Calculation of capacitance between two cylinders

Question

I'm trying to calculate the capacitance of two circular cylinders (it's a coil). I'm ok with the physics but I'm stuck in a point of the calculation. I have a complex function which contains the electrostatic potential ($U$) and the force lines ($V$), also, $a$ is the distance between the two cylinders . Here are the functions:

$x=\frac{a\sin V}{\cosh U -\cos V}$ is the real part and $y=\frac{a\sinh U}{\cosh U-\cos V}$ is the imaginary part.

So I get to this point and after this the book ("William R. Smythe - Static and Dinamic electricity") tells: "Eliminating V from these equiations gives:"

$x^{2}+y^{2}-2ay\coth U +a^{2}=0$

So here is my problem, I can't figure out how to get to this equation using $x$ and $y$. I know that I have to obtain a circle equation, because this is the shape of the equipotential lines around the cylinders but I don't see how to get to the final expression.

Many thanks

Felipe.

Answer 1

We start from finding $sinV$ and $cosV$:

$\displaystyle y=\frac{a\sinh U}{\cosh U-\cos V} \to cosV=coshU-\frac ay sinhU$

$\displaystyle x=\frac{a\sin V}{\cosh U -\cos V}=\frac{a\sin V}{\cosh U -coshU+\frac ay sinhU}=\frac{y\sin V}{sinhU} \to sinV=\frac x y sinhU$


We know that $\displaystyle sin^2V+cos^2V=1 \\\displaystyle\to (coshU-\frac ay sinhU)^2+(\frac x y sinhU)^2=1 \\\displaystyle \to cosh^2U+\frac{a^2}{y^2}sinh^2U-2\frac a y sinhU coshU + \frac{x^2}{y^2}sinh^2U=1 $


On the other hand, we also know that:

$cosh^2U-sinh^2U=1 \to cosh^2U=1+sinh^2U$

So we can rewrite the equation to:

$\\\displaystyle \to 1+ sinh^2U+\frac{a^2}{y^2}sinh^2U-2\frac a y sinhU coshU + \frac{x^2}{y^2}sinh^2U=1 \\\displaystyle \to sinh^2U+\frac{a^2}{y^2}sinh^2U-2\frac a y sinhU coshU + \frac{x^2}{y^2}sinh^2U=0$

By multiplying the equation to $\dfrac{y^2}{sinh^2U}$:

$\\\displaystyle \to y^2 + a^2 - 2aycothU +x^2 =0$ (Proved).

Answer 2

From $$ x^{2}+y^{2}-2ay\coth U +a^{2}=0 $$ write $$ -2ay\coth U = -a^2 - x^2 - y^2 $$ and then $$ \coth U = \frac{+a^2 + x^2 + y^2}{2ay} $$ Now hold $U$ to be a constant, so that $\coth U = k$, some constant. You have $$ k = \frac{+a^2 + x^2 + y^2}{2ay} $$ so \begin{align} 2aky &= a^2 + x^2 + y^2\\ 0 &= a^2 + x^2 + y^2 - 2aky \\ (ak)^2 &= a^2 + x^2 + y^2 - 2aky + (ak)^2 \\ (ak)^2 &= a^2 + x^2 + (y - ak)^2 \\ 0 &= a^2 (1-k^2) + x^2 + (y - ak)^2 \\ \end{align} That's the equation of a circle with center $(0, ak)$, and some radius which you can work out.