin{equation}
\vf E_j  = 0
\end{equation}
Thus the $\jj$  component disappears and results in the elliptic
integral with only an $\ii$  component
\begin{eqnarray}
\vf E = {1\over 4\pi\epsilon_0} {Q\over 2\pi} \int\limits_0^{2\pi}
{{(r - R\cos\phi')\,\ii\,d\phi'}\over{(r^2 - 2rR\cos\phi' +
R^2)^{3/2}}}
\end{eqnarray}
 \vfill\eject
\end{document}