Power radiated by point charge Calculus

1. Nov 27, 2009

HPRF

It is found that Poyntings vector gives

P = ExH = (mu0q2a2sin2(theta)/6pi2cr2)r

Total Power = (mu0q2a2/6pi2c)$$\int$$(sin2(theta)/r2)(2pir2sin(theta)d$$\theta$$)

What I am unsure of is where the

(2pir2sin(theta)d$$\theta$$)

appears from. Can anyone help?

2. Dec 7, 2009

caleb20k

actually the true equation is that S=1/2Re{ExH*} thats the real part of ExH* and H* is the conjugate of H.

this leads to Pwr = Int[S.dA]. this is the eqn for power passing thru a surface of area dA (assumed to be spherical). in it we integrated over the surface area of the sphere because the poynting vector is defined as the power per area. So if we multiply S by the area we obtain the power

where dA is r-squared times the solid angle == r^2*d(Omega). Expressed another way, dA=r^2*sin(theta)d(theta)d(phi) the r^2 sin(theta) comes out of the jacobian because you are changing from cartesian coordinates to spherical coordinates.

so that makes the Power=Int[1/2*Re{ExH*} * (r^2*sin(theta)d(theta)d(phi)),

here theta is integrated over 0 to Pi, Phi is integrated from 0 to 2*Pi

upon integrating you get the 2*Pi from the Phi integral.
which leaves you with the d(theta)*r^2*sin(theta)

hope that the response isnt too late.

3. Dec 14, 2009

caleb20k

in all truth you dont need that 1/2 on the Poynting vector S. Just a bad habit of mine from dealing with equations using the Poynting vector in E&M. So omit that 1/2 from what I stated above

4. Dec 15, 2009

caleb20k

sorry, had a brain fart when I posted that. I meant keep the 1/2 and dismiss the Real part. that is only when calculating the power/solid angle.

S = 1/2 (ExH*)