1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Electromagnetism problem using Gauss's Law
  1. Proof using Gauss' Law (Replies: 5)

Loading...