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Field due to coaxial cylinders

  1. May 12, 2010 #1
    1. Charge of uniform density 76nC/m^3 is distributed throughout a hollow cylindrical region formed by two coaxial cylindrical surfaces of radii 1.5 mm and 3.7 mm. Determine the magnitude of the electric field at a point which is 2.6 mm from the symmetry axis.



    I tried denistyxr/3epsilon but id dont think it is the correct equation to use
     
  2. jcsd
  3. May 12, 2010 #2
    The correct equation to use is gauss's law.

    [tex]\int\int \mathbf{E} \cdot d\mathbf{a}=\int\int\int \rho/\epsilon_0 dV[/tex]
     
  4. May 12, 2010 #3
    i dont understand how to do the triple integration however i remember that E=k int(dq/r^2)
    E.dA is phi i want E
     
  5. May 12, 2010 #4
    The density is uniform so you don't actually have to worry about triple integration, you just have to get the right volume. Same for the double integration, you just have to get the right Gaussian surface area. In other words

    [tex]EA=\frac{\rho}{\epsilon_0}V[/tex]

    Try to work it out and we can help you where you go wrong.
     
  6. May 13, 2010 #5
    i tried this rule and EA=densityXv/epsilon

    so E=Q/Aepsilon =Q/4pir^2Xepsilon ?????????? doesnt give an answer
     
  7. May 13, 2010 #6
    i got it .it will be EA=densityXv/epsilon then EX2*pi*r*h=densityXpi*r^2*h/epsilon
    then 2E=density*r/epsilon then E=density*r/2epsilon
     
  8. Sep 15, 2012 #7
    how to set it up:
    a=gaussian radius
    h=height of cylinder
    r=inner radius

    E(2∏ah)=[ρ∏h(a2-r2)]/ε0

    E=[ρ∏h(a2-r2)]/[(2∏ah)ε0]

    E=[ρ(a2-r2)]/[(2a)ε0]
     
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