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\begin{center}
\textbf{Vector Fields}

\textit{\textbf{Conservative Fields}}
\end{center}

\textbf{Question}

Two sources of strength $m$ are placed at $(0,0,\pm l)$. Find the
velocity due to these sources, and state where the velocity is
zero. Determine the velocity at the point $(x,y,0)$ in the $xy$-plane,
and state where in the $xy$-plane the speed is greatest.


\textbf{Answer}

For the two-source system
$$\phi(x,y,z) = \phi(\un{r}) = -\frac{m}{| \un{r} - l \un{k}|} -
\frac{m}{|\un{r} + l \un{k}|}.$$
This gives the velocity field
\begin{eqnarray*}
\un{v}(\un{r}) & = & \nabla \phi(\un{r})\\
& = & \frac{m(\un{r}-l\un{k})}{|\un{r}-l\un{k}|^3} + \frac{m(\un{r} +
l\un{k})}{|\un{r} + l\un{k}|^3}\\
& = & \frac{m(x\un{i}+y\un{j} + (z-l)\un{k})}{[x^2+y^2+(z-l)^2]^{3/2}]}
+ \frac{m(x\un{i} + y \un{j} + (z+l)\un{k})}{[x^2+y^2+(z+l)^2]^{3/2}}.
\end{eqnarray*}
Notice that $v_1=0$ if and only if $x=0$, and $v_2=0$ if and only if
$y=0$.

$$\un{v}(0,0,z) = m \left ( \frac{z-l}{|z-l|^3} + \frac{z+l}{|z+l|^3}
\right ) \un{k}$$
is only $\un{0}$ if and only if $z=0$. So $\un{v}=\un{0}$ at the
origin only.

For points in the $xy$-plane
$$\un{v}(x,y,0) = \frac{2m(x\un{i} +y \un{j})}{(x^2+y^2+l^2)^{3/2}}.$$
So the velocity is radially away from the origin in the plane, as is
required by symmetry. The speed at $(x,y,0)$ is given by
\begin{eqnarray*}
v(x,y,0) & =& \frac{2m\sqrt{x^2+y^2}}{(x^2+y^2+l^2)^{3/2}}\\
& = & \frac{2ms}{(s^2+l^2)^{3/2}} = g(s)\\
\textrm{with }s & = & \sqrt{x^2+y^2}\\
& & \\
\textrm{For max }g(s) & & \\
0 & = & g'(s) = 2m\frac{(s^2+l^2)^{3/2} - \frac{3}{2}s(s^2+l^2)^{1/2}
2s}{(s^2+l^2)^3}\\
& = & \frac{2m(l^2 -2s^2)^{3/2}}{(s^2+l^2)^{5/2}}.
\end{eqnarray*}
So the speed in the $xy$ plane is at its greatest when it the points of
the circle $x^2+y^2 = l^2/2$.
 
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