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Force on a uniformly charged solid sphere

  1. Dec 20, 2006 #1
    1. The problem statement, all variables and given/known data
    Griffiths Example 8.2
    Determine the net force on the northern hemisphere of a uniformly charged solid sphere of radius R and charge Q. use maxwell's stress tensor to calculate the force. (Not Gauss/Coulomb's law)


    2. Relevant equations
    The stress tensor is defined by
    [tex] T_{ij} = \epsilon_{0} (E_{i}E_{j} - \frac{1}{2} \delta_{ij} E^2) + \frac{1}{\mu_{0}} (B_{i}B_{j} - \frac{1}{2} \delta_{IJ} B^2) [/tex]
    where i and j are coordinates
    3. The attempt at a solution
    ok For the bowl part
    [tex] da = R^2 \sin \theta d\theta d\phi \hat{r} [/tex]

    [tex] \vec{E} = \frac{1}{4 \pi \epsilon_{0}} \frac{Q}{R^2} \hat{r} [/tex]

    so
    [tex] T_{zx} = \epsilon_{0} \left(\frac{Q}{4 \pi\epsilon_{0} R^2}\right)^2 \sin \theta \cos\theta \cos\phi [/tex]

    [tex] T_{zy} = \epsilon_{0} \left(\frac{Q}{4 \pi\epsilon_{0} R^2}\right)^2 \sin \theta \cos\theta \sin\phi [/tex]

    [tex] T_{zz} = \frac{\epsilon_{0}}{2} \left(\frac{Q}{4 \pi\epsilon_{0} R^2}\right)^2 (\cos^2 \theta - \sin^2 \theta)[/tex]

    The net froce is in the Z direction (taking the Z axis to point stragiht up, teh flat part of this bowl lying on the XY plane...

    [tex] (T da)_{z} = T_{zx}da_{x}+T_{zy}da_{y}+T_{zz}da_{z} = \frac{\epsilon_{0}}{2} \left(\frac{Q}{4 \pi\epsilon_{0} R^2}\right)^2 \sin\theta \cos\theta d\theta d\phi[/tex]

    I dont understand what happened in that last step...
    ... i know this question seems silly.... i should know all of this thoroughly in and out at this point...

    anyway why does it turn into that??

    what are dax and day and daz??
    did they convert into spherical coords and then 'expand' ??

    thanks for your help
     
  2. jcsd
  3. Dec 21, 2006 #2

    dextercioby

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    Science Advisor
    Homework Helper

    The projections of the [itex] d\vec{a} [/itex] vector onto the 3 cartesian axis.

    He made all multiplications and additions and ended up with that result.

    Daniel.
     
  4. Dec 21, 2006 #3
    i didnt see it till i actually looks at da and then i saw it

    thanks for your help
     
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