- #1

HPRF

- 32

- 0

P = ExH = (mu

_{0}q

^{2}a

^{2}sin

^{2}(theta)/6pi

^{2}cr

^{2})

__r__This apparently leads to

Total Power = (mu

_{0}q

^{2}a

^{2}/6pi

^{2}c)[tex]\int[/tex](sin

^{2}(theta)/r

^{2})(2pir

^{2}sin(theta)d[tex]\theta[/tex])

What I am unsure of is where the

(2pir

^{2}sin(theta)d[tex]\theta[/tex])

appears from. Can anyone help?