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Surface integral to Lateral integral?

  1. Feb 27, 2012 #1
    1. The problem statement, all variables and given/known data
    I have some working out my lecture gave me to a problem and I don't think I understand part of it. Hoping you could help me.


    It's using Gauss' Law to find the capacitance of a cylindrical capacitor of length L but this information shouldn't matter for my question.

    [itex]\lambda=\frac{∂q}{∂l}[/itex]

    [itex]\oint E \bullet \Delta A = \frac{\sum q}{\epsilon}[/itex]

    He then jumps to.

    [itex]E \int \Delta A = \frac{\lambda L}{\epsilon}[/itex]

    First question, how does he go from the surface integral to the normal integral? Has it got anything to do with removing the dot product?

    He then changes
    [itex]E \int \Delta A[/itex] to E(2πrL)

    I understand this is taking the integral of dA which becomes A, which is the area of a circle, hence 2πr, but how does he then bring the L into play...would this not be the Volume, not the area?

    Thanks for any help.
     
  2. jcsd
  3. Feb 27, 2012 #2

    lanedance

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    could it be that E is constant in the integral and always makes the same angle with the surface normal?

    This is generally the best way to make use of Guass's theorem - exploit the symmetry of the problem to simplify any integrals
     
  4. Feb 27, 2012 #3
    So when an angle doesn't change a surface integral can be turned into a lateral integral?
    Sorry I haven't done much study on surface integrals so I don't know a lot about them. I'll look into them.
     
  5. Feb 27, 2012 #4

    lanedance

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    so if the relative angle, and field strength both don't change the scalar result of the vector dot product is just EdA, everywhere over the surface. As E is constant, the integral is just E.A
     
  6. Feb 28, 2012 #5
    Perfect sense, thanks so much for that
     
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