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Time period of small oscillations of the point dipole

  1. Jul 19, 2017 #1
    1. The problem statement, all variables and given/known data
    In an infinite flat layer of thickness 2d, volume charge density is given according to the law: ρ=(ρ°)(x)/d and (-d≤x≤d). Here, x is the axis perpendicular to the plane. In the layer, there is a thin channel in which a point dipole of mass m and dipole moment p is placed. Calculate the time period of small oscillations of the dipole.

    2. Relevant equations
    Differential form of electric field:

    div E = 4πρ and
    div E = ∂Ex/∂x + ∂Ey/∂y + ∂Ez/∂z

    3. The attempt at a solution
    "Attempt to the solution has been made in CGS system"

    div E = 4πρ
    ∂Ex/∂x + ∂Ey/∂y + ∂Ez/∂z = 4πρ

    ∂Ey/∂y and ∂Ez/∂z will be zero.
    Therefore,
    ∂Ex/∂x = 4πρ =(4πρ°x)/d
    Ex(x) = (4πρ°)/d ∫x dx = [(4πρ°)/d]*[x2]/2 = (2πρ°x2)/d + C

    How will we find the constant C?
    What exactly will be the electric field at point d from the midpoint of the layer? Will it be 2πρ°(2d) = 4πρ°d?

    I am confused how to proceed from here.
     
    Last edited: Jul 19, 2017
  2. jcsd
  3. Jul 21, 2017 #2

    rude man

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  4. Jul 22, 2017 #3
    Yes, as I have already mentioned, I am not using SI system.
    Here, I have used CGS system
     
  5. Jul 22, 2017 #4

    rude man

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    OK, I might look at it from SI & will post if I'm successful, in which case maybe you can "translate" into cgs.

    Funny, my introductory Physics course was also in cgs but I never mess with it anymore since it's totally at variance with electrical engineering.
     
  6. Jul 22, 2017 #5
    Wow! CGS?
    That is very rare to hear. Which country did you graduate from?
     
  7. Jul 22, 2017 #6

    rude man

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    U.S. of A.
    Cambridge, MA
    1962
    So it was a while ago & maybe even Harvard has "reformed"! :smile:
     
  8. Jul 24, 2017 #7
    Oh, that was long back ago. My university in Russia still uses CGS system for Electromagnetism course.
     
  9. Jul 24, 2017 #8

    rude man

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    I'm not sure but I think pure physics here is still taught using cgs but applied physicists and certainly engineers all use the Système Internationale.
     
  10. Jul 24, 2017 #9

    rude man

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    How about a Gaussian right cylinder running from x=0 to x = ∞? What can you say about Ex at x = ∞?
    EDIT: since the sheets are infinite in the y and z directions you don't have to put the far end of the Gaussian cylinder at infinity. What is the E field for all |x| > d?
     
    Last edited: Jul 25, 2017
  11. Jul 24, 2017 #10

    haruspex

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    It is not clear to me what this means. Whereabouts in the layer is the dipole? Is it in the middle, at one extreme, or an unknown point between the two?

    Anyway, consider just a thin infinite sheet of the layer at distance x from the middle, thickness dx. What is the field due to that?
     
  12. Jul 25, 2017 #11

    rude man

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    I assume the dipole is located at x=0, is that right @sid0123?
    (By means of the Gaussian surface, any other parallel layer within -d<x<d can also be easily chosen.)
     
  13. Jul 25, 2017 #12
    Sorry for the late reply.
    I think the centre of the line joining the dipole is located at x=0 i.e. the middle of the plate.
    There was a small hint to the problem given and there, they have solved for C and they finally came up with an equation:
    E(x)=(2πρ°(x2-d2))/h and that means, they have taken C to be -h.

    How I am trying to view this problem is that electric field at x=0 should be 0.
    Let electric field at x=0 be E(0) and at x=x be E(x)

    Now, solving the integral ∂Ex/∂x =(4πρ°x)/d i.e. ∫dEx=(4πρ°/d) ∫x dx
    Taking limits of LHS from E(0) to E(d) and of RHS from x=0 to x=d, we get,
    E(x)-E(0)=((2πρ°)/d)[x2-02] and we get E(x)-E(0)=((2πρ°x2)/d) and here, E(0) would be 0
    So, the equation should finally become, E(x)=((2πρ°x2)/d)
    Is it right till here?
     
  14. Jul 25, 2017 #13

    haruspex

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    Certainly not.
    I do not see how you conclude that. It's not what I get if I substitute -h for C in your expression for the integral.
     
  15. Jul 25, 2017 #14
    Yes, we don't get -h. I skipped 2πρ°
    Yeah, so we get -2πρ°h
    But I didn't understand how?
     
  16. Jul 25, 2017 #15
    But why? The volume charge density is related according to the relation ρ=(ρ°)(x)/d
    At x=0, won't the volume charge density at the centre become zero?
    And if it would be zero, how would there be any electric field at that point?
     
  17. Jul 25, 2017 #16

    rude man

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    Because there is + charge above x=0 and - charge below x=0 so a test charge at x=0 will be repelled by the + charges and attracted by the - charges, giving a negative, non-zero E field at x=0.

    Why are you using h instead of d all of a sudden?
    The derived expression E(x)=(2πρ°(x2-d2))/h with a - sign in front of it and the condition -h <= x <= h looks right.
     
  18. Jul 25, 2017 #17
    My apologies for this. The hint that I was referring to had used h and I forgot to change it to d.

    But there is no negative sign in front of this expression and also, I failed to understand how did you reach even close to this expression. Can you please throw some more light on this?
     
  19. Jul 25, 2017 #18

    rude man

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    There is another problem with the statement of the question: in order to solve for frequency of oscillations it is insufficient to specify just the dipole moment. The separation distance of the dipole charges is also needed.
     
  20. Jul 25, 2017 #19
    And if there were no point charges, i.e. no point dipole inside the infinite plate, would the electric field at x=0 then be 0?
     
  21. Jul 25, 2017 #20
    No, the answer is independent of the distance between the charges of the dipole.
    The answer to this problem is √[(πmd)/(ρ°p)]
     
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