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Triple integral in eletromagnetism: prove force doesnt depend radius.

  1. Mar 30, 2013 #1
    1. The problem statement, all variables and given/known data
    This problem may be dull, I know, but maybe there is a hidden math trick that i don't know of. This picture sums up the problem.


    2013-03-31 03.36.07.jpg
    So, you should prove by simplifing the integral that [itex]F^e[/itex], the eletric force applied between two spheres, onde with a charge [itex]q_1[/itex] and the other with the charge [itex]q_2[/itex] (distributed evenly in volume, with a charge density [itex]\rho[/itex]) doesn't depend on [itex]R_2[/itex], it only depends on [itex]q_1, q_2[/itex] and [itex]d[/itex]


    2. Relevant equations
    [itex]F^{e}=\int_V \frac{q_1\rho }{4\pi \delta^{2}} cos(\phi ) dV [/itex]


    3. The attempt at a solution
    [itex]F^{e}=\int_V \frac{q_1\rho }{4\pi \delta^{2}} cos(\phi ) dV =\frac{q_1q_2 }{4\pi\frac{4}{3}\pi R_2^{3}}\iint_{0}^{R_2}\frac{cos(\phi )sin(\phi )r^2}{\delta^{2}}drd\phi=\frac{q_1q_2 }{4\pi\frac{4}{3}\pi R_2^{3}}\iint_{0}^{R_2}\frac{cos(\phi )sin(\phi )r^2}{(d-rcos(\phi )^2+(rsin(\phi )^2))} drd\phi=\frac{q_1q_2 }{4\pi\frac{4}{3}\pi R_2^{3}}\iint_{0}^{R_2}\frac{cos(\phi )sin(\phi )r^2}{(d^2+r^2-2drcos(\phi ))} drd\phi[/itex]

    How do you simplify this integral or at least show that the expression doesn't depend on [itex]R_2[/itex]? I tried to derivate with respect to [itex]R_2[/itex] but it didn't helped...
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 31, 2013 #2

    mfb

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    If I remember correctly, it is easier to integrate over a spherical shell only. If this is equivalent to a point-charge, the remaining part is easy.

    Do you have to use integration? Gauß' law would be so nice...
     
  4. Mar 31, 2013 #3
    You are saying that first we could show that a sphere with some charge density is equal to a spherical surface with a surface density choosen to have the same total charge and then simplify the integral. Yeah that could work but still get a pretty dull integral. I just wanted to know if there was some hidden math trick but seems like Guass's law is the way to go.
     
  5. Mar 31, 2013 #4

    mfb

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    No, the idea is to get rid of the r-integration: Show that a spherical shell has the same potential (at some specific point) as a point-charge in the center of the shell. The filled sphere just consists of "many" spherical shells.
     
  6. Mar 31, 2013 #5
    yeah but i was asking for the eletric force not the potencial...
     
  7. Mar 31, 2013 #6
    I got the many shperical shells=sphere with some volume density though
     
  8. Apr 1, 2013 #7

    mfb

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    If the potential is the same, the electric field and the force are the same as well.

    Oh, I think I see the problem: The force is a vector (with variable orientation), you have to add vectors and not their magnitude. That gets messy - calculate the potential instead, this should be easier.
     
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