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Electric Field Intensity

  1. Feb 6, 2005 #1
    Hi, I would appreciate is someone could help me figure out this problem:

    The electric Field Intensity in free space from a very long, uniformly charged cylinder of radius 5cm is 100 kV/m at a radius of 1m. What must be the surface charge density on the cylinder?

    I've tried using different equations, but the only one that seems to get me somewhere, yet nowhere is D = eoE. please, can someone help??
     
  2. jcsd
  3. Feb 6, 2005 #2

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    Since you're using D, I'm guessing this is a linear dielectric problem, and you're looking for the bound surface charge. If that's right, just use the gauss law for D. Use D= [tex]\epsilon_0[/tex] E to get E in the vacuum outside the cylinder.

    Edit:

    and then use [tex] \sigma_b = P \cdot \hat n[/tex] to get the bound charge.
     
    Last edited: Feb 6, 2005
  4. Feb 6, 2005 #3
    I suggest you to use the Gauss' law for a portion [tex]L[/tex] of the cylinder:

    [tex]E \cdot 2 \pi r L=\frac{Q}{\epsilon_0}[/tex]

    and the definition of [tex]\sigma[/tex] for your geometry

    [tex]\sigma=\frac{Q}{2\pi R L}[/tex]

    (R - cylinder radius anr r - distance from the cylinder axis to the point where E is evaluated)
     
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