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hosein Group Title

Hi A question?

  • 2 years ago
  • 2 years ago

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  1. hosein Group Title
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    can explain how this two sides are equal?\[\cos \theta da=r^2d \Omega\]

    • 2 years ago
  2. hosein Group Title
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    where da is surface element in spherical coordinate & omeg is solid angle

    • 2 years ago
  3. JamesJ Group Title
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    It looks wrong. If you integrate over the entire sphere, the right hand side is \[ 4 \pi r^2 \] but the left hand side is \[ 8 \pi r^2 \]

    • 2 years ago
  4. hosein Group Title
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    it's in classical electrodynamics (by Jackson)page28

    • 2 years ago
  5. TuringTest Group Title
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    I think this is the same formula we are finding in figure 4.2.4 on this page http://ocw.mit.edu/courses/physics/8-02sc-physics-ii-electricity-and-magnetism-fall-2010/conductors-and-insulators-conductors-as-shields/MIT8_02SC_notes9.pdf though the area is not represented as a differential here. I trust that James' math is right though, so that must have some change in the problem.

    • 2 years ago
  6. quarkine Group Title
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    r square* d(sigma) = r*d(theta) *r*d(phi) =projection of area element=dA cos(theta) ....in azimuthal coordinates..|dw:1334694350771:dw|

    • 2 years ago
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