anonymous
  • anonymous
A charged sphere of radius R has its origin at the origin of coordinates. The upper half of the sphere (z>0) has a positive uniform volume charge density ρ, while the lower half (z<0) has a negative uniform charge -ρ. Find the electric field produced at an arbitrary point on the z-axis above the sphere (z>R).
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Your best bet is to calculate the electric potential at that point and then find the electric field from there, although you could use a symmetry argument and work with the electric field all by itself -- I would advise against that though.
anonymous
  • anonymous
Recall that \[ \Phi(\vec{r}) = \int \frac{\rho}{|\vec{r}-\vec{r}'|} dV' \] Where |dw:1378263082680:dw|
anonymous
  • anonymous
|dw:1378263335816:dw| Oops, should be a full sphere...

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anonymous
  • anonymous
Calculate \[\Phi_1=\frac{ \rho }{ 4 \pi \epsilon }\int\limits_{\phi=0}^{2 \pi}d \phi \int\limits_{r=0}^{R}r^2\int\limits_{\theta=0}^{\frac{ \pi }{ 2 }}\frac{ \sin(\theta) }{ \sqrt{z^2+r^2-2rzcos(\theta)}}d \theta dr\]and \[\Phi_2=-\frac{ \rho }{ 4 \pi \epsilon }\int\limits\limits_{\phi=0}^{2 \pi}d \phi \int\limits\limits_{r=0}^{R}r^2\int\limits\limits_{\theta=\frac{ \pi }{ 2 }}^{\pi}\frac{ \sin(\theta) }{ \sqrt{z^2+r^2-2rzcos(\theta)}}d \theta dr\]then you get, applying superposition:\[\Phi=\Phi_1+\Phi_2\]and finally:\[E_z=-\frac{ d \Phi }{ dz }\]If you need help to calculate the integrals, shout
anonymous
  • anonymous
why can't we hit this will full on Gauss !?
anonymous
  • anonymous
ok i get it why you can't :P
anonymous
  • anonymous
Isn't this right??!
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anonymous
  • anonymous
adn then
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anonymous
  • anonymous
No, there isn't full spherical symmetry in this problem, only azimuthal symmetry. The field lines will not be radial.
anonymous
  • anonymous
Yes, I need help to calculate the integrals ;/
anonymous
  • anonymous
ok Jemmurray.. thanks for clearly.. this is clearly beyond my kin then !
anonymous
  • anonymous
@ebc here goes my proposal for integrals calculation. I hope I have not made any mistake. If you find something inconsistent, let me know
anonymous
  • anonymous
Carlos and Jemurray, thank you so much!

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