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Maxwell stress tensor to calculate force (EM)

  1. May 5, 2015 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    A sphere with dielectric constant ##\varepsilon## and radius R is placed inside a homogenous external electric field ##\vec E_0##. The sphere is divided in 2 hemispheres such that their common interface is orthogonal to the external field. Using the energy-momentum tensor, calculate the attraction force between the 2 hemispheres.

    2. Relevant equations
    I think they meant "Maxwell stress tensor" instead of "energy-momentum tensor" since the latter seems to be related to special relativity and has nothing to do with this part of the course which is based on the book of Zangwill.
    In page 81 in Zangwill's book, there's a formula to calculate the force via Maxwell stress tensor which eventually reduces to ##\vec F = \varepsilon_0 \int_S dS[(\hat n \cdot \vec E ) \vec E -\frac{1}{2}(\vec E \cdot \vec E)\hat n]##. The thing is that I'm not really understanding this formula, especially the part about "S" the surface of integration. There's an example of a charged balloon in the book but it's not clear to me either.
    3. The attempt at a solution
    I tried to brainlessly apply the formula in hope of falling over the solution so that I could hope for a reverse engineering in order to understand it, but I failed.
    Here's my attempt:
    The electric field inside the sphere is ##\vec E_{\text{int}}=\frac{3}{2+\varepsilon}\vec E_0##.
    I choose spherical coordinates ##(r, \theta , \phi)##.
    So ##\hat n=r\cos \phi \sin \theta \hat i + r \sin \theta \sin \phi \hat j + r \cos \theta \hat k##.
    Also, ##\vec E _{\text{int}}=\frac{3}{2+\varepsilon}|\vec E_0|\hat k##.
    I picked the surface of integration as a hemisphere, so ##r=R##, ##\theta:[0, \pi/2]## and ##\phi :[0,2 \pi]##.
    So I plugged and chugged ##\vec E_{\text{int}}## and ##\hat n## inside of the formula given above. As soon as I obtained a non vanishing result in the ##\hat i## direction, I stopped my calculations, I know something is wrong but I can't avoid this result unless I change the surface of integration but I've no idea which one to pick... So I'm stuck here.
    I expect the force to be in the ##\hat k## direction only due to the symmetry of the problem, of course.
    Any help is appreciated, thanks!
     
  2. jcsd
  3. May 7, 2015 #2
    I don't know how you are getting a non vanishing result in the i-hat direction. All but the k-hat direction should vanish after integrating with respect to phi. What is the vector expression you got for your integrand?
     
  4. May 7, 2015 #3

    fluidistic

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    Yeah that's what I was expecting.
    [tex]\vec F = \varepsilon_0 \int_S dS [\left ( \frac{3R\cos(\theta)|\vec E_0|}{2+\varepsilon} \cdot \frac{3|\vec E_0|\hat k}{2+\varepsilon} \right ) -\frac{1}{2}\left ( \frac{3}{2+\varepsilon} \right ) ^2 |\vec E_0|^2R(\cos \phi \sin \theta \hat i + \sin \theta \sin \phi \hat j + \cos \theta \hat k) ][/tex]
    Now focusing only on the term with ##\hat i## (due to length of the expression), I get that it's worth a constant times ##\int_0^{2\pi}\int_0^{\pi/2} \cos \phi \sin ^2 (\theta) d\theta d\phi##.
    Oops my bad you're right, it's worth 0... same for the term with ##\hat j##... Thanks a lot!
     
  5. May 7, 2015 #4

    fluidistic

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    Ok I reach ##\vec F=\left ( \frac{3}{2+\varepsilon} \right )^2\varepsilon_0 |\vec E_0|^2 \pi \frac{a^3}{2}\hat k##.
    I integrated over a single hemisphere. I am guessing that I should integrate over the 2 hemispheres separately and then add the result in order to get the total force. Overall I guess that the total force is actually twice the one I reached (assuming the math is ok).
    Therefore my answer would be ##\vec F=\left ( \frac{3}{2+\varepsilon} \right )^2\varepsilon_0 |\vec E_0|^2 \pi a^3 \hat k##.
     
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