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Homework Help: Electromagnetism problem using Gauss's Law

  1. Sep 19, 2011 #1
    1. The problem statement, all variables and given/known data
    Using Gauss's law calculate the electric field everywhere for the infinitely long insulating cylinder of radius R and charge density of rho = 3r^2(nc/m^3). SHOW ALL YOUR WORK INCLUDING DIAGRAMS.

    2. Relevant equations

    I am lost.

    3. The attempt at a solution

    I dont know where to start
  2. jcsd
  3. Sep 19, 2011 #2
    Well...what is Gauss's Law? Write out. You should have had that in your "relevant eqtn"!!
  4. Sep 19, 2011 #3
    Electric Flux = qin/Eo
  5. Sep 19, 2011 #4
    In integral form...?
  6. Sep 19, 2011 #5
    I'm not sure what you mean.
  7. Sep 19, 2011 #6
    Have you seen this before?

    [tex]\oint \vec{E} \cdot d\vec{A} = \frac{\sum Q_{en}}{\epsilon_0}[/tex]

    I pretty much drilled this equation into the back of my head...
  8. Sep 19, 2011 #7
    No. I'll take it the surface integral of E dot dA is equal to flux. I can remember that. I am still confused about how to use that equation to solve this problem.
  9. Sep 19, 2011 #8
    Well, the cylinder is infinitely long. What is the E-field emitting? what happens if the cylinder is a wire?
  10. Sep 19, 2011 #9
    Is it emitting charge? I don't know, what happens?
  11. Sep 19, 2011 #10
    You have a long wire (forget being cylinder for now), what is the E-field? What is the direction?
  12. Sep 19, 2011 #11
    If it helps, draw a Gaussian Surface over the wire/cylinder.
  13. Sep 19, 2011 #12
    I have absolutely no idea.
  14. Sep 20, 2011 #13
    Do you know what Gauss's Law means...? What it states? Don't just direct meback to the formula. What is "dA"?
  15. Oct 30, 2011 #14
    This was obviously posted a while ago, answering may help others though so:

    Begin by constructing a cylindrical Gaussian surface of radius r and length l around the cylinder. Then the electric field outside of the cylinder is, by Gauss's law:

    [itex]\oint[/itex]E.da = Q / ε0 where Q is the total ENCLOSED charge.

    The da on the left hand side refers to the area that the electric field passes through the Gaussian surface, not of the cylinder itself:

    E x 2∏rl = Q/ε0

    Now, the Q enclosed is the volume charge density multiplied by the volume in which Q is contained (i.e. the volume of the cylinder):

    Q = ρ x ∏R^2 x l (R - the radius of the CYLINDER not the Gaussian surface is used)

    Substitute back in:

    E x 2∏rl = ρ x ∏ R^2 x l / ε0
    E = ρ x R^2 / 2rε0

    Then substitute ρ = 3r^2:

    E = 3rR^2/2ε0 (outside; r>R)

    For inside the enclosed charge has "the same r":

    E x 2∏rl = ρl∏r^2 / ε0
    E = ρr / 2ε0

    Substitute ρ:

    E = 3r^3 / 2ε0 (inside; for r < R)
    Last edited: Oct 30, 2011
  16. Jun 27, 2012 #15
    May I join the discussion?

    Will this line of reasoning apply to calculating (estimating) the electromagnetic field (in Gauss) that one would expect to measure at a given distance from a high voltage power line? Since I would not have the charge available, but may have the voltage and line current, could the field be calculated in Gauss (or Teslas and then multiply by 10^4)? Will you show me the formula?
  17. Jun 27, 2012 #16
    Hello Steve, welcome to PF!
    It would be better if you start a new discussion, do you see the date of the last post? :smile:
  18. Jun 28, 2012 #17
    Yes, the exact same method can be applied. A typical "text-book example" gives an infinitely long wire with a linear charge density λ. Just as with a cylinder a cylindrical Gaussian surface can be used for a wire (a wire in reality is really just a thin cylinder so this makes sense).
    For your question however if you have a scalar potential (voltage, V) there is a simple formula connecting this to the E field:

    E= -∇V - δA/δt , where A is the vector potential

    If ∇xE = 0 (i.e. B field not changing with time) then this reduces to:

    E= -∇V

    Which in the one dimensional case of a wire can be treated as:

    E= -dV/dx
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