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Finding Electric Flux and Electric Fields

  1. Oct 5, 2011 #1
    I have two questions that I cannot seem to figure out.

    1. The problem statement, all variables and given/known data
    1)A uniform electric field E=30 N/C exists parallel to the axis of a square pipe of side length l=5 cm as
    shown in the figure. Calculate the total electric flux for the slanted face of the pipe

    2) A very long metal rod, radius R, has a uniform surface charge density σ. (a) Ignoring end effects, find
    the electric field E at a distance R along the radial direction from the surface of the cylinder. Also draw a
    diagram show the cross-section of the cylinder and Gaussian surface in your solution. (b) Find the speed
    v such that an electron could travel in a circular orbit about the rod at a distance R from the rod’s

    2. Relevant equations

    1) psi= EA cos theta

    3. The attempt at a solution

    1) EA cos theta = (30 N/C) * (A) cos 30

    A = (5.00cm)(w) = (5.00cm)*(5.00cm/cos30) = 0.2887m^2

    psi EA = (30 N/C) (0.2887)cos30 = 7.50 m^2/C

    I am pretty sure this is incorrect, but this is the closest I can get. Can anyone help me out?

    I have no idea how to do #2. If someone could lead me in the right direction I would greatly appreciate it.

    Thanks guys,

    Last edited: Oct 5, 2011
  2. jcsd
  3. Oct 5, 2011 #2
    For #1, it would be helpful if we could see the figure being referenced.

    For #2, there are a few things that will help push you in the right direction:
    - The radial direction just refers to the direction that moves outward from the center of, in this case, the circular cross-section.
    - The Gaussian surface that encloses the charge will have a similar geometry to the conductive surface that bears the charge.
    - Remember the Faraday conventions of electric field lines: Density of field lines through an area should be proportional to the relative strength of the field over a given area, each charge of equal magnitude should be emanating an equal number of field lines in a symmetric fashion (for example if you have a proton that you have emanating 5 field lines, they should be spaced evenly and an alpha particle in the same diagram would have to have 10 symmetrically spaced lines), field lines always emanate perpendicular to the surface of the charge, and field lines never cross nor kink.

    Also, you can solve for v such that the motion of an electron would be perfectly circular (using the equations for uniform circular motion) and applying Coulomb's force.

    I hope this is both helpful and not spoonfeeding. :)
  4. Oct 5, 2011 #3
    I totally forgot to include the supplied figure. Thanks for the heads up. I have edited the first post and included it.
  5. Oct 5, 2011 #4
    Okay, so for #1, my approach would frankly just be to say that whatever the flux is through the back end, the flux through the slanted face would be the negative of that: the net flux through any object in a uniform electric field should be zero (because the uniformity of the electric field implies that there is no net electric field caused by the charges, if any, inside the surface, hence no net flux). Further, since the field runs parallel to the axis of this square pipe, the only surfaces whose normal-area vectors have a component parallel to the field are the "back" surface and the slanted "front" surface. (Surfaces whose normal-area vectors are perpendicular to the electric field would see a flux of 0 because perpendicular vectors always have a dot product of 0.)

    This is an easier method, and while it's not universally applicable it's a handy thing to keep in mind through much of your work in electricity and magnetism: symmetry is absolutely your friend.

    To do it the other way, though, you would basically just use some trigonometry to determine what the parallel component of the electric-field vector (relative to the normal-area vector of your slanted surface) is, and take phi = the closed integral of E *dot* dA. That is, if we choose E to be in the positive x direction, you would multiply E by the x-component of A to get your flux.
  6. Oct 9, 2011 #5
    I am having problems with this one as well. Cant seem to get it started.
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