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Total volumetric charge distribution of the universe

  • Thread starter Zee Prime
  • Start date
  • #1
Zee Prime
Greetings! I'm new here and I think about this place as soon as I see what the statement asks.

Homework Statement


Considering the volumetric density ρv=(e-2r/r2), figure the total charge (ℚ) of the universe.

Homework Equations


[/B]
ρv=ΔQ/ΔV -> (ΔQ ∝ ΔV)
ℚ=∫v ρv dxdydz

The Attempt at a Solution


I know you can figure it out ℚ when you've a pack of coordinates (bounds of the volume) in which you can calculate the total charge if you find some kind of symetry or not (i.e: cilindric, spherical coordinates and so on); but my mind just stacked overflow when the book asks the total charge of the universe... I wonder what system of coordinates and values should I use for the triple integral?

I've read that the shape of the universe —or known one— is flat; but I'm pretty sure I haven't the proper knowledge and mathematical understanding to realize that; so I assume for early problems, the shape is spherical, so I would use the following:

ℚ=∫vρv dv = ∫∫∫ρvr2Sin(Φ) drdΦdΘ

Jacobian Determinant.

I've found this problem at the second chapter of the book Electromagnetic Theory - Hayt. I'd appreciate some help with this problem. Thank you for your attention and keep this pantheon of physics alive! Congrats on this forum. :woot:
 

Answers and Replies

  • #2
phyzguy
Science Advisor
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I think you've set it up properly. Go ahead and put limits on the integrals and evaluate them.
 
  • #3
Zee Prime
I think you've set it up properly. Go ahead and put limits on the integrals and evaluate them.
But buddy, how would you put the limits of a spherical-universe (some rate of change?) The book suggest 6.28[C] as a result.

Ty for reply! :oldbiggrin:
 
  • #4
phyzguy
Science Advisor
4,372
1,352
But buddy, how would you put the limits of a spherical-universe (some rate of change?) The book suggest 6.28[C] as a result.
Ty for reply! :oldbiggrin:
Well, r will go from 0 to ∞. What about θ and φ?
 

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