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Gauss' Law - Sphere with spherical cavity

  1. Oct 5, 2008 #1
    Gauss' Law - Sphere with spherical cavity - Please help

    1. The problem statement, all variables and given/known data

    A sphere of radius 2a is made of nonconducting material that has a uniform volume charge density . (Assume that the material does not affect the electric field.) A spherical cavity of radius a is now removed from the sphere, as shown in Figure P19.62. Show that the electric field within the cavity is uniform and is given by Ex = 0 and Ey = Pa/3Eo. (Hint: The field within the cavity is the superposition of the field due to the original uncut sphere, plus the field due to a sphere the size of the cavity with a uniform negative charge density -.)

    http://greenlanternbattery.googlepages.com/p19-62.gif

    2. Relevant equations

    P= Q/V


    3. The attempt at a solution

    Esphere=Q/Eo

    I'm not really sure if the Esphere is right, and I'm not sure where to go next if it is. In my notes from the prof, it says "for each point P in the cavity need to consider the contribution to the E field from the positive charge distribution (let the vector be r+) and from the negative charge distribution (r-)." I am not really sure what this means though.
     
    Last edited: Oct 5, 2008
  2. jcsd
  3. Oct 5, 2008 #2
    Anyone?
     
  4. Oct 5, 2008 #3
    draw it. make a point for the center of the sphere and another point for the center of the cavity (negative sphere). choose any other point. draw the vectors involved..theres a simple and obvious solution.
     
  5. Oct 5, 2008 #4
    I have drawn it. I have made points at the centre of both the sphere and the cavity.

    Choose any other point? Do you mean any point inside the cavity? Or a point outside the circle?

    Draw the vectors involved? You mean E and A right? If so, it says prove the E component vectors are equal to what they give. How then do you know which way the vectors are going?

    What does my prof mean, by what he put?

    I'm hoping the solution will become obvious to me soon.
     
  6. Oct 5, 2008 #5
    I mean r+ and r-.
    they point toward the centers of their respective spheres.


    yes a point in the cavity
     
  7. Oct 5, 2008 #6
    On my prof's thing, he chose a point P near the top right of the cavity. He had r+ and r- pointing to them, away from the center. So, sorry about the confusion, I have those drawn in, as I used the diagram he made. The obvious solution is not coming to me though, and I feel completely lost right now.
     
  8. Oct 5, 2008 #7
    one should be away and one should be toward.

    do you understand what the vectors represent? I cant make it any more obvious without just handing you the answer. you're probably making it more complicated than it really is. its quite simple.
     
  9. Oct 5, 2008 #8
    the vectors represent the charge distribution, whether positive or negative, depending on the r vector. right?
     
  10. Oct 5, 2008 #9
    no. you should already have done a problem in which you calculate the electric field at distance d from the center of a sphere of radius r with uniform charge density. where d<r.
     
  11. Oct 5, 2008 #10
    umm...i did electric flux where R>d...i am not finding one with electric field
     
  12. Oct 5, 2008 #11
    good enough. the electric field at a point is equal in magnitude to the flux at that point and it is a vector so it points in the direction of the flux lines. (thats the whole point of flux lines. to visualize the electric field)
     
  13. Oct 5, 2008 #12
    Umm..I did a completely different question with the flux. This entire question was in my original post.
     
  14. Oct 5, 2008 #13
    So how do you know which way the electric field vector is pointing?
     
  15. Oct 5, 2008 #14
    by whether its a positive or a negative charge.
     
  16. Oct 5, 2008 #15
    but it doesn't say if there is a positive or negative charge? Or does this have to do with r+ and r-?
     
  17. Oct 5, 2008 #16
    the trick here is to think of the cavity as a sphere of opposite charge superimposed on the other sphere.
     
  18. Oct 5, 2008 #17
    After you left, I think I may have got it. Just had to look at the example problems for the 15th time I guess. Hopefully it's good, because I was able to prove Ey for sure.

    Thanks very much for all your help! :D
     
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