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Gauss' Law: Charge enclosed zero = E field zero?

  1. Sep 5, 2015 #1
    1. The problem statement, all variables and given/known data
    A cylindrical shell of radius R and length H has its charge uniformly distributed on its curved surface

    Find the electric field at a point P from the axis, a distance r away, measured radially outward from the midpoint of the shell such that R>r

    2. Relevant equations

    φ = ∫E⋅dA = Qenc / ∈o


    3. The attempt at a solution

    I constructed a gaussian surface (cylinder) inside the larger cylinder of radius R.

    I realize, my chosen gaussian surface encloses no charge ∴ φ & Qenc are both zero.

    I have been told I am not allowed to simply jump and say the E field must also then be zero if the flux is zero, but for this case, since the charge on the outer cylinder is uniformly distributed, does that also mean the net electric field at point P must be zero?

    If not, please shed some light on this!

    Thank you
     
  2. jcsd
  3. Sep 5, 2015 #2

    SammyS

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    Is there sufficient symmetry to conclude that the field is zero everywhere inside to charge distribution?
     
  4. Sep 6, 2015 #3
    Yes.

    I think that answers my question, so then, we should be examining at the symmetry of the chosen surface with respect to the original surface.
     
  5. Sep 6, 2015 #4

    SammyS

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    It also has a lot to do with what symmetry implies regarding the electric field and whether you an find a surface so the field is normal to the surface over regions where the field's magnitude is constant and in all other regions the field is parallel to the surface.
     
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