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Electrostatics: a question regarding a charged insulated sphere

  1. Aug 25, 2008 #1
    the question that i have encountered was:
    "given an insulated charged sphere, what is the force applied by one half of the sphere on the other half?"

    i have tried calculating the integral but it seemed a little too complicated for me (plus i am not absolutely confident what the correct integration actually is)
    is there a better way approaching this question?
  2. jcsd
  3. Aug 25, 2008 #2
    what is the force on each individual charge dq?
  4. Aug 25, 2008 #3
    the question is what is the entire force "sensed" by half a sphere
    the only parameter given is the charge density p (forgot to mention)
  5. Aug 25, 2008 #4
    imagine the sphere is filled with fluid under pressure. that should give the same result for the force at the surface (where the charge is). it should be easy to calculate the force on a plane down the center.

    you understand that all the charge is evenly distributed at the surface right?
  6. Aug 25, 2008 #5
    why would the charge be distributed at the surface if the sphere is insulated?
  7. Aug 25, 2008 #6
    its easy to see if the sphere is a conductor. there cant be any electric field within the conductor. I dont think it makes any difference its an insulator.

    what class is this for?
  8. Aug 25, 2008 #7
    wait. charge density? so its asking about a hypothetical sphere with uniform charge density throughout? well that would be different. I'm not sure how to solve that.
  9. Aug 25, 2008 #8

    if the sphere was a conductor then i would have used the pressure sensed by each half a sphere in order to calculate the force (as you suggested), but since the sphere is insulated i can't seem to come up with an idea on how to solve this ...
  10. Aug 25, 2008 #9
    insulated sphere with a uniform charge density p, correct.
    thanks for your help, i hope someone else has an idea though.
  11. Aug 25, 2008 #10
    insulated? I dont see what that has to do with anything. it might help if you just type in the problem as it is stated in the book.
  12. Aug 25, 2008 #11
    I dont see any easy solution but you can always just use the pressure technique to calculate the force from each infinitesimally thick shell and integrate over all r. maybe someone can come up with something better though.
  13. Aug 25, 2008 #12
    the fact that it is insulated and not conducting makes it possible for the charge to be uniformly distributed throughout the entire sphere. if it was not insulated then charges would not have been bounded and there would have been no electric field inside.
    the question is originally stated in a different language, i merely translated it into english.
    anyway, thanks again.
  14. Aug 25, 2008 #13
    well I dont know what 'insulated' means in your language but in English it means that the sphere is surrounded by an insulating material. and even if it were itself an insulating material the charge would ordinarily still be at the surface.
  15. Aug 25, 2008 #14
    would it help if you reverse it and think in terms of gravity and pressure as a function of depth?
  16. Aug 25, 2008 #15
    i am not completely sure how that helps... maybe if you could be more specific?
  17. Aug 25, 2008 #16
    the force of attraction of the 2 halves must equal the pressure across a plane through the center. simply reverse everything and you have your original problem. attraction becomes repulsion, etc...
  18. Aug 25, 2008 #17


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    Do you have Mathematica or something? You could do it in Cartesian coordinates, or spherical coordinates, but since you're dealing with two hemispheres, I'll prefer Cartesian. But then again I'm not familiar with volume integration using spherical basis vectors since [tex]\mathbf{a_{\phi}}[/tex] and the other basis vectors are not constant for varying values of [tex]\phi[/tex] and [tex]\theta[/tex].

    If you got this question from a textbook, there's a very good chance that you can exploit some law or theory to cut down on a lot of integration and vector calculus.

    But otherwise I'll start with [tex]\mathbf{E}(x,y,z) = \int_{vol} \frac{\rho}{4\pi \varepsilon} \frac{\mathbf{r}-\mathbf{r'} }{|\mathbf{r}-\mathbf{r'}|^3} dv'[/tex]. Limits of the integration is confined to the left half of the hemisphere, so y,z ranges from -r to r, x from -r to 0. Once you get the field (by Mathematica or by tedious hand calculations), you'll need to integrate for the force F using [tex]d\mathbf{F} = p\mathbf{E}dv[/tex] which is a volume integration over the other half of the hemisphere.

    As said before, this is really tedious without Mathematica or some symmetrical considerations.

    [quote="granpa]the force of attraction of the 2 halves must equal the pressure across a plane through the center. simply reverse everything and you have your original problem. [/quote]Hi, I'm not sure what force of attraction you're talking about. The charge is uniform throughout without any change in sign, is it not?
  19. Aug 25, 2008 #18
    see post 14
  20. Aug 25, 2008 #19
    the electric field is simply proportional to r (up to the surface) and always directed outward. just like gravity.

    and the pressure below ground is simply equal to the weight of material in a column (not wedge) directly above it.
  21. Aug 25, 2008 #20
    so a plane in between the two halves will experience pressure due to the force of repulsion, that makes sense, but i can't figure out how to calculate the pressure without finding the electric field (or force for that matter)
    if you meant something else then i probably just didn't get it
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