Question of a problem on Poynting vector

[yungman]

I have been working on this problem and I cannot get the solution as shown in the book. I scan the question and solution of the book and my work. It is too difficult for me to type it in, please open the attach scanned pdf file. THe top part is the copy of the book, the lower part is my work. You can see the solution on my part does not agree with the book and I check over my work many times already. Please tell me what did I do wrong
Thanks

 

[gabbagabbahey]

We can't see your attachment until it is approved by admin. It would be quicker if you uploaded your pdf to imageshack.us and then posted a link to it.

 

[yungman]

We can't see your attachment until it is approved by admin. It would be quicker if you uploaded your pdf to imageshack.us and then posted a link to it.

It can be opened now.
Thanks

 

[gabbagabbahey]

It can be opened now.
Thanks

Not by me it can't. It still says that it is pending approval.

 

[yungman]

Not by me it can't. It still says that it is pending approval.

I just opened already! I wonder why!

 

[gabbagabbahey]

You can open it before it has been approved since you are the one that attached it. Everyone else has to wait 'til its approved.

Again, you can save a lot of time by uploading it to a free imagehosting site and then just posting a link to it.

 

[Defennder]

I don't think your expression for the real part of \mathbf{E}(R,\theta,t) and \mathbf{H}(R,\theta,t) are correct. Remember that the original phasor expression for E and H has 'j' inside, meaning that you have to multiply throughout by j, so that makes the formerly imaginary part real and the formerly real part imaginary.

 

[yungman]

I don't think your expression for the real part of \mathbf{E}(R,\theta,t) and \mathbf{H}(R,\theta,t) are correct. Remember that the original phasor expression for E and H has 'j' inside, meaning that you have to multiply throughout by j, so that makes the formerly imaginary part real and the formerly real part imaginary.

Thanks again. It is so obvious that I just miss it all together. "j" translate to e to the power j(pi/2) which change the cos() to -sin().

Thanks again.