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Help with potentials!

  1. Mar 22, 2009 #1
    1. The problem statement, all variables and given/known data
    A uniform line charge of density pl = 1nC/m is arrange in the form of a square 6m on a side. (In the z plane) Find the potential at (0,0,5) m

    Answer:35.6 V

    2. Relevant equations
    R = The distance from the line to the point (0,0,5)
    dL = ax dx + ay dy + az dz

    3. The attempt at a solution
    Well any point on the uniform line charge can be selected for this problem so I picked (0,3,0)
    There for the distance is R= Sqrt((5-0)^2+ (0-3)^2)) Then I know that the potential of a charge distribution is equal to the integration of the whole volume of (p dV)/(4pi*Eo the Permittivity of free space*R)

    Therefore, the x components dont matter because at both points they are zero. So I figured I would just integrate in terms of y get an answer and then integrate in terms of z and get an answer then the sum of those would be the total potential. However it doesn't seem to be coming out to the right answer.

    Integral [0 to 3] pl/4pi*E0*R dy = 4.6241
    Integral [0 to 5] pl/4pi*E0*R dz = 7.7068

    does anyone know what I am doing wrong??
    thanks for your help
     
  2. jcsd
  3. Mar 22, 2009 #2
    I'm sorry I don't know the answer. I'm just wondering about the z-plane. what does that even mean. a palce has two axis
     
  4. Mar 22, 2009 #3

    turin

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    Homework Helper

    I think that your problem statement is missing some information. I assume that "the z plane" means the xy plane (i.e. the z=0 plane). I assume that the square of line charges is centered on the origin.

    This problem is not symmetric under exchange of any two points on the square. There is some symmetry, but that's not it.
     
  5. Mar 22, 2009 #4
    Yes, I was wrong the uniform line charge is in the x-y plane where z = 0.... I apologize.

    So I really cant just pick a point on the line charge...i thought it wouldn't matter because it is uniformly distributed. Do you have any idea on how I could attack this problem??
     
  6. Mar 23, 2009 #5

    turin

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    Homework Helper

    The uniformity makes the integrand simpler than if the distribution were generally nonuniform. Your hints are: integration and symmetry. Oh, and you may have a formula in your book for the potential due to a finite length of line charge, so think about that as a possibly alternative approach.
     
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