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musemonkey
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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. Homework 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).
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. Homework 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).
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!