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Electric Field in a cylindrical Coaxial Capacitor

  • Thread starter redz
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  • #1
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Homework Statement



Two infinite coaxial metal cylindrical tubes of radius a and b (a < b) are charged
with charge per unit length (unit [C/m]) [tex]\lambda[/tex] and [tex]-\lambda[/tex] respectively.

Calculate the electric field between the tubes (i.e. for a < r < b)

Where [tex]\epsilon_{0}[/tex] is the permittivity of free space
2. The attempt at a solution

Considering a Gaussian surface with radius R & length L where a < R < b, we can use Gauss' law to find the enclosed charge;

[tex]\oint \vec{E}. d\vec{a} = \frac{ q_{enc}}{\epsilon_{0}}[/tex]

This can then be rewritten as;

[tex]\oint \vec{E}. d\vec{a} = \frac{\lambda L}{\epsilon_{0}}[/tex]

Then I have no clue of where to go. Can take the magnitudes of the vectors and take the E outside or do i have to do something different?
 
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Answers and Replies

  • #2
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Well you defined a Gaussian surface (a cylinder with radius R). Since it has the same symmetry as the charges, you can claim that the electric field will be constant along your Gaussian surface. That way you can pull it out of the integral, and you will be left with E*A, where A is the area of your Gaussian surface.

The direction of the electric field will point from positive to negative charges.
 
  • #3
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That makes perfect sense thank you very much.
 

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