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I cant understand the meaning of this root

  1. Jun 15, 2009 #1
    http://i42.tinypic.com/2yju26p.gif

    i know that the root supposed to be a distance
    but i cant see the distance from where to where??

    the charges are not a single point

    i dont know how they calculate this distance

    ??
     
  2. jcsd
  3. Jun 15, 2009 #2

    tiny-tim

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    Hi transgalactic! :smile:

    It's the distance from the centre of the cube …

    5 + 1.25, 1.25, and 1.25

    (and later 1.252 + 1.252 = 3.125)
    Yes, you're right :smile:

    but I think they're hoping it will give a good enough approximation. :wink:
     
  4. Jun 15, 2009 #3
    so if its threw the center of each cube
    and in the red part
    they calculate from face A

    we need two distances only

    can you draw in the drawing what triangles they are doing for each face
    how they calculate the hipotenuse ??

    and why they treat every cube in the same way??

    the center point of each small cube has a different distance to point p
    ??
     
  5. Jun 15, 2009 #4

    tiny-tim

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    No, you find the "hypotenuse" (it's really the diagonal of a cube :wink:) from three distances: √(x2 + y2 + z2) …

    x = y = 1.25,

    z = 5 (I assume: it's not marked) + 1.25 = 6.25 :smile:
     
  6. Jun 15, 2009 #5
    this is the original question:
    http://i41.tinypic.com/2ytzr60.gif

    ok if we calculate the center point of cube A
    this is the axes
    http://i44.tinypic.com/4ibl8g.gif
    on the x axes its:
    3.75

    on y axes it:
    1.25

    the distan
    on z axes its:
    6.25

    p is distant by 5 cm and i dont have any where the [tex]5^2[/tex]
    i cant see the triangle with 90 angle(so i could calculate its hipotenuse )


    ??
     
  7. Jun 15, 2009 #6

    tiny-tim

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    No, that's B (or C) … A is 1.25 1.25 and 6.25
    that's 5cm to the centre of the surface of cube A …

    to the centre of the volume of cube A, you must add another 1.25, to make 6.25
    (btw, it's "hypotneuse", with a "y" :wink:)

    it's not really a triangle, more the diagonal of a box, with sides 1.25 1.25 and 6.25 :smile:
     
  8. Jun 15, 2009 #7
    i cant imagine it in the given drawing,
    can you draw it with axes

    ??
     
  9. Jun 15, 2009 #8
    where is the center point in this drawing?
     
  10. Jun 15, 2009 #9

    tiny-tim

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    the centre point is the centre of the cube marked A.
     
  11. Jun 15, 2009 #10
    where is our (0,0,0)
    point
    ?
     
  12. Jun 15, 2009 #11
    why the multiply the denominator by 10^-2
    ??
     
  13. Jun 15, 2009 #12
    its only one cube we have 64 of them

    how they sum 64 potentials without any integral?
     
  14. Jun 15, 2009 #13

    Mark44

    Staff: Mentor

    They sum the 64 potential charges by adding them up. Remember that integration is a glorified way of adding things.
     
  15. Jun 15, 2009 #14

    Mark44

    Staff: Mentor

    I believe what they are doing is changing from cm to m. 1 cm = 10^(-2) m.
     
  16. Jun 15, 2009 #15

    Mark44

    Staff: Mentor

    It seems that both you and Tiny-tim have been using the lowest point shown on the cube (near the dimension 1.25 cm) as the origin. You can have it wherever you want it - just be consistent in using the same point for calculating distances.
     
  17. Jun 16, 2009 #16
    even if i calculate the potential of one cube

    how they calculate the sum of 64 potentials without making 64 equations
    and without making any integral?
     
  18. Jun 16, 2009 #17

    HallsofIvy

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    That has already been answered. They are treating each small cube as a "point" and summing. That gives an approximation to the integral. Do you remember how "Riemann sums" approximate the integral when you are first learning integration?
     
  19. Jun 16, 2009 #18
    they say
    "this part contributes ..."
    "this part contributes ..." etc..

    but then they do in a sum "..."


    i dont know Riemann approximation

    i know to sum 64 potentials
    or doing an integral

    they didnt do neither
     
  20. Jun 16, 2009 #19

    tiny-tim

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    The original question says "Use a numerical method" …

    that means a numerical approximation,

    in which you use a finite sum instead of an integral, to get an approximate result. :wink:

    (and yes, you do need 64 of them, but there will be a lot of quadruplicates, which will shorten the task)
     
  21. Jun 16, 2009 #20
    i dont know this method

    i see that they calculate the contribution only of some parts and not all 64
    and then they jump to the final solution
    without any explanation
     
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