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Symmetry Argument for Cylinder

  1. Jan 30, 2017 #1
    1. The problem statement, all variables and given/known data
    I want to check my understanding of the symmetry arguments that allow for E to come out Gauss's Law and the symmetry arguments that allow for E vector *dA to become EdA. Specifically for an infinitely long cylinder.

    2. Relevant equations
    ∫EdA=q/ε

    3. The attempt at a solution
    So, for an infinitely long cylinder, we know that the orientation does not change when translated or rotated about an axis. The E field also is perpendicular to the curved surface of the cylinder, and so there is cylindrical symmetry (obviously). We know that the E field always points in the same direction, so we can essentially 'get rid of' the vector. Next, we assume that the E field is the same magnitude at all points (constant) on the cylinder (there is nothing to suggest otherwise), we can pull it out of the integral, so the integral becomes E∫dA.
    But why can the E field at the ends of the cylinder be ignored (which allows us to make all the previous assumptions)? Is it because they are in different directions, so they cancel out?
     
  2. jcsd
  3. Jan 31, 2017 #2

    Orodruin

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    Hint: Reflection symmetry.
     
  4. Jan 31, 2017 #3
    I am sorry but I am a bit confused as to how to apply reflection symmetry as an argument...
     
  5. Jan 31, 2017 #4

    Ray Vickson

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    An infinitely long cylinder does not have ends.
     
  6. Jan 31, 2017 #5
    That makes sense; it's just a bit odd to think about! Thank you.
     
  7. Feb 1, 2017 #6

    Orodruin

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    Wait a minute here, the charged cylinder no. But he is talking about applying the divergence theorem to find the electric field. The cylinder you integrate over should be finite and have end caps.The OP is implicitly talking about two cylinders. The source and the integration surface.


    What happens to the field if you make a reflection in a plane perpendicular to the cylinder?
     
  8. Feb 1, 2017 #7

    Ray Vickson

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    I agree that the integration surface is a finite cylinder, and if the OP is wise he/she will take its ends to be perpendicular to the cylindrical axis. That way, the electric field points along the integration surface at the end, and so ##\vec{E} \cdot d\vec{A}## vanishes on the ends.
     
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