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Electric field of a long round bar

  1. Nov 14, 2014 #1
    1. The problem statement, all variables and given/known data
    I have to prove that the electric field round a long conductive round bar of radius R is equal to that of a long thin wire, meaning the charge is on the center line

    2. Relevant equations
    The Gauss theorem:
    ##E\times A=q##
    Where A is the area of the surface and q is the charge enclosed in it

    3. The attempt at a solution
    To find the field E when the charge is concentrated on the axis i take a round long volume with radius r:
    ##E\times 2\pi rl=\lambda l\rightarrow E=\frac{\lambda}{2\pi r}##
    Where [itex]\lambda[/itex] is the charge per unit length.
    The field near a round bar:
    ##E\times 2\pi rl=\sigma\cdot 2\pi R l##
    Where [itex]\sigma[/itex] is the charge per unit surface, since the charge is concentrated only on the surface.
    But [itex]\sigma\cdot 2\pi R l=\lambda l[/itex]
    They both are the charge per unit length. and so is if i use the charge per unit mass.
    I feel uncomfortable
     
  2. jcsd
  3. Nov 16, 2014 #2

    ehild

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    It is true if you use cgs system of units where ε0=1.

    Correct.

    Simplifying, you get that E=σ R/r and σ =λ/(2πR). Substitute the second equation into the first, what do you get? Is E the same as in case of a thin long wire at the centre?
     
  4. Nov 16, 2014 #3
    I get ##E=\frac{\lambda}{2\pi r}## like for a long thin wire at the center, and that's good, but i feel uncomfortable with that, is that the method to prove it?
     
  5. Nov 16, 2014 #4

    ehild

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    Gold Member

    You do not need to feel uncomfortable, it was the right proof. :) Good job!
     
  6. Nov 16, 2014 #5
    Thanks
     
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