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Homework Help: Electric Flux Through a Cone

  1. Mar 8, 2012 #1
    1. The problem statement, all variables and given/known data

    A cone is resting on a tabletop as shown in the figure with its face horizontal. A uniform electric field of magnitude 4550 N/C points vertically upward. How much electric flux passes through the sloping side surface area of the cone?


    2. Relevant equations

    flux = ∫ E*dA

    3. The attempt at a solution

    I'm really just confused given the geometry of the situation. In general, I understand that to find the electric flux, you choose a Gaussian surface (exploiting symmetries when possible), and the flux will be equal to the electric field times the area of the Gaussian surface, which is equal to the charge enclosed divided by ε_0.

    But what I'm unsure about is what kind of Gaussian surface to use here?

    As for finding the charge, since the electric field is uniform and vertically upward, should I treat it as a line of charge?

    This cone business is confusing me!

    Any help would be greatly appreciated.
    Thank you!
  2. jcsd
  3. Mar 8, 2012 #2
    well you can treat cone itself as the gaussian surface. Since there is only constant electric
    field, no charges are present inside the cone. so by gauss's law, total flux is zero. but the total flux is flux through the slanted surface + the flux through the flat surface. since E points vertically upwards, its easy to calculate the flux through the flat surface. that will then give you the flux through the slanted surface.
  4. Mar 8, 2012 #3
    Turns out the total flux is not 0, so I'm assuming the charge that emits the electric field is to be enclosed in the gaussian surface, even though they don't mention any such charge. And so that's where I'm confused...how do I find the charge. From that point, finding the flux should be easy enough, but as it is, I don't have a q(enc). Also, since E is constant and vertically upwards, then I'm assuming it comes from a line of charge, so maybe I ought to be using lamba as line charge density, and use q=(lamba)l, but these are all unknown values. I don't see how to compute this gives what I have?
  5. Mar 8, 2012 #4
    well this constant E is created by charges which are far away. so don't worry about them. at the location of this cone, you just have a constant E. problem is simple
  6. Mar 8, 2012 #5
    Ok, wow, so mathematically, the problem is very simple. Thank you!

    However, I don't understand why there is a flux. I thought that in order for there to be a flux, there must be a charge enclosed...in this case there's not.
  7. Mar 8, 2012 #6
    the net flux through the entire cone is zero. its like water flowing through some closed imaginary surface in the river. if we only look at some part of the surface, then there is water going through it. But on the whole, the net water flowing out is zero, since water coming in is same as water going out
  8. Mar 8, 2012 #7
    I see. So, attention to detail is key, here.

    Thank you so much for the help! I really appreciate it, Mr. Newton, as well as all of your laws. ;)
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