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Why my school method doesnt work

  1. Jun 17, 2009 #1
    the question is :
    http://i39.tinypic.com/ll89s.gif

    the solution is here:
    http://i39.tinypic.com/bej341.gif

    my question is about part b when calculating the potential between [tex]r_1[/tex] and [tex]r_2[/tex]:
    r1<r<r2
    we take the potential from the outer sphere (-150)
    which is constant inside of it,and we sum with the potential from the inner sphere
    which changes by r.

    so its -150+( 9*10^9 *10* 10^-9)/r

    they get a different expression in another way

    my question is,where is my mistake in my way of solving?
     
  2. jcsd
  3. Jun 17, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The mistake is that you haven't used "with V= 0 at r= [itex]\infty[/itex]".

    Any potential is "relative" to some base value. If you take V= 0 at [itex]\infty[/itex]
    you will should find that the constant potential inside [itex]r_2[/itex] is 0.
     
  4. Jun 17, 2009 #3
    i did take it that way.
    and the potential in r1<r<r2
    is not zero
    the answer in the book doesnt say that too


    ??
     
  5. Jun 17, 2009 #4
    i cant understand how i didnt use that the potential in infinity is 0
    what part of my solution says other wise?
     
  6. Jun 18, 2009 #5
    and its not legat lo put the integration variable in the interval

    ??
     
  7. Jun 18, 2009 #6
  8. Jun 18, 2009 #7

    Mark44

    Staff: Mentor

    Sure it's legal. It just isn't done very much. When you say that something isn't legal, you'll be taken more seriously if you cite a reason for your assertion.

    I think what you're having a problem with is this integral:
    [tex]\int_{r_1}^r \frac{89.9}{r^2}dr
    [/tex]
    To evaluate this definite integral, find the antiderivative of the integrand (which will be a function of r), and then evaluate this antiderivative at r (which involves exactly zero work), and then subtract its value at r1, which is 30 cm. What they're doing is finding the potential at a distance r, where r1 < r < r2.
     
  9. Jun 18, 2009 #8

    Mark44

    Staff: Mentor

    simplify!!
     
  10. Jun 18, 2009 #9
    -150+ 90/r

    and they have

    -450+ 90/r

    ??
     
  11. Jun 18, 2009 #10

    Mark44

    Staff: Mentor

    When you integrated, did you put in r1 (which is 30 cm)? I can see from the answer you posted how they got their answer. You didn't show how you got your result, so I have to guess at what you might have done wrong.
     
  12. Jun 18, 2009 #11
    but i am using a totaly different method
     
  13. Jun 19, 2009 #12

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    And we don't know what method you are using because you have refused to show what you have done! How can we possibly say what, if anything, you are doing wrong when you don't show what you have done?
     
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