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Insulating sphere

  1. Feb 12, 2007 #1
    1. The problem statement, all variables and given/known data
    An insulating sphere with a radius of 0.100 m has a charge of 0.965 nC uniformly distributed throughout its volume. The center of the sphere is a distance 0.295 m above a large uniform sheet that has a charge density of -7.90 nC.

    Find the distance from the center of the sphere to the point inside the sphere at which the electric field is zero.


    2. Relevant equations
    r < R

    E = 1/4pi*E_o * Qr/R^3


    Is there only 1 sphere that is directly above the sheet that is 0.295m ? I'm not exactly sure how the picture looks, so I wanted to verify this and then what to do next.
     
  2. jcsd
  3. Feb 12, 2007 #2

    Doc Al

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    Yes. Picture the large sheet as being in the x-y plane at z = 0. The sphere of charge can be pictured as being centered at point (0, 0, 0.295).
     
  4. Feb 12, 2007 #3
    Okay so having that out of the way we have the following givens:

    r = 0.100m
    +Q = 0.965 nC
    R = 0.295
    -Q = -7.90

    The field lines out of the sphere is going outward since its a positive charge.

    We want to find the distance where the electric field is zero.

    How should I start this problem based on a Solid insulating sphere with radius R and charge Q distributed uniformly throughout the volume?

    Since the question asks inside the sphere it should be the case that:

    r < R

    E = 1/4pi*E_o * Qr/R^3

    We want E = 0 ?

    1/4*pi*E_o * Qr/R^3 = 0 ?
     
  5. Feb 12, 2007 #4

    Doc Al

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    Good. That's the field from the uniform sphere of charge.

    No. You want the point where the total electric field due to all the charge is zero, not just the field from the sphere of charge.

    The field at any point is the sum of the field from the plane plus the field from the sphere.
     
  6. Feb 12, 2007 #5
    I currently have the following equation:

    (-7.90 / 2*E_0) + (1/4*pi*E_0) = .965/r^2

    I sum up the charge of the sheet with the sphere and solve for r I get r = 0.45 but thats incorrect and I was wondering if I was attempting this correctly?
     
  7. Feb 12, 2007 #6

    Doc Al

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    I don't understand this equation. You should have something that looks like: Field from sheet + Field from sphere = 0.

    Also, you should be able to point to the spot on your diagram (assuming you've made one) where this condition would be met. That will help you visualize what's going on.
     
  8. Feb 12, 2007 #7
    Yes that is what that equation represents field from sheet + field from sphere = 0. I know the distance should be above the center point of the sphere.
     
  9. Feb 12, 2007 #8

    Doc Al

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    I'm not seeing it. Your first term looks like the field from the sheet. But what's the second and third terms? And why isn't the sum set equal to zero?

    Right. (Just checking.)
     
  10. Feb 12, 2007 #9
    Is this the electric field for a sphere = 1/4pi*E_o * Qr/R^3
     
  11. Feb 12, 2007 #10

    Doc Al

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    Yes, as I thought you had stated several posts ago.
     
  12. Feb 12, 2007 #11
    okay I'm a little confused on which "r" we're looking for, here's my math so far:

    [tex] \frac {\sigma} {2*e_o} = \frac {1} {4*pi*e_o} * \frac {Qr} {R^{3}} [/tex]

    = [tex] \frac {-7.9} {2*(8.85*10^{-12})} = \frac {1} {4*pi*(8.85*10^{-12})} * \frac {.965*r} {.295^{3}} [/tex]

    Is big R = 0.295 the distance from center of sphere to the sheet ?

    = [tex] 3.49*10^{-11} = \frac {6.70*10^{-12}*r} {.025672} [/tex]

    = [tex] \frac {3.49*10^{-11}} {6.70*10^{-12}} * (.025672) = r => 1.3*10^{-25} [/tex]

    and I don't think that is correct =/
     
    Last edited: Feb 12, 2007
  13. Feb 13, 2007 #12

    Doc Al

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    No, big R is the radius of the sphere. (The formula for the field within the uniform sphere of charge has nothing to do with the distance to the sheet.)

    Also: Careful with your arithmetic. Simplify as much as possible before plugging in numbers. (For example, the [tex]\epsilon_0[/tex] cancels from both sides.)
     
  14. Feb 13, 2007 #13
    I think it's easier to do this problem using Gauss's law to simplify spherical E in terms of charge density. Then set that equal to Eplate and after some simple algebra you have your answer.
     
  15. Feb 13, 2007 #14

    Doc Al

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    That part's already solved. :wink:
    That's exactly what he's doing.
     
  16. Feb 13, 2007 #15
    Okay let's try again from the start:

    [tex] \frac {\sigma} {2*e_o} = \frac {1} {4*pi*e_o} * \frac {Qr} {R^{3}} [/tex]

    = [tex] R^3*\frac {\frac {\sigma} {2*e_o}} {\frac {1} {4*pi*e_o}} [/tex]

    = [tex] R^3*\frac {\frac {\sigma} {\frac {1} {2*pi}}} {Q} [/tex]

    Is this correct, I think I might have done something wrong, but it's pretty late. Thank you
     
    Last edited: Feb 13, 2007
  17. Feb 13, 2007 #16

    Doc Al

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    Yes, that's correct. Rewrite it more simply as:
    [tex]r = \frac{2 \pi \sigma R^3}{Q}[/tex]
     
  18. Feb 13, 2007 #17
    thank you.
     
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