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Flux of an Electric field with a constant charge density

  1. Nov 18, 2009 #1
    This is actually a problem in my Vector Calc class, but is very fitting for this section.

    I have been trying to understand this problem, and cannot seem to figure out where to go.

    The z-axis carries a constant electric charge density of λ units of charge per unit length with λ > 0. The resulting electric field is given by

    E=5λ*xi+yj/x^2+y^2

    Compute the flux of outward through the cylinder x^2 + y^2 ≤ R^2, for 0 ≤ z ≤ h. Use lambda for λ.


    I have been having pretty good success with most of the problems from this section, but I cannot seem to set this guy up.

    Since both the normal vector, and the electric field are pointing radially away from the z axis, the angle between them is 0.

    Can someone give me a step in the right direction please? I have a few problems that are driving me nuts.
     
  2. jcsd
  3. Nov 18, 2009 #2

    gabbagabbahey

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    Hi want2learn, welcome to PF!:smile:

    Since you are asked to calculate the electric flux, a good place to start might be to write down the definition (in equation form) of electric flux....don't ya think?:wink:
     
  4. Nov 19, 2009 #3
    Ok, when I said I don't know where to start, I did not mean that basic.

    I know Gauss's electric flux equation. I know that it involves the Electric field, the normal vector to the area, and the area of the cylinder (2*pi*r*h).

    How can I find the normal vector in this problem?
     
  5. Nov 19, 2009 #4

    gabbagabbahey

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    Draw a picture...the cylinder has 3 surfaces, what are they? How does one normally find the unit normal to a surface?...Do that.
     
  6. Nov 19, 2009 #5
    I drew a picture, and have everything set up. the Surface Area of the cylinder would be 2pi*r^2+2pi*r*h. Since the charge is located along the z axis, and the cylinder is oriented around the z axis, the top and bottom of the cylinder should not be used in the SA. So, the surface area would just be 2*pi*r*h, right?

    And, normally I would cross two vectors to find the normal vector, and I am not getting how you find it on a cylinder like this. Since the angle between the normal and the electric field is 0, does it end up just being the Electric field*Surface Area ?
     
  7. Nov 19, 2009 #6

    gabbagabbahey

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    How do you know that the angle between the electric field and the surface normal is zero if you can't find the surface normal?

    Also, the equation for flux involves an integral doesn't it?...What makes you think the result of that integral will just be electric field*surface area?
     
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