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Uniform Charge Distribution. Find total charge.

  1. Sep 6, 2010 #1
    1. The problem statement, all variables and given/known data

    We have a rectangular box with diagonal corners at (0,0,0) and (-1,2,3)cm. We place charge inside this box with distribution, ρ(r)=2x-3(z-1) with units of nC/m3. What is the total charge inside this box?

    2. Relevant equations

    3. The attempt at a solution

    I have an answer, but I am not sure if I thought about it correctly. I would greatly appreciate confirmation on my thinking, and my final answer if possible. Thanks a lot.

    I started with a triple integral of the charge distribution over the bounds for all three dimensions. After integrating, I ended up with a result of 15. I finally converted that into 15E-15 C after factoring in the cm to m conversion and the nC to C conversion. Thanks again!
  2. jcsd
  3. Sep 7, 2010 #2
    Your idea ist correct, but you must be careful with the direction of integrating. The 15 is wrong. If you do the integration correctly you will get a different number.

    If you don't see what you have to change, do the same box (i.e. the same integration) for a constant charge density [tex]\rho=1[/tex]. You should get a positive total charge but I suspect you will not.
  4. Sep 7, 2010 #3
    Is it supposed to be 3? I believe my mistake was on the bounds of integration for the x- integral. I had it going from 0 to -1, but perhaps it is supposed to be reversed?
  5. Sep 7, 2010 #4
  6. Sep 7, 2010 #5
    Hmmm ok. So would the proper answer, after accounting for units, be 3E-15 C ?
  7. Sep 7, 2010 #6
    Yes, or 3 fC (femto-Coulomb)
  8. Sep 7, 2010 #7
    Ok great. If I have any other questions, I will ask, but I think that takes care of it! Thanks!
  9. Sep 7, 2010 #8
    Wait I am confused again. Even if I reverse the bounds, won't the answer remain unchanged. A negative sign will go out front, but won't it be countered by another negative sign when solving the integral?
  10. Sep 7, 2010 #9
    Reversing the bound and putting a minus is valid mathematical operation so it will not change your wrong result.

    If you define the first integral from 0 to -1 you will always get the wrong answer.

    If you look up the definition of the total charge, you will see that it is defined as the integral over the whole space, nothing in terms of boundaries, etc.
    So by definition the integral over x has to go from -1 to 0.
    This is the reason why the box had so strange coordinates.

    You can see it easier if you compute just the volume of the box. The Volume obviously has to be positiv, but with your definition it isn't.
  11. Sep 7, 2010 #10
    Ok I changed the integral to go from -1 to 0 for the x component. I'm still getting 15 though. I'm getting 6+27-18=15. Any ideas?
  12. Sep 7, 2010 #11
    Wait ok I think I got it. I'm getting -6+27-18=3. That's the right answer, but I hope I did it correctly
  13. Sep 7, 2010 #12
    Sorry. I was a bit too quick before. I just checked the 3, which is correct, but it should be -3.

    Post your calculations if I you want me to help find the error. Those numbers could be anything.
  14. Sep 7, 2010 #13
  15. Sep 7, 2010 #14
    wrong... too late today. Tomorrow again....
    Last edited: Sep 7, 2010
  16. Sep 7, 2010 #15
    Sorry I am confused. Do you mean you are going to help me tomorrow? Thanks.
  17. Sep 8, 2010 #16
    Good morning. My late night calculations were a bit wrong. Everything is fine with your +3. Sorry for the confusion.
  18. Sep 8, 2010 #17
    Ok great. So all my work was correct and my final answer of 3E-15 is correct as well?
  19. Sep 8, 2010 #18
  20. Sep 8, 2010 #19
    Ok thanks
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