# Potential of parallel infinite charged wires

1. Nov 19, 2008

### musemonkey

1. Two infinitely long wires running parallel to the x-axis carry uniform charge densities $$+\lambda$$ and $$-\lambda$$. Find the potential at any point. Show that the equipotential surfaces are circular cylinders, and locate the axis and radius of the cylinder corresponding to a given potential $$V_0$$

2. Relevant equations.

Potential of one wire charge density $$\lambda$$ I calculated to be :

$$V_a(b) = \frac{-\lambda}{2\pi\epsilon_0} \ln (b/a)$$ where a,b are shortest distances to the wire, and distance a is the reference.

I let the wires be equidistant from the x-axis, denoting the distance d. I put the $$-\lambda$$ wire at (-d,0) and the $$+\lambda$$ at (d,0).

3. The attempt at a solution

Setting the reference at the origin, the reference distance for both wires is $$a = d$$. Letting $$r_+, r_-$$ be the distance from a point (x,y) to the positive and negative wires, respectively, the potential at (x,y) is then

$$V(x,y) = \frac{\lambda}{2\pi\epsilon_0}( -\ln(r_+/d) + ln(r_-/d) )$$
$$= \frac{\lambda}{2\pi\epsilon_0}\ln(\frac{r_-}{r_+})$$.

The distances are

$$r_+ = \sqrt{(x-d)^2 + y^2}$$
$$r_- = \sqrt{(x+d)^2 + y^2}$$.

So the potential can be written

$$V(x,y) = \frac{\lambda}{4\pi\epsilon_0}\ln(\frac{(x+d)^2 + y^2}{(x-d)^2 + y^2})$$.

If so, then the eqn. of a $$V_0$$ equipotential surface is

$$k = \frac{(x+d)^2 + y^2}{(x-d)^2 + y^2})$$ where $$k = \exp( \frac{4\pi\epsilon_0 V_0}{\lambda} )$$.

This simplifies to

$$x^2 + y^2 + d^2 - 2xd\frac{k+1}{k-1} = 0$$.

and I don't see how this eqn. could define a cylinder, and moreover, on intuitive grounds, I don't see how a cylinder could possibly be the geometry of the equipotential surfaces. By symmetry, an equipotential cylinder would have to be centered at the origin -- right? -- but that would means that the potential at $$(d + \varepsilon, 0 )$$ a bit to the right of the positive wire would be the same as the potential at $$(-d - \varepsilon, 0 )$$ a bit to the left of the negative wire!

2. Nov 19, 2008

### gabbagabbahey

Strictly speaking, $$x^2 + y^2 + d^2 - 2xd\frac{k+1}{k-1} = 0$$ defines a circle, not a cylinder. However, the potential is independent of z, so these circles are the same for a given $V_0$ all the way along the z-axis, and together this infinite collection of circles (of the same radius and center) define a cylindrical surface.

To see that your equation represents a circle, just "complete the square" in 'x', an you can rewrite your equation into the form $$(x-a)^2+y^2=R^2$$ which of course, represents a circle centered at $(a,0)$ with radius $R$... What do you get for your $a$ and $R$ values?....Clearly, your cylinder is not centered at the origin, but rather runs parallel to the z-axis and is centered at $(a,0)$.

3. Nov 19, 2008

### musemonkey

Now that you mention it, that was the plain-as-day thing to do. Thanks!