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hosein Group TitleBest ResponseYou've already chosen the best response.0
can explain how this two sides are equal?\[\cos \theta da=r^2d \Omega\]
 2 years ago

hosein Group TitleBest ResponseYou've already chosen the best response.0
where da is surface element in spherical coordinate & omeg is solid angle
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.0
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

hosein Group TitleBest ResponseYou've already chosen the best response.0
it's in classical electrodynamics (by Jackson)page28
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
I think this is the same formula we are finding in figure 4.2.4 on this page http://ocw.mit.edu/courses/physics/802scphysicsiielectricityandmagnetismfall2010/conductorsandinsulatorsconductorsasshields/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

quarkine Group TitleBest ResponseYou've already chosen the best response.0
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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