1. Not finding help here? Sign up for a free 30min 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!

Gauss' Law; long, thin wire

  1. Nov 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Calculate the electric field at a point 2.79 cm perpendicular to the midpoint of a 2.02 m long thin wire carrying a total charge of 5.04 uC.

    You could integrate BUT if the wire is very long compared to the distance from the wire to where you are calculating the electric field, then the electric field will be radial and Gauss's law might be easier.


    2. Relevant equations

    ER=λ/2piε0R

    3. The attempt at a solution
    First I tried doing it by integral (getting E=kexQ/(x2+R2)3/2) but then it gave me the hint to use Gauss' Law, so I treated it like a cylinder, using charge divided by length for λ and 0.0279 m for R. I came out with 5.76x107 N/C which I am told is wrong.

    THEN I worked that to E=2keλ/R after having a "duh" moment but I still seem to be getting the wrong answer (2.58x105)

    Edit: calculator/math error. I solved it.
     
    Last edited: Nov 4, 2012
  2. jcsd
  3. Nov 5, 2012 #2
    The situation is very easy if you use assume really an infinitely long wire. Then, the electric field can only depend on the distance to the wire, say ρ in cylindrical coordinates.
    Gauss's law [itex]\int_{\partial V}\mathbf{E}d\mathbf{A}=\frac{1}{\varepsilon_0} \int \rho dV[/itex]
    becomes:
    [itex]2\pi \rho E_\rho(\rho)\int dz=\frac{1}{\varepsilon_0} \lambda \int dz[/itex]​
    with charge density [itex]\lambda[/itex] [C/m]. This is a rather dirty trick to see that
    [itex]E_\rho(\rho)=\frac{\lambda}{2 \pi \varepsilon_0 \rho }\ . [/itex]​
    This result is correct in the infinite-length-limit of a line charge. Here you can find the full calculation.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook