**Required math: calculus, algebra**

**Required physics: electrostatics**

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Chapter 2, Post 47.

Two infinite wires lie in the x-y plane, parallel to the axis. One carries a charge density of and lies at location while the other carries a charge density of and lies at location . Find the potential at a location in rectangular coordinates.

The field due to an infinite wire can be found using Gauss’s law in cylindrical coordinates. For the wire carrying charge density we have, using a Gaussian cylinder of unit length centred on the wire (see Example 3 in this post)

where is the cylindrical distance from the wire, and for the wire with charge density

The potentials from the two wires add (according to the superposition principle), so we get

where the limits on the integral arise from taking the origin as the zero point for potential. The distance is the cylindrical distance from the wire at which we want the potential.

Similarly, the potential for the positive wire is

so the total potential is

In terms of we have

so we get

We can find the equipotential surfaces, that is, the surfaces where for some constant . First, note that if then , so the plane is the equipotential surface for . To find the other surfaces, we can consider so we have

We can now define

so we get

The fourth line is obtained by dividing through by , which requires , or . We saw above that is a special case, giving the plane as a surface.

This equation has the form of a circle in the plane, with centre at and , and with a radius of . Thus the equipotential surfaces are circular cylinders, with axes given by the lines running through the centres of the circles. Note that since was defined as an exponential, it is always positive, so the argument of the square root in the equation for the radius is always positive.

We can revert back to expressions containing to see the relation between the surfaces and the potentials.

The radius becomes

The axis is (with ):

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## Comments

Please explain this point ,

” the radius becomes

(2ae^2pi,epsilon,k/ lambda)/(e^4pi,epsilon,k/lambda -1)

I’ve got it

A minor correction. In the introduction you state “wires lie in the x-y plane, parallel to the x axis”. Shouldn’t they be lying in the z-y plane?

No – the wires are in the plane. Since they are parallel to the axis, the potential does not depend on .

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[…] solution to this problem relies on an earlier example in which we worked out the potential due to two charged wires on their own. We saw there that the […]

[…] exploit the uniqueness of solutions to Laplace’s equation. If we look back at the problem of two charged wires, we found that if we had a linear charge density of on a wire at and a density of at , the […]