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Homework Help: Energy of a system

  1. Oct 8, 2013 #1
    1. The problem statement, all variables and given/known data

    There are 2 long coaxial insulating cylinder. The inner and outer cylinders have radii of a and b and charge densities λ and -λ uniformly distributed on the surface. Calulcate energy per unit length 2 ways (equations below)

    2. Relevant equations

    W = [itex]\frac{1}{2}[/itex][itex]\int σ V da[/itex]
    W = [itex]\frac{ε0}{2}[/itex][itex]\int E^2 dτ[/itex]

    3. The attempt at a solution

    So using gausses law, with s (radius of cyliunder) < a there is no electric field as there is no enclosed charge. Outside both cylinders there is no E because there is no net charge. SO the only E and energy is between the 2 cylinders.

    So I have to do it 2 ways, one with the first and one with the second. I am unsure though how to set up gauss's law to find the E field. Also since we are using line charge how does that change things?

    So i wrote |E|* 2*pi*l*s =1/ε0[itex]\int λ*2*pi*s*dl[/itex]

    but this doesn't seem right. I am unsure how to deal with the line charge. Do i use like surface charge still but write it as σ = 2*pi*r*dl but what are the limits of integration? I am just a little confused at this part. Help would be nice! Thanks
  2. jcsd
  3. Oct 8, 2013 #2
    sorry I wiorked a little bit, is the electric field E = λ/(2*ε0) * (s^2-a^2) where s is the radius on between a and b?


    I did some looking around and worked a little more and here is where I am at. The electric field in between the two cylinders is [itex]\frac{λ}{2*pi*ε0 * s}[/itex] where s is the radius of the gaussian surface. So then the potential in this area is then

    v = -[itex]\int E ds[/itex] and the limits of integration go from b to a (reference at infinity).
    so then plugging in E we get

    [itex]\frac{λ}{2*pi*ε0}[/itex]* ln(b/a) (negative was used to switch division).

    Then do I plug this into the first formula for work? Does σda = λ*l*2*pi*s*ds? What are the limits of integration? From a to b? I think I made a mistake somwehre.
    Last edited: Oct 8, 2013
  4. Oct 9, 2013 #3
    Hi guys, so I figured out my main problem with this question. I am having troubles coming to terms with the fact that were using a cylinder but the charge is given in terms of the linear charge density. So when using gauss's law, what is the charge enclosed? Is it just lamba * l?
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