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Sketching the electric field of a bar electret

  1. Aug 8, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-8-8_9-12-21.png

    2. Relevant equations
    ##σ_b = \vec P ⋅\hat n

    3. The attempt at a solution
    upload_2017-8-8_9-20-45.png

    The polarized system could be transformed into two charged discs S1 and S2 of radius a kept at a distance L.
    The discs S1 and S2 have surface charge densities -P and P respectively.

    A) For L>>a,
    In this case, the system could be approximated as two equal and opposite charges kept at a distance L.
    So, the Electric field can be sketched as
    upload_2017-8-8_10-43-11.png

    B) For L << a,
    The system could be approximated as a capacitor.
    So, the electric field inside the two discs will be uniform.
    And the electric field outside the two discs will be in the direction from the positive plate to the negative plate. I can't say anything more in this case.
    upload_2017-8-8_10-55-39.png
    C) For L ≈ a,
    The electric field is perpendicular to the plates in the direction from positive to negative plate in the vicinity of the center of the plates.
    But, in the region far from the center, I can't say anything.
    Here, I need help.
     
  2. jcsd
  3. Aug 8, 2017 #2
    upload_2017-8-8_20-21-55.png

    ##\vec P## is along the axis of the cylinder.
    ##\vec E ## can be sketched as a curve line from positive plate to negative plate.
    How to sketch ## \vec D ## as there is no free charges?
     
  4. Aug 9, 2017 #3

    rude man

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    How to sketch nothing? :smile:

    Another view: put a very thin right circular Gaussian cylindrical surface coaxially with the electret. One end goes to infinity, the other sticks inside the electret. What is D integrated over the whole surface equal to? So then what must D be for the end that sticks in the electret?
     
  5. Aug 9, 2017 #4
    The system has "frozen - in " polarization. This means : initially the cylinder was placed in the presence of external electric field, it got polarized with polarization up to ## \vec P ## and then the external field was removed. Now , this cylinder with Polarization ##\vec P ## is our system.

    So, I have to calculate ## \vec D ## due to this polarization.
    Considering a very thin right circular Gaussian cylindrical surface coaxially with the electret. One end goes to infinity, the other with the surface named ##S_{in }##sticks inside the electret.

    Since, the cylilnder is very thin , I can take ##\vec D ## constant over ## S_{in} ##.
    ## \vec D ## on the other end in the infinity is 0.
    ## \vec D⋅ d \vec a ## on the curved surface is 0 as ## \vec D ## is perpendicular to this surface.

    Hence, applying Gauss's law , we get,
    ## \{\int _{S_{in}} \vec D⋅ d \vec a = ± DS_{in}\} = \{Q_f =0\}
    \\ D = 0##
    Hence , ##\vec D ## in the vicinity of the axis is 0.

    If I assume that ##\vec D ## is 0 in the whole system cylindrical region, then it means that there are free charges at infinite distances from the cylinder and lines of ## \vec D ## originate from the free positive charge (kept at infinite distance from the cylinder ) and terminate at negative charge ( kept at infinite distance from the cylinder). Some of these lines may pass through the cylinder if these opposite charges are kept on the opposite sides of the axis.
    Is this correct?
     
    Last edited: Aug 9, 2017
  6. Aug 10, 2017 #5

    rude man

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    Yes, but it's the free charge INSIDE the gaussian surface that counts for D! You can have any number of free charges outside the surface anywhere; these lines of D however when integrated over the gaussian surface will always = 0 & it does not change the maxwell equation ∇⋅D = ρfree so ∫∫ D⋅dA = ∫∫∫ D dV = qfree.
     
  7. Aug 10, 2017 #6
    The free charge inside the Guassian Surface is 0. So, ##\vec D ## is 0. Hence, I don't have to draw anything for ## \vec D ##. Right ?
     
  8. Aug 10, 2017 #7

    rude man

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    Right. Had they placed free charge outside the surface you would have had to draw D lines between free charges entering and leaving the surface but since they didn't you can't draw anything.
     
  9. Aug 10, 2017 #8

    TSny

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    For the long, thin cylinder D would be fairly strong inside the cylinder. There is no free charge in the system, which implies ##\nabla \cdot \vec D = 0## everywhere. But that doesn't mean D = 0 everwhere.

    Keep in mind that ##\vec D = \varepsilon_0 \vec E + \vec P##.
    ##\vec E## will be very small inside the cylinder except near the ends.
     
  10. Aug 10, 2017 #9

    rude man

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    T, how come? D integrated over the entire gaussian surface = 0 (no free charge inside): D = 0 at infinity, D integrates to zero along the sides since any field will be in the axial direction so D dot A = 0 everywhere along the sides; so how can D integrated over the inside end face be nonzero?
     
  11. Aug 10, 2017 #10

    TSny

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    I don't think D is in the axial direction for all points on the gaussian surface.

    As an analogy, replace the electret by a solenoid of the same shape. The divergence of B is zero everywhere. So, the flux of B through any closed surface must be zero. In particular, if you choose a gaussian surface as you did for the electret cylinder, then the total flux of B through this surface is zero. But, we know that B is not zero inside the solenoid.

    The relation ##\nabla \cdot \vec D = \rho_{\rm free}## does not mean that ##\rho_{\rm free}## alone determines ##\vec D##. A vector field is determined by both its divergence and its curl (and maybe boundary conditions). From ##\vec D = \varepsilon_0 \vec E + \vec P## and the fact that the curl of ##\vec E## is zero for an electrostatic field, you can see that ##\nabla \times \vec D = \nabla \times \vec P## for the electret. So, ##\vec P## also acts as a "source" for ##\vec D##.
     
