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Electric Potential Problem

  1. Jun 9, 2013 #1
    1. The problem statement, all variables and given/known data
    A very long cylinder of radius 2.00 cm has a uniform charge density of 1.50 nC/m. Taking the reference level for the zero of potential to be the surface of the cylinder, find the radius of equipotential surfaces having potentials of 10.0 V, 20.0 V, and 30.0 V.

    2. Relevant equations
    Va - vb = ∫E * dr
    Lower limit: a
    Upper limit: b
    λ = 1.50 nC/m

    3. The attempt at a solution
    I have solved, using Gauss' Law, the electric field of this cylinder.

    E = λ/(2[itex]\pi[/itex]ε0)(r)

    Now to integrate, I removed the λ/(2[itex]\pi[/itex]ε0) factor out of the integrand and did this:

    Vr - V.02 cm = λ/(2[itex]\pi[/itex]ε0) * ∫(1/r)dr

    Lower limit for this integral: r (the distance from the axis of the cylinder)
    Upper limit for this integral: 2.00 cm (the radius of the cylinder)

    So Vr - V.02 cm = Vr =(λ/(2[itex]\pi[/itex]ε0) * ln(.02/r)

    Since the first part of the problem states that the potential difference is 10 V, I plugged 10 V in for Vr and just solved for r, since that should be the equipotential surface radius. I retrieved an answer of 1.38 cm, but this is incorrect. I can't seem to find my error. Thanks in advance.

    The correct answer is 2.90 cm.
     
    Last edited: Jun 9, 2013
  2. jcsd
  3. Jun 9, 2013 #2
    You missed a negative sign somewhere!
     
  4. Jun 9, 2013 #3
    Where am I missing a negative sign?
     
  5. Jun 9, 2013 #4

    haruspex

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    The charge is positive, and you're looking for a location where the potential is higher than at the surface of the cylinder? Doesn't sound right to me. Check the question.
     
  6. Jun 9, 2013 #5
    V= -∫E.dr. I thought you have missed the negative sign in this. But you have interchanged the integration limits so it is the same thing.

    Didn't notice the question carefully, i totally agree with harupex, there is no r where potential would be greater than that of cyllinder. As the potential keeps on deceasing with r. Maybe the negative sign is missed in the question itself. Check it for -10, -20.
     
  7. Jun 9, 2013 #6
    Yeah when I plug in -10V for Vr I do get the correct answer, 2.90 cm. However, how can there be a negative potential from a positively charged cylinder? Or does the -10V just mean that the potential at 2.90 cm is 10 volts lower than at the surface of the cylinder?

    Thanks for the help :)
     
  8. Jun 9, 2013 #7

    haruspex

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    Yes.
     
  9. Jun 10, 2013 #8
    Finding potential is all about reference. If you take potential at infinity as the reference potential(zero potential) than the potential due to positive charged cyllinder would be positive at for every r as it keeps on on increasing as we r decreases(or as we move towards cyllinder). In the question reference potential is the potential of the cyllinder so the potential around it would be negative.

    You have calculated for the expression V(r) - V(0.02), which is self explaining!
     
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