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Potential between Two Wires

  1. May 5, 2008 #1
    Two wires (radius a) with charge densities +λ and -λ are placed parellel to on another a distance d away. Find an expression for the potential difference between the wires and simplify with the assumption the d>>a.
    From that I need to find the potential difference at the midpoint between the two wires.

    I have figured out the E field and potential for one wire (the first part of the problem). For outside one wire, E= λ /((2pi)(eo)(r). I'm not really sure how to approach this - find the E field of this first?) Any tips would be appreciated.
  2. jcsd
  3. May 6, 2008 #2
    I think you want the assumption d<<a so that you can use Gauss's law and treat them as parallel plates. Yes, you would want to find the E field first, then go to potential, and then use the definition of capacitance.
  4. May 6, 2008 #3
    The page definetly says d>>a. Perhaps a typo?
    This is the work I have so far...anyone mind telling me if I'm on the right track or not?

    V(on -) = [λ/(2pi)(e0)] x ln[x/a]
    V(on +) = [-λ/(2pi)(e0)] x ln[(d-x)/a]
    then add them together?
  5. May 6, 2008 #4
    Ohhh, I think I see the picture now. Sorry, I thought it was that there was some gap between two wires butted up against each other. Instead, they are two wires above and below each other, some distance away?

    Are the wires very long? In other words, can we exploit Gauss' Law, or do we have to do the integrals? It looks like you have used Gauss's law, so I assume it is fine to assume "very long."

    So then we just want to use

    [tex]V = - \int \mathbf{E} \cdot d \mathbf{l}[/tex]

    So, yes, to get the total field E you want to add the two together. You will see that they cancel outside the gap and add inside the gap.
  6. May 6, 2008 #5
    Oh wait, you added the potentials, no you don't want to do that. You want to get the total electric field, and then do the integral.
  7. May 6, 2008 #6
    the wire is "very long", and they are parellel wires...
    so I can use Gauss's law for finding E?
  8. May 6, 2008 #7
    Yes, you can use Gauss' law. You already said the result of Gauss' law in your first post, so just use that. Then you need to integrate, which is where the d>>a comes into play.
  9. May 6, 2008 #8
    Electric potential is a scalar quantity (U/q), and it is simply equal to negative the path integral of the electric field. The functions for electric field are continuous at the outer edge of the wires, so you simply have to find the net potential at the two wire edges and use that to find the potential difference. If one rod is +L (L being landa) and the other is -L, then they should technically counter each other.
  10. May 7, 2008 #9
    pidoctor 314, you wouldn't happen to be taking a Physics 217 take-home final exam at Cornell, would you? Keep in mind that your email address is linked to this forum account.
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