# Solving Poyntings Vector for a point charge

1. Nov 27, 2009

### HPRF

1. The problem statement, all variables and given/known data

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. Nov 27, 2009

### jdwood983

[strike]The power is over the whole of the volume, and not just one component of the volume.[/strike] When you are coming from Cartesian coordinates to http://en.wikipedia.org/wiki/Spherical_coordinate_system#Cartesian_coordinates", we find

$$\int dV=\int_{-\infty}^\infty dx\int_{-\infty}^\infty dy\int_{-\infty}^\infty dz=\int_0^\infty r^2dr\int_0^\pi \sin\theta \,d\theta\int_0^{2\pi} d\phi$$

Last edited by a moderator: Apr 24, 2017
3. Nov 27, 2009

### gabbagabbahey

Errr... not quite...Poynting's theorem tells us that the power crossing a surface $\mathcal{S}$ is given by

$$P(\textbf{r})=\int_{\mathcal{S}}\textbf{S}\cdot d\textbf{a}$$

(using $\textbf{S}$ to represent the Poynting vector and $d\textbf{a}$ to represent the infinitesimal vector area element of the surface)

To find the power radiated, one usually uses a spherical surface of radius $r$ (so that $d\textbf{a}=r^2\sin\theta d\theta d\phi\mathbf{\hat{r}}[/tex])) and then takes the limit as [itex]r\to\infty$ (since radiation escapes to infinity).