Supplemental Notes

Today (Monday, October 16), we didn’t quite get to show that the field lines for two infinite oppositely-charged paraxial cylinders are circles. The way to do this is to write the equation for the equipotentials so that the equipotentials are level sets (surfaces if we consider the z-direction, contours if we restrict ourselves to the plane. In any event, the field lines will be circles.)

So, define a function g(x, y) by

g(x, y) = k^{2} = \frac{r_{1}^{2}}{r_{2}^{2}} = \frac{(x − a)^{2} + y^{2}}{(x + a)^{2} + y^{2}} .

We have seen that the equipotentials are circles centered on the x-axis at x = ±a\frac{k^{2}+1}{k^{2}−1} and radius \frac{2a}{k^{2}−1}.

Anyway, you can show (okay, it took me a while) that

\frac{∂g}{∂x} = 4a\frac{(x − a)^{2} − y^{2}}{r_{2}^{4}} , \frac{∂g}{∂y} = 8a\frac{xy}{r_{2}^{4}} .

We now wish to express the curve of the field lines by y in terms of x. (We could look for some y(x) explicitly, or y as an implicit function of x. In this case, it turns out not to matter.) The relation we need is from physics; the field lines will be tangent to the gradient of g (note that g(x, y) in not the potential, but the gradient of g is still tangent to the field lines). Thus,

\frac{dy}{dx} = \frac{\frac{∂g}{∂y}}{\frac{∂g}{∂x}} = \frac{2xy}{x^{2} − a^{2} − y^{2}} .

There are many ways to solve such an item; the way presented below is one of the most direct, but it does use some knowledge of 18.03 and a bit of 18.02 that may not have been presented. That is, we rewrite the above as

2xy dx − (x^{2} − a^{2} − y^{2}) dy = 0,

and see if (remember, that’s an if at this stage) this can be rewritten as

df = \frac{∂f}{∂x} dx + \frac{∂f}{∂y} dy = 0.

Well, it can’t. If it could, the term 2xy would be \frac{∂f}{∂x} so that \frac{∂^{2}f}{∂y∂x} = 2x, whereas the term −(x^{2} − a^{2} − y^{2}) would be \frac{∂f}{∂y} so that \frac{∂^{2}f}{∂x∂y} = −2x; close but no cigar.

So, what we do is look for an integrating factor; the details of how to do this take a while and don’t add much to the physics (and they’re hard to type), so I’ll just say that we mulitply by a factor of y^{−2}, and this works, in that we now have

\frac{2x}{y} dx − (\frac{x^{2}}{y^{2}} − \frac{a^{2}}{y^{2}} − 1) dy = 0.

Fill in the details yourself, but the result is

\begin{aligned}\frac{2x}{y} & = \frac{∂}{∂x} (\frac{x^{2}}{y} − \frac{a^{2}}{y} + y) \\ − (\frac{x^{2}}{y^{2}} − \frac{a^{2}}{y^{2}} − 1) & = \frac{∂}{∂y} (\frac{x^{2}}{y} − \frac{a^{2}}{y} + y) , \end{aligned}

and so the field lines follow a curve given by

(\frac{x^{2}}{y} − \frac{a^{2}}{y} + y) = c,

where c is a constant of integration. Rearranging gives

x^{2} + (y − c/2)^{2} = a^{2} + c^{2}/4.

In this form, remember that c could be positive or negative; this might help you decide what the circles look like.