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Surface integral/point charge question

  1. May 15, 2009 #1
    1. The problem statement, all variables and given/known data
    A uniform surface charge lies in the region z = 0 for [tex]x^2 + y^2 > a^2[/tex], and z = [tex]\sqrt{a^2-x^2-y^2}[/tex] for [tex]x^2 + y^2 \leq a^2[/tex]. Find the force on a unit charge placed at the point (0,0,b)

    2. Relevant equations
    dF = G ([tex]\delta[/tex])(vector of point to surface) / (magnitude of surface)^3
    [tex]\delta[/tex] is constant because it's a uniform surface.

    surface area = [tex]\int\int{\left\|{\frac{\partial f}{\partial x}} \times {\frac{\partial f}{\partial y}}\right\| dx dy}[/tex] = [tex]\int\int{{\sqrt{1 + ({\frac{\partial f}{\partial x}})^2 + ({\frac{\partial f}{\partial y}})^2} dx dy}[/tex]

    3. The attempt at a solution
    So basically, I know that this is a hemisphere with radius a centred at the origin, with an infinitesimal surface spanning the rest of the xy-plane. I know how to establish the vector from (0,0,b) to a sphere because a sphere is just radius a all around. I can identify any point on the sphere as (a sin [tex]\varphi[/tex] cos [tex]\theta[/tex], a sin [tex]\varphi[/tex] sin[tex]\theta[/tex], a cos[tex]\varphi[/tex])

    But this isn't a sphere; it's a really big sheet with a bump at the origin. I don't know how to identify the vector.

    The integral bounds seem simpler, but are just as complicated for me. In cylindricals, for the hemisphere, z goes from 0 to [tex]\sqrt{a^2-r^2}[/tex], r goes from 0 to 1, and [tex]\theta[/tex] goes from 0 to 2 [tex]\pi[/tex]. But isn't this a surface integral? Why do I have three variables? I really have no idea what I'm doing.
  2. jcsd
  3. May 15, 2009 #2
    With my very, very limited knowledge of E&M...

    It sounds to me like you need to solve 3 separate problems here:

    1.) Force due to an infinite sheet
    2.) Force due to a finite circular sheet... a disk, if you will
    3.) Force due to a hemisphere

    Individually, each of these shouldn't be too hard. And thanks to the superposition principle, you know that once you have the answer to these three questions (in the same coordinate system, naturally... probably x-y-z is easy because you know the force is going to be in the z-direction only, due to symmetry) you're done.

    If this is moronic, please ignore.
  4. May 15, 2009 #3
    Why would I need to find the force due to a finite circular sheet/disk though? Only 1 and 3 are the protruding surfaces.
  5. May 15, 2009 #4
    Anyone else have any ideas?
  6. May 15, 2009 #5
    You have to do the circular sheet / disk because I assume the infinite sheet doesn't cover the circular area where the hemisphere overlays it.

    By putting a sheet with an equal but opposite surface charge density down, you effectively make it look like an infinite sheet with a circular hole cut out of it. By adding the hemisphere to that, you get exactly the geometry you describe.

    Your geometry is:
    an infinite sheet with surface charge density D
    a disk with surface charge density -D
    a hemisphere with surface charge density D

  7. May 15, 2009 #6
    Yes, but then it leads to my next question: are forces on the hemisphere equal to forces on the surface? (If so, I can probably just equate the surface integrals to each other. But then what happens to the hidden disk?)
  8. May 15, 2009 #7
    What do you mean by "are forces on the hemisphere equal to forces on the surface"?

    You can calculate the force resulting from each of the three pieces separately and independently. These have no interactions with each other if we're talking about electrostatics.

    Once you calculate the force on a unit charge at (0,0,b) from each individual piece, you just have to add these answers together to the net force. The interactions between the hemisphere, the imaginary disk, and the rest of the infinite sheet don't really enter in.

    Am I just misunderstanding what you're asking?
  9. May 15, 2009 #8
    To answer your question literally:

    In cylindrical coordinates, the force will exhibit symmetry in the radial and angular directions, but not in the vertical (or z) direction.

    And the net force on the hemisphere will equal the net force on the sheet (which would also be the net force on the imaginary disk) which is zero for electrostatics.
  10. May 15, 2009 #9

    That was what I was looking for, yes. >_> Because I wasn't sure if trying to get away with three separate answers would be a good idea. (One for each region.)
  11. May 15, 2009 #10
    Yes, yes. It's a neat trick called the Superposition Principle. I'm sure you used it when you were doing easier stuff, like Coulomb's law. Basically, since the Superposition Principle works for Coulomb's law (it says that if you have two charges then the potential at an arbitrary point is just the sum of the two potentials from each of the constituent points) and since all electrostatic forces arise due to charge distributions exerting the Coulomb force, it has to work for abitrary potentials and distributions.

    Ergo, whenever you get a complicated geometry, just come up with enough simple geometric pieces to glue together and solve the problem on each one.

    Also, it's vital to notice the symmetry inherent in these problems. You know that if you get a force with anything other than z-component, you're in trouble. If you were looking for the general force vector field, you would expect cylindrical symmetry in this case. Why? The geometry does not discriminate directions.
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