# Equipotential lines around two parallel charged wires

I'm studying electric potential, and I'm trying to follow the derivation of equipotential lines surrounding two parallel, oppositely charged wires, found here. But I feel like the source is skipping a few critical steps in the derivation right at the end, and it's got me lost.

The derivation arrives at the expression $$\frac{(x+a)^2 + y^2}{(x-a)^2 + y^2} = e^{2 \eta}$$. They next say,

Then, introducing the hyperbolic trigonometric functions

$$2 \sinh \eta = e^\eta - e^{- \eta}$$

and

$$2 \cosh \eta = e^\eta + e^{- \eta}$$

where

$$e^\eta = \cosh \eta + \sinh \eta$$

and using

$$\cosh^2 \eta - \sinh^2 \eta = 1$$

we obtain

$$(x - a \coth \eta)^2 + y^2 = \left( \frac{a}{\sinh \eta} \right)^2$$

All of this makes perfect sense to me right up to the very last line. I'm struggling to understand how they combine all this information to arrive at the final expression. Any insight would be greatly appreciated!

• Maybe it would be useful to provide bibliographic details of the book and indicate the chapter in which this is found? – ZeroTheHero Feb 12 '19 at 22:54
• This is a very good idea. I've updated the question to include the source. Thanks! – 1Teaches2Learn Feb 12 '19 at 23:01
• Might Mathematics be better suited for this maths question? – Kyle Kanos Feb 13 '19 at 0:24

Let $$\dfrac{(x+a)^2 +y^2}{(x-a)^2 +y^2}= b^2$$ where $$b=e^{\eta}$$ to delay the use of the sinh and cosh functions.

You can use algebra to get

$$\left [ x - a \left( \dfrac{b^2+1}{b^2-1}\right) \right ]^2 + y^2 = a\left ( \dfrac{2b}{b^2-1}\right ) = \left [ x - a \left( \dfrac{b+b^{-1}}{b-b^{-1}}\right) \right ]^2 + y^2 = a\left ( \dfrac{2}{b-b^{-1}}\right )$$

and then use the definitions of sinh and cosh remembering that $$b=e^{\eta}$$.

Perhaps the use of the sinh and cosh functions directly makes the algebra easier?

Thanks @Farcher!! With your suggestion, I got it! I'm feeling really pumped right now. Since I had to type this up, I figured I'd share here as well. The following is the set of algebraic steps to reach the final answer.

Let $$e^\eta = b$$. Then \begin{align*} \frac{(x+a)^2 + y^2}{(x-a)^2 + y^2} &= e^{2\eta} \\ \end{align*}

becomes \begin{align*} \frac{(x+a)^2 + y^2}{(x-a)^2 + y^2} &= b^2 \\ \end{align*}

Start by expanding the binomials, bringing the denominator to the RHS, and distributing $$b^2$$: \begin{align} \frac{(x+a)^2 + y^2}{(x-a)^2 + y^2} &= b^2 \\ \frac{x^2 + 2xa + a^2 + y^2}{x^2 - 2xa + a^2 + y^2} &= b^2 \\ x^2 + 2xa + a^2 + y^2 &= b^2x^2 - b^22xa + b^2a^2 + b^2y^2 \end{align}

Now, group similar terms, resulting in the factors of $$(b^2+1)$$ and $$(b^2 -1)$$. Note that the parentheses in the first line below aren't necessary, but are used to help visualize the groupings of similar terms: \begin{align} (b^22xa + 2xa) &= (b^2x^2 - x^2) + (b^2a^2 - a^2) + (b^2y^2 - y^2) \\ 2xa(b^2+1) &= x^2(b^2 - 1) + a^2(b^2 - 1) + y^2(b^2 -1 ) \\ 2xa(b^2+1) &= (x^2 + a^2 + y^2)(b^2 -1 ) \end{align}

Divide both sides by $$(b^2-1)$$ and do some regrouping: \begin{align} 2xa\frac{b^2+1}{b^2 - 1} &= x^2 + a^2 + y^2 \\ x^2 - 2xa\frac{b^2+1}{b^2 - 1} &= -a^2 - y^2 \end{align}

And now the ninja trick. Complete the square, and then bring $$y^2$$ to the LHS: \begin{align} x^2 - 2xa\frac{b^2+1}{b^2 - 1} + a^2 \left(\frac{b^2+1}{b^2 - 1} \right)^2 &= a^2 \left(\frac{b^2+1}{b^2 - 1} \right)^2 -a^2 - y^2 \\ \left(x - a \frac{b^2+1}{b^2 - 1} \right)^2 + y^2 &= a^2 \left(\frac{b^2+1}{b^2 - 1} \right)^2 -a^2 \end{align}

At this point, the LHS is in the desired form, so I will suppress it for the moment as we continue to manipulate the RHS. First, pull a factor of $$a^2$$ from both terms in the RHS. Then, combine the terms. Expand the binomials. And notice that most of the terms cancel in the numerator, except the two $$2b^2$$ terms which end up adding. Then, since each individual factor is squared, we can "pull out" the square: \begin{align} \left(x - a \frac{b^2+1}{b^2 - 1} \right)^2 + y^2 &= a^2 \left( \frac{(b^2+1)^2}{(b^2 - 1)^2} -1 \right) \\ &= a^2 \frac{(b^2+1)^2-(b^2 - 1)^2}{(b^2 - 1)^2} \\ &= a^2 \frac{(b^2)^2+2b^2+1 - ((b^2)^2 - 2b^2 + 1)^2}{(b^2 - 1)^2} \\ &= a^2 \frac{4b^2}{(b^2 - 1)^2} \\ &= \left(\frac{2ab}{b^2 - 1} \right)^2 \end{align}

Both LHS and RHS are in the desired form at this point. So we observe that $$(b^2 +1) = b(b + b^{-1})$$ and $$(b^2 -1) = b(b - b^{-1})$$: \begin{align} \left(x - a \frac{b(b+b^{-1})}{b(b - b^{-1})} \right)^2 + y^2 &= \left(\frac{2ab}{b(b - b^{-1})} \right)^2 \\ \left(x - a \frac{b+b^{-1}}{b - b^{-1}} \right)^2 + y^2 &= \left(\frac{2a}{b - b^{-1}} \right)^2 \end{align}

Replace each $$b$$ with the original $$e^\eta$$, and then apply the hyperbolic trig identities, and simplify: \begin{align} \left(x - a \frac{e^\eta+e^{-\eta}}{e^\eta - e^{-\eta}} \right)^2 + y^2 &= \left(\frac{2a}{e^\eta - e^{-\eta}} \right)^2 \\ \left(x - a \frac{2 \cosh \eta}{2 \sinh \eta} \right)^2 + y^2 &= \left(\frac{2a}{2 \sinh \eta} \right)^2 \\ \left(x - a \coth \eta \right)^2 + y^2 &= \left(\frac{a}{ \sinh \eta} \right)^2 \end{align}