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Force on a point charge

  1. Dec 21, 2006 #1
    Griffiths problem 8.4
    1. The problem statement, all variables and given/known data

    Consider 2 eual point charges q, separated by a distance 2a. Construct the plance equidistnace form the two charges. By integrating Maxwell's stress tensor over the plane determine the force of one charge on the other.


    2. Relevant equations
    [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
    This does bear some similarities to the case for a uniformly charged solid hemisphere.. except in this case the bowl part is at z = infinity

    suppose we had a poin charge located at z= a and another at z=-a and the plane is the XY plane.
    the hemisphere shouldnt contribute to anything E=0 at z=infinity... right??

    i am just wondering... how does this stress tensor 'work' for this problem

    do we calculate the stress tensor on the plane (infinite plane) but how would that tell us anything about the electric field at a point beyond this plane??

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

    Meir Achuz

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    1. You can show that the integral of T over the hemisphere vanishes.
    The area of the hemisphere ~R^2, wjhile E^2~1/R^4.
    2. Just do the integral over the plane: [tex]\int{\bf\hat k\cdot T}[/tex].
     
  4. Jan 2, 2007 #3
    The total force of a collection of charges enclosed in a volume [tex] V [/tex] is given by

    [tex]\mathbf{F} = \frac{d\mathbf{P}}{dt} = \oint_V T_{ij} dS_j [/tex].

    This means that integrating over a closed volume containing one of the particles in your problem yields the force on that particle. The integral you are asked to calculate is the integral over the plane between the two particles, but you can allways enclose the integral by adding the integral over the infinite hemisphere (which is zero, as Meir Achuz pointed out above).

    I hope this is clear.
     
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