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Electric Field - 1/8 of Sphere

  1. Feb 20, 2014 #1
    1. The problem statement, all variables and given/known data
    Calculate the intensity of the electric field (in the origin of cartesian axes) created by 1/8 of a certain sphere charged with density of charge equal σ.

    opxmh3.jpg

    Answer:
    [tex]\frac{{\sigma \sqrt 3 }}{{16{\varepsilon _0}}}[/tex]

    2. Relevant equations
    E = KQ/R²
    Flux = Q/ε
    E.A = Q/ε

    3. The attempt at a solution

    Well, I tried to calculate the electric flux in 1/8 of the sphere (see figure). Then, knowing the area of 1/8 sphere, we could calculate the electric field using gauss law. But, after all, the answer of the problem is different.
     
  2. jcsd
  3. Feb 20, 2014 #2
    You'll have to explain your process more -- what values you used, and how you fit those into your equations -- since I don't quite understand it. What flux is going through the sphere?
     
  4. Feb 21, 2014 #3
    The flux in total sphere is x, so flux now is x/8.. No?
     
  5. Feb 21, 2014 #4
    Flux from what, specifically? That's true if there's a point charge at the center of the sphere, but not the case if there's only an 1/8 of the sphere. I don't think Gauss's law is going to be very helpful.

    There's really no symmetry to exploit, so I think this is just going to be a full out vector integration. It will look something like this:

    [itex]\int^{R}_{0}\int^{\frac{\pi}{2}}_{0}\int^{\frac{\pi}{4}}_{0}d\stackrel{\rightarrow}{E}dV[/itex]

    Where [itex]dV = \rho^{2}sin\theta d\rho d\phi d\theta[/itex], and [itex]\stackrel{\rightarrow}{r} = \left\langle \rho sin\phi cos\theta , \rho sin\phi sin\theta , \rho cos\theta \right\rangle[/itex]
     
  6. Feb 21, 2014 #5
    But, this problem is from a middle school book, so don't think spherical coordinate is the unique solution. Did anyone knows other way to solve this problem ?
     
  7. Feb 22, 2014 #6

    rude man

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    Using spherical coordinates, the potential at the origin seemingly must be kσA/r since all charge is r away from the observation point. A = area of 1/4 of hemisphere, k = 1/4πε0.

    Computing the E field as - del V seemingly gives |E| = kσπ/2
    which differs from the given answer by about 15%.

    So, sorry, can't get your answer.

    Need some expert help here ...

    EDIT: I realize my computations are overly simple - the gradient at the origin must be evaluated with excursions from the origin in all three coordinates - and when you do that you get dV a function not only of dr but of r, dθ and dø as well. Not that that helped me solve the problem.

    mfb, tsny et al ... ?
     
    Last edited: Feb 22, 2014
  8. Feb 22, 2014 #7
    Using cartesian coordinates?
     
  9. Feb 22, 2014 #8

    rude man

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    The integrations using cartesian would be a lot messier. This problem is a natural for spherical.
     
  10. Feb 22, 2014 #9
    Hum... If we draw a cube over the sphere of picture? We can calculated something?
     
  11. Feb 22, 2014 #10

    TSny

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    There is some symmetry that you can exploit. For 1/8 of a sphere occupying one octant of the xyz coordinate system, the net E vector at the origin will have equal x, y, z components. So, you just need to evaluate one component, say the z-component.
     
  12. Feb 22, 2014 #11
    Thanks, but could you be more specific? Because how to calculate this component without using spherical coordinates.. I'm confused..
     
  13. Feb 22, 2014 #12

    TSny

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    I was just saying that if you are going to integrate to find E, then you can just do the integration for the z-component since the other components will be the same. Using spherical coordinates, the integral is not difficult.

    However, if you want to avoid integration altogether, then you can resort to subterfuge.

    Some hints:

    (1) First consider finding the electric field at the center of a uniformly charged hemisphere. You should be able to relate the net E field at the origin of the hemisphere to the z-component of the E field of the 1/8 sphere.

    (2) To handle the hemisphere, imagine putting a point charge q at the origin. The net force of the hemisphere on q equals qE where E is the net field at the origin due to the hemisphere. So, if you can get the force on q you can get E at the center of the hemisphere.

    (3) The force that the hemisphere exerts on q equals the force that q exerts on the hemisphere.

    (4) See if you can get the net force of q on the hemisphere without any detailed integration.
     
  14. Feb 22, 2014 #13
    Thank you very much TSny !!! I found the electric field created by hemisphere is σ/4.e0, so, my problem is 1/4 of a hemisphere, so, axis z = y = x = σ/16.e0, now, using pitagoras, we can find the answer !!!
    :)
     
  15. Feb 22, 2014 #14

    TSny

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    Sounds good!
     
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