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Method of images in Griffiths' text

  1. Jan 1, 2009 #1

    KFC

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    In Griffiths' text "Introduction to electrodynamics", p.125, how does the author get (3.15) and (3.16)? I try for an hour to deduce the relation with geometry, but still can't get that.
     
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  3. Jan 1, 2009 #2

    olgranpappy

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    in my version of Griffiths', (3.15) and (3.16) are definitions (of q' and b)... so, could you write down the question without reference to Griffiths'?
     
  4. Jan 1, 2009 #3

    KFC

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    Thanks for reply.

    (3.15) is [tex]q' = -\frac{R}{a}q[/tex]

    (3.16) is [tex]b = \frac{R^2}{a}[/tex]

    I don't know why this is definition, I think we could deduce those expression with geometry; otherwise, what's the physical significance of defining q' like that?
     
  5. Jan 1, 2009 #4

    olgranpappy

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    Right. So you want to know how to come up with those values. This is discussed by Griffiths in the chapter. To show that these values indeed do give a zero potential on the sphere is shown in problem 3.7, I think. Give it a try. Cheers.
     
  6. Jan 2, 2009 #5

    KFC

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    Thanks for reply. I will do it myself again.

    Now, I just found another problem in using method of images in Grifiths' text. In page 124, there the author trying to calculate the work done by moving the charge from infinity to d and get the result

    [tex]W=\int_\infty^d\vec{F}\cdot d l = \frac{1}{4\pi\epsilon_0}\int_\infty^d\frac{q^2}{4z^2}dz = \frac{1}{4\pi\epsilon_0}\left.\left(-\frac{q^2}{4z}\right)\right|_\infty^d = -\frac{1}{4\pi\epsilon_0}\frac{q^2}{4d}[/tex]

    But according to the preceding calculation, the force on the charge due to the image charge is (eq. 3.12)

    [tex]\vec{F} = -\frac{1}{4\pi\epsilon_0}\frac{q^2}{(2d)^2}\hat{z}[/tex]

    The force is negative so it seems that in the work calculate we should plug the minus sign there, the result should be positive??
     
  7. Jan 3, 2009 #6

    olgranpappy

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    Yes, since the force and the displacement are in the same direction the work should be positive. The 2nd equality after 'W' is wrong. The sign should be
    [tex]
    \frac{-1}{4\pi\epsilon_0}\int_{\infty}^{d}\frac{q^2}{4z^2}dz\;.
    [/tex]

    Maybe you got the sign wrong by double counting. E.g., [itex]d\vec\ell[/itex] should be [itex]\hat z dz[/itex] not [itex]-\hat z dz[/itex] because the direction is already accounted for in the limits of integration.

    But the important point is that you recognized that the answer should be positive, which is right. So, however many minus sign errors you make you can always put the sign correct at the end.
     
  8. Jan 3, 2009 #7

    KFC

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    Thanks. But the following expression is given by Griffiths' text not me, that's why I am wondering the sign

    [tex]W=\int_\infty^d\vec{F}\cdot d l = \frac{1}{4\pi\epsilon_0}\int_\infty^d\frac{q^2}{4z ^2}dz = \frac{1}{4\pi\epsilon_0}\left.\left(-\frac{q^2}{4z}\right)\right|_\infty^d = -\frac{1}{4\pi\epsilon_0}\frac{q^2}{4d}[/tex]
     
  9. Jan 3, 2009 #8

    olgranpappy

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    So, either he is considering two charges of the same sign in which case the expression you gave for the force is wrong, or he is wrong.
     
  10. Jan 3, 2009 #9

    olgranpappy

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    Or he is considering the work done by "you" in bringing the charge into position rather than the work done by the field.
     
  11. Jan 3, 2009 #10

    KFC

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    Oh. I see. How careless I am LOL
     
  12. Jan 3, 2009 #11

    gabbagabbahey

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    That seems like a more reasonable conclusion than "he is wrong" :smile:
     
  13. Jan 4, 2009 #12

    olgranpappy

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    not necessarily... Griffiths may write sacred texts, but he is far from infallible.
     
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