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Gauss's law

  1. Feb 20, 2006 #1
    Two long, charged, concentric cylinders have radii of 3.0 and 6.0 cm. The charge per unit length is 4.8 10- 6 C/m on the inner cylinder and -8.0 10-6 C/m on the outer cylinder. Find the electric field at

    (a) r = 4.0 cm and

    (b) r = 7.1 cm

    I know how to find the electric field for each individual cylinder AT 3 and 6 cm... using this equation:

    E (electric field) = Q (total charge) / (2*pi*radius*(Q/lambda)*epsilon0)

    but i haven't a clue as to how to find the electric field when you include another charged surface using gauss's law...

    any help would be much appreciated - as you can see i'm pretty lost.

    some equations i know that may or may not help:

    Lambda (sp?... linear charge density) = Q / L where Q is the total charge of the surface and L is the length of the gaussian surface

    E*A = q/e0 where e0 = 8.85E-12 and q is the charge within the gaussian surface... and A is the surface area of the cylinder (not including the two circles on either end).

  2. jcsd
  3. Feb 20, 2006 #2
    If you know how to find the electric field for one cylinder you can find it for both. Just include the charge from both cylinders in your Q.
  4. Feb 20, 2006 #3
    And remember that the charge due to the outside cylinder does not factor into the r=4 cm calculation because you are "inside of the outside" cylinder.

  5. Feb 20, 2006 #4
    so does that mean for A and B i'm only finding the electric field for the inside and outside cylinders, respectively?
  6. Feb 20, 2006 #5
    In part (b) your Gaussian surface encloses both the inner and outer cylinders, so you need to include the charge for both. The electric field you calculate will be the electric field due to both cylinders.
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