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# 2.6: Two Semicylindrical Electrodes

This section requires that the reader should be familiar with functions of a complex variable and conformal transformations. For readers not familiar with these, this section can be skipped without prejudice to understanding following chapters. For readers who are familiar, this is a nice example of conformal transformations to solve a physical problem.

$$\text{FIGURE II.5}$$

We have two semicylindrical electrodes as shown in figure $$II$$.5. The potential of the upper one is 0 and the potential of the lower one is $$V_0$$. We'll suppose the radius of the circle is 1; or, what amounts to the same thing, we'll express coordinates $$x$$ and $$y$$ in units of the radius. Let us represent the position of any point whose coordinates are (x , y) by a complex number $$z = x + iy$$.

Now let $$w = u + iv$$ be a complex number related to $$z$$ by $$w=i\left (\frac{1-z}{1+z}\right )$$; that is, $$z=\frac{1+ iw}{1- iw}$$. Substitute $$w = u + iv \text{ and }z = x + iy$$ in each of these equations, and equate real and imaginary parts, to obtain

\begin{align}\label{2.6.1}u&=\frac{2y}{(1+x)^2+y^2};\quad\quad &&v=\frac{1-x^2+y^2}{(1+x)^2+y^2};\\ x&=\frac{1-u^2-v^2}{u^2+(1+v)^2}; &&y=\frac{2u}{u^2+(1+v)^2}.\label{2.6.2}\end{align}

In that case, the upper semicircle $$(V = 0)$$ in the $$xy$$-plane maps on to the positive $$u$$-axis in the $$uv$$-plane, and the lower semicircle $$(V = V_0)$$ in the $$xy$$-plane maps on to the negative $$u$$-axis in the $$uv$$-plane. (Figure $$II$$.6.) Points inside the circle bounded by the electrodes in the $$xy$$-plane map on to points above the $$u$$-axis in the $$uv$$-plane.

$$\text{FIGURE II.6}$$

In the $$uv$$-plane, the lines of force are semicircles, such as the one shown. The potential goes from 0 at one end of the semicircle to $$V_0$$ at the other, and so equation to the semicircular line of force is

$\label{2.6.3}\frac{V}{V_0}=\frac{\text{arg}\,w}{\pi}$

or

$\label{2.6.4}V=\frac{V_0}{\pi}\tan^{-1}(v/u).$

The equipotentials ($$V$$ = constant) are straight lines in the $$uv$$-plane of the form

$\label{2.6.5}v=fu.$

(You would prefer me to use the symbol $$m$$ for the slope of the equipotentials, but in a moment you will be glad that I chose the symbol $$f$$.)

If we now transform back to the $$xy$$-plane, we see that the equation to the lines of force is

$\label{2.6.6}V=\frac{V_0}{\pi}\tan^{-1} \left (\frac{1-x^2-y^2}{2y}\right ).$

and the equation to the equipotentials is

$\label{2.6.7}1-x^2-y^2=2fy,$

or

$\label{2.6.8}x^2+y^2+2fy-1=0$

Now aren't you glad that I chose $$f$$ ? Those who are handy with conic sections (see Chapter 2 of Celestial Mechanics) will understand that the equipotentials in the $$xy$$-plane are circles of radii $$\sqrt{f^2 + 1}$$, whose centres are at $$(0 , \pm f )$$, and which all pass through the points $$(\pm 1 , 0)$$. They are drawn as blue lines in figure $$II$$.7. The lines of force are the orthogonal trajectories to these, and are of the form

$\label{2.6.9}x^2+y^2+2gy+1=0$

These are circles of radii $$\sqrt{g^2 −1}$$ and have their centres at $$(0 , \pm g)$$. They are shown as dashed red lines in figure $$II$$.7.

$$\text{FIGURE II.7}$$