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Homework Help: Electric Field HELP

  1. Jan 28, 2010 #1
    1. The problem statement, all variables and given/known data
    A long hollow cylinder has an inside radius a, an outside radius b and is uniformly charged with a uniform charge density ρ.
    What is the electric field as a function of radius r? Include answers for r < a, a<r<b, and r>b.

    2. Relevant equations

    3. The attempt at a solution

    I'm not sure where to even start with this equation.
  2. jcsd
  3. Jan 28, 2010 #2
    so when is says long, that means infinitely long. which means that your electric field will be in a radial outward direction.

    my advice is put a point p at (0,0,d) where d is the distance from the centre of the cylinder. The concept here that will make your life easier is as follows:
    inside a ring of any width, there is no electric field. Outside the ring however, the electric field acts as though the sum of all the charge on that ring is at the centre of the ring. Let me explain further. If i have a ring with charge 1C(stupid large charge, it wont happen). And i want to figure out the electric field at a point d away from that ring on the same plane, It will be E = KQ/d². And E in terms of r will obviously be E(r)=kQ/r².

    So what you want to do here, is "assemble" all the charge on the cylinder(an infinite amount of rings placed side by side) to an infinitely long line running through the centre of the cylinder. Once you have this, it will be an easy question to solve.

    If you have any questions dont hesistate althought i might not get back to you until tomorrow
  4. Jan 28, 2010 #3
    Start with Gauss' Law. It will provide great insight by just looking at q(enclosed) for the different radii. Wisely choose your Gaussian surfaces (in this case cylinders) in order to maximize symmetries.
  5. Feb 9, 2010 #4
    HI huntingrdr
  6. Feb 9, 2010 #5
    i am imran rouf from Pakistan.
    i am sutudent of Bsc physics in a Govt college i want to know about Ring of charge and Disk of charge.
    i am waiting for your reply.
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