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Mathematical method in electrical potential ?

  1. Jul 14, 2013 #1
    Mathematical method in electrical potential ??

    Hi All,
    I need mathematical help from the topic electrical potential for lectures on physics by Richard Feynman.

    Need some help to understand mathematical method used here.
    question :
    1) From electrical potential.png
    ( how this partial differential of ( -p/4∏ε (z/r^3)) with respect to z
    ⇔ -p/4∏ε (1/r^3 - 3z^2 / r^5) ?
    similarly how to do partial differentiation with respect to x

    2) From perpendicular_field.png
    ( how this E = p/4∏ε * 3z/r^5 √(x^2 + y^2) ⇔ p/4∏ε * (3 cosθsinθ/r^3)



    Please look for attachments
     

    Attached Files:

  2. jcsd
  3. Jul 15, 2013 #2

    tiny-tim

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    hi igraviton! :smile:

    (try using the X2 button just above the Reply box :wink:)
    you use both the product rule (or the quotient rule) and the chain rule

    for the chain rule, use ∂/∂z (f(r)) = ∂f(r)/dr ∂r/∂z

    (and ∂r/∂z = … ? :smile:)
     
  4. Jul 22, 2013 #3
    Thank you tiny-tim,
    after doing partial differentiation and chain rule i got following.
    d (z/r^3)/dz = 1/[r][/3] - 3z/[x^4] * dr/dz
    but actual answer is 1/[r^3] - 3z^2/[x^5]
     
  5. Jul 22, 2013 #4

    SteamKing

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    And what is the definition of 'r' in terms of x, y, and z?
     
  6. Jul 22, 2013 #5

    tiny-tim

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    (you mean ∂r/∂z)

    ok … and now what is ∂r/∂z ? :smile:
     
  7. Jul 22, 2013 #6
    yes Tiny-tim,

    actual answer is

    [itex]\frac{\partial (z/r^3)}{\partial z } = \frac{1}{r^3} - \frac{3z^2}{r^5} [/itex]


    tiny-tim, as you told me i did partial differentiation using quotient rule and chain rule

    and i got

    [itex]\frac{\partial (z/r^3)}{\partial z } = 1/(r^3) - (3z)/r^4 \frac{\partial r}{\partial z} [/itex]


    where
    [itex]
    r = √(x^2 + y^2 + z^2)
    [/itex]
     
    Last edited: Jul 22, 2013
  8. Jul 22, 2013 #7

    tiny-tim

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    yes, but what is ∂r/∂z ?

    (you can't just leave it there like that! :rolleyes:)
     
  9. Jul 22, 2013 #8
    I don't know !!! :confused:
    what is [itex]\frac{\partial r}{\partial z}[/itex]

    I did following :

    [itex] \frac {\partial \frac {z}{r^3}}{\partial z} = \frac {1 * r^3 - z * \frac {\partial r^3}{\partial z}}{(r^3)^2}[/itex]

    = [itex] \frac {1}{r^3} - \frac {z \frac {(\partial r^3)}{\partial z}}{r^6} [/itex]

    where
    [itex] \frac {\partial r^3 }{\partial z} = (\frac {\partial r^3 }{\partial z}) (\frac {\partial r}{\partial z}) [/itex]

    => [itex] 3 * r^2 * \frac {\partial r}{\partial z} [/itex]

    putting back I got

    => [itex] \frac {1}{r^3} - \frac {3z}{r^4} * \frac {\partial r}{\partial z} [/itex]
     
  10. Jul 22, 2013 #9
    Let me know if i am doing wrong.. thanks in advance
     
  11. Jul 22, 2013 #10

    tiny-tim

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    ?? :confused:
    … so what is ∂r/∂z ? :smile:
     
  12. Jul 22, 2013 #11

    SteamKing

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    Think of r as f(x,y,z) = (x^2+y^2+z^2)^(1/2)

    What is ∂f/∂z?
     
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