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Electric field in a cylindrical configuration

  1. Aug 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Immediately outside a very long, cylindrical wire of radius R1=0.6mm, the electric field is 50 kV/m directed towards the wire's surface. The wire is in air.

    What is the charge per unit along length the wire? (take care with sign)

    What is the field at R2=3.0mm? (i.e. 2.4mm from the wire's surface)

    A hollow cylindrical metal tube with inner radius 3.0mm is now placed around the wire, to form a coaxial cable, what will be the charge per unit length on the inner surface of this tube? Explain.

    2. Relevant equations

    Really not too sure,

    3. The attempt at a solution

    even just a headstart would be much appreciated if anybody knows how to solve these.

  2. jcsd
  3. Aug 14, 2009 #2


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    Can you tell if the charge on the wire is positive or negative? Why do you think so?

    Consider using Gauss's Law for this problem. There might even be an example in your textbook treating something similar that you can use to pattern your solution after. Look for a cylindrical gaussian surface.
  4. Aug 16, 2009 #3
    I think the charge on the cylinder/wire is negative, as all the electric field is directed toward it's surface. If it was +vely charged then the electric field would be directed away from the wire, as the field represents the movement of +ve charges.

    From here I am still confused, except for knowing that the charge per unit length on the wire will be negative. How do i use Gauss? how do i tackle the other parts of the question?

    For part (b), what is the field at R2 = 3.0mm, can I use the formula for field due to a point charge and just use any point along the cylinder?
  5. Aug 16, 2009 #4


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    You got the sign of the charge right. Now for Part (a)

    Imagine a Gaussian surface of radius that is a cylinder of length L (the length doesn't really matter in the end) and radius 0.6 mm, i.e. it essentially looks like a piece of the wire. Note that the electric field is constant on the surface and points towards the axis.

    To answer part (a), you need to complete the following steps. Once you understand how part (a) works, part (b) should be easy.

    Find an expression for the electric flux through this Gaussian surface. (Note that there is no flux through the two flat caps of the cylinder because the field is parallel to the surface).
    Assuming that charge Q is enclosed by the surface, write Gauss's Law.
    The answer to part (a) is Q/L which you can get from the previous step.
  6. Aug 16, 2009 #5
    I got -1.67x10-9 C/m for Q/L.

    used a formula which was E field for an infinite line of charge, is this the right value?

    and for (b) I just put the values in another formula to see what sort of answer i got - field at 3.0mm = 1.7x106 C/m.

    how am i doing?
  7. Aug 17, 2009 #6


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    You are doing OK if getting a number for an answer is what you are after. However, if the problem is, as I suspect, about Gauss's Law, plugging number in this formula will not help you understand Gauss's Law and how to use it.
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