  12. Aug 10, 2017 #11

    rude man

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    I have to agree. Pushoam, I'm sorry for the confusion. TSny is very dependable; I, not always. :frown:

    EDIT: I'm wondering myself which direction the D field lines point in the electret.

    I put a dielectric slab between two parallel plates charged to +/-Q. The D, E and P vectors all point from + plate to - plate. So inside the dielectric the P vector points locally from - to + dielectric surface charge. I then remove the slab & the polarization stays.

    Does the P vector then still point from - to + in the slab? And if so, and since the E vector still points from + to -, then if D = P that means the E and D vectors point in opposite directions?
     
    Last edited: Aug 10, 2017
  13. Aug 11, 2017 #12
    In the vicinity of the axis, the electric field and polarization are along the axis. Consequently, ##\vec D ## will be along the axis. And the Gaussian infinite cylinder encloses only this region.
    So, using the argument given in post #4, ##\vec D## is 0.
    How do you get to know this?
    The cylinder has the opposite charge densities at the two ends respectively. So, how can one know it before solving it?

    BTW, if I assume what you said, then multiplied by ε_0, ## \vec E ## becomes very very small compared to ## \vec P ##. Consequently, ##\ hat D≈ \hat z##.





    If you choose a gaussian surface as you did for the electret cylinder, then the total flux of B through this surface is zero.

    Yes, I choose so as I think there is nothing wrong in choosing this Gaussian Surface. So, the flux through the end surface inside the solenoid is 0.
    As the magnetic field is along the axis and constant, I can take it outside the integration, which gives zero magnetic field.
    So, something is wrong with this argument. This I understood.
    But what is wrong?

    Yes, so,I have:
    ##\nabla \cdot \vec D = \rho_{\rm free} =0 ##
    ##\nabla \times \vec D = \nabla \times \vec P =0##
    Boundary Conditions: At infinity, ## \vec D ## is 0.
    Now, ## \vec D = 0 ## satisfies all three.
    So, using uniqueness theorem, ##\vec D = 0##.

    I have listed above all the arguments which make me think ##\vec D = 0##. Are these correct?
     
  14. Aug 11, 2017 #13

    rude man

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    @T, I'm interested in the hint given in post 2, to wit, "D lines terminate on free charges".. With no free charges, is the hint simply wrong?
     
  15. Aug 11, 2017 #14

    TSny

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    For the very long, thin dielectric cylinder the E field is essentially a dipole field where the opposite charges are at the ends of the cylinder. So, unless you are near one of the ends, the E field will be weak. In the central region of the cylinder, you are far from both ends.


    Yes
    Yes. And ##D \approx P## inside the long cylinder (except near the ends).


    upload_2017-8-11_11-57-42.png


    No, it is not true that you can assume ##\nabla \times \vec P =0## everywhere. If that were true, then the line integral of ##\vec P## around any closed loop would have to be zero. Can you construct a loop for the long cylinder where it is clear that the line integral is not zero?
     
  16. Aug 11, 2017 #15

    TSny

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    No, it is right. So, it means that the D lines will not start or stop anywhere.
     
  17. Aug 11, 2017 #16

    rude man

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    If they "terminate on free charges" they don't start or stop anywhere?? Sounds self-contradictory. I guess I'm mystified. Life's like that sometimes I guess.
     
  18. Aug 11, 2017 #17

    TSny

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    :oldsmile:
    Sorry, I should have been clearer. D field lines can only start or stop on free charge. In the problem at hand, there is no free charge. So, when I stated that "the D lines will not start or stop anywhere", I meant for the specific problem at hand. The D field lines either form closed loops or go to infinity for the polarized cylinder. Think of the B field of a solenoid for an analogy.
     
  19. Aug 11, 2017 #18

    rude man

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    OK, after some review of my Resnick & Halliday, I conclude that P = σbound is correct as Pushoam stated in post 1.

    I also agree D = P inside the electret. I suppose that the D lines continue outside the electret, forming closed paths just like a B field.

    Not sure about the E field. Modeling the electret as a huge dipole, q = πr2σbound, d = L, it would seem that the E field follows more or less as Pushoam depicted it in his post 1.

    I also still don't know about the lines of P. My belief is they are confined to inside the electret and point from the - end to the + end, just like a dipole moment does.

    Further comments welcome. This is obviously not an area of expertise for me.
     
  20. Aug 11, 2017 #19

    TSny

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    Yes
    Yes, D = P to a good approximation inside, except near the ends. This is because E is "small" except near the ends and D = εoE + P.
    Yes.
    Yes.
     
  21. Aug 12, 2017 #20
    Approximating the system as a dipole with dipole moment ## \vec p= qd ##

    The electric field due to dipole is given by
    ## \vec E = \frac k {r^3 } [ \vec p⋅ \hat r \hat r - \vec p ]##
    Here ## \vec r = r \hat p ##
    So, ## \vec E = \frac k {r^3 } 2 \vec p##
    r in the central region i. e. ## r_c ##< r at ends i.e. ## r_e##
    Thus, ## \vec E ## in the central region is greater than that at the ends.
    Sorry, I couldn't understand the answer given in the picture.
    While applying Gauss's law ,
    I took the flux through the curved surface to be 0. This is the mistake. The flux through the curved surface is not zero.
    If the solenoid is infinitely long and I take a small cylindrical gaussian surface inside the solenoid, then the flux through the curved surface is 0. This gives us the information that the magnetic field inside the solenoid is constant.
    Right?

    Inside the cylinder, ## \vec P ## is uniform. So, ## \nabla \times \vec P = 0 ## . Outside the cylinder, there is vacuum. So, ## \vec P = 0##.
    Hence, ## \vec P ## is 0 everywhere. Isn't it?
     
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