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Posted:
6 years ago
30 avr. 2019, 18:12 UTC−4
It depends on what your problem looks like.
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ZHANG, Pu
School of Physics,
Huazhong University of Science and Technology
It depends on what your problem looks like.
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Posted:
6 years ago
30 avr. 2019, 19:17 UTC−4
It depends on what your problem looks like.
Thanks for your reply. I illuminate a nanosphere with a plane wave and I want to find the total power scattered in the far field. When I integrate ewfd.(normE^2)/(2Z0const) over the main sphere I get an extremely small value.
>It depends on what your problem looks like.
Thanks for your reply. I illuminate a nanosphere with a plane wave and I want to find the total power scattered in the far field. When I integrate ewfd.(normE^2)/(2Z0const) over the main sphere I get an extremely small value.
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Posted:
6 years ago
1 mai 2019, 12:25 UTC−4
You can simply integrate the Poynting vector over a closed surface.
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ZHANG, Pu
School of Physics,
Huazhong University of Science and Technology
You can simply integrate the Poynting vector over a closed surface.
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Posted:
6 years ago
1 mai 2019, 18:43 UTC−4
You can simply integrate the Poynting vector over a closed surface.
Thanks alot for your reply.
I'm interested in the total power scattered to the far field.
I made a typo error in the expression in my previous power.. I meant it is
ewfd.(normEfar^2)/(2Z0const)
At what distance should I make the surface integration?
The variable normEfar is calculated in comsol at 1[m] so I think it is constant with the distance..it only differs with the angle as shown in the attached figure.. so why my result differ when I integrate over closed surfaces with different radius?
>You can simply integrate the Poynting vector over a closed surface.
Thanks alot for your reply.
I'm interested in the total power scattered to the far field.
I made a typo error in the expression in my previous power.. I meant it is
ewfd.(normEfar^2)/(2Z0const)
At what distance should I make the surface integration?
The variable normEfