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Coaxial Cylindrical Conductors

  1. Feb 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Two coaxial cylindrical conductors are shown. The inner cylinder has radius a = 2 cm, length 10 m, and carries a total charge of Q inner = +8nC. The outer cylinder has an inner radius b = 6 cm, outer radius c= 7 cm, length 10 m, and carries a total charge of Q outer = -16nC. What is Ex, the x-component of the electric field at point P which is located at the midpoint of the length of the cylinders at a distance r= 4 cm from the origin and makes an angle of 30 degrees with the x-axis?


    2. Relevant equations
    Surface charge density = 2pi * radius a * length
    Linear charge density = 2pi * radius a * surface charge density
    Electric field = 2 * k * linear charge denisty / r


    3. The attempt at a solution
    I drew a Gaussian sphere at the radius r = 4 according to where point P lies and tried to determine the charge inside, which I think may be where I went wrong. There is an enclosed cylinder of +8 nC that has a radius of 2 that falls completely within the Gaussian sphere; however, I cannot see that the outer charge affects this particular sphere since it is within the 2nd cylinder. I calculated the surface charge density to be 2.5133 using the radius of 0.04 m, then plugged that into the linear charge density formula to receive an answer of 0.63165. I then tried plugging that into the formula for the electric field given by my prof and received an answer of 2.84245 e 11. I took that and mulitplied by cos 30 degrees to account for the diagnol line upon which P lies. I cannot get the right answer (interactive online problem) and I'm getting the feeling that I'm either using the wrong charge, wrong radius, or wrong formulae...or a combination of all 3!!!! Please help me!!!! Thanks! :uhh:
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 5, 2008 #2
    So what direction does the electric field, from Gauss's law, travel in? I'm guessing this is where are getting tripped up.

    Also, do you have a picture? I can't tell why they even give you an outer cylinder, something you mentioned also, maybe the problem creator is just trying to throw you off. Could you repost your work but in terms of symbols rather than numbers?
     
  4. Feb 6, 2008 #3
    I'll try attaching the image, but as for the symbols, I don't have anything that will display them. My computer is a little too old to have packages like Microsoft Equation or anything like that. If it's really a problem, I'll try seeing if I can download something. Also, I'm really not sure what direction the electric field travels in!
     

    Attached Files:

    Last edited: Feb 6, 2008
  5. Feb 6, 2008 #4

    Doc Al

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    Staff: Mentor

    Your Gaussian cylinder at (radius r, say) totally encloses the inner conductor. What's Gauss's law? Set up the equation and solve for E. (No need to compute surface charge!)

    The charge on the outer conductor is irrelevant. The field at r = 4 cm will be radially outward.

    You can just type in your equations without using symbols, or you can use Latex. (See Introducing LaTeX Math Typesetting .)
     
  6. Feb 6, 2008 #5
    The interactive help that comes with the problem stated that I had to calculate the surface charge in order to calculate the linear charge. Also, do I have the right formula for E? And if I don't need the surface charge, from where do I get the linear charge?
     
  7. Feb 6, 2008 #6

    Doc Al

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    Staff: Mentor

    That makes no sense to me. Using the data given, you can calculate the linear charge density directly.
    Yes. (Use Gauss's law to verify it, if you're unsure.)
    You are given the total charge and the length. (Assume it's uniformly distributed.)
     
  8. Feb 6, 2008 #7
    YEA! It worked!!!!!! Thanks so much!
     
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