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Equipotential surfaces for the given charge distribution

  • Thread starter mjordan2nd
  • Start date
  • #1
178
1

Homework Statement



Two infinitely long wires running parallel to the x axis carry uniform charge densities [tex] \lambda [/tex] and [tex] - \lambda[/tex].

a.) Find the potential at any point (x, y, z) using the origin as your reference.

b.) Show the equipotential surfaces are circular cylinders, and locate the axis and radius of the cylinder corresponding to a given potential V_0.

Homework Equations



In part a I oriented my axes such that both wires lay in the x-y plane a distance, a, away from the x-axis. I found the potential to be

[tex] V(x,y,z) = \frac{\lambda}{4 \pi \epsilon_{0}} ln \left[ \frac{(y-a)^{2}+z^{2}}{(y+a)^{2}+z^{2}} \right][/tex]

The Attempt at a Solution



I know that I can find the equipotential surfaces by simply allowing the argument of the logarithm to be constant. I was wondering, however, if I could find the answer using partial derivatives as well. I feel that I should be able to find the equipotential surfaces to this potential by allowing the partial derivative with respect to y go to 0, and the partial derivative with respect to z go to 0. When both partials are 0, shouldn't I find an equipotential surface? I worked out the math, and my answer is inconsistent, so I believe that this method is incorrect, but I can't quite figure out why. If someone could point out the error in my logic, I would appreciate it. Thanks.
 

Answers and Replies

  • #2
213
8
well thats just bad mathematics

think of a 2d circle with radius r

[tex] r = \sqrt{x^2 +y^2}[/tex]

now from analysis we know that

[tex] dr = \frac {\partial r}{\partial x} dx + \frac {\partial r}{\partial y} dy [/tex]

since you want r to be constant this reduces to

[tex] -\frac {\partial r}{\partial x} dx = \frac {\partial r}{\partial y} dy [/tex] *

from this you can see that by eliminating one of the partials you reduce the other variable to a constant or you have something very weird i.e. [tex] 0 \times dx = 0 \times dy[/tex]. So the only sensible way to get 0's into * is to have x and y constants but then that's not a circle anymore its a triangle
 

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