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Charge at a distance from a non-conducting hemisphere

  1. Apr 24, 2013 #1
    1. The problem statement, all variables and given/known data
    A non-conducting hemisphere of radius R centered at the origin has a total charge Q spread uniformly over its surface. The hemisphere is oriented such that its base is in the (y,z) plane. Find the electric field anywhere along the x axis for x > 0. Give explicitly the value of the electric field at x = 0.
    Here is an image of the problem as it was given to me:
    http://i.imgur.com/GZ5Edcc.jpg


    2. Relevant equations
    ERing = (1/4∏ε0)*Qx/(x2+R2)3/2
    Q = 2ρ∏r2

    3. The attempt at a solution
    I attempted to sub-divide the hemisphere in to separate rings and integrate the sum, but the integral I ended up with was pretty ugly and I'm not even sure it's correct. I added a variable r, representing the radius of each individual ring and integrated with respect to it.
    My integral:
    (R2ρ/2∏ε0)*∫(x+(R-r))dr/((x+(R-r))2+r2)3/2
    From 0 → R
    I could probably solve this integral given enough time, but I'm pretty sure I will be expected to do this on a test and was hoping there was a less complicated solution.

    Thank you
     
  2. jcsd
  3. Apr 24, 2013 #2
    You are integrating k dq/(d^2) over the charged surface. What do you know about dq? Well, you know that since Q = 2ρπR^2, you know that ρ = Q/(2πR^2) and that dq = ρ dA, which is a differential unit of area.

    How will you integrate over these differential units? Since the radius doesn't change, you have two angles, and you know that dA = R^2 sin(θ) dθ dφ. This lets you integrate over the surface.

    You still need to define d, though. d is the distance from the charge to the point you're measuring the electric field at. Try to draw the distance to a differential unit of area on the sphere. You'll see it forms a triangle: the distance to the unit of area, the radius of the shell, and the length from the origin. You'll also have the angle φ. Using the law of cosines, you can find out the distance.

    Now that you know your parameters of integration, you can take it from here.
     
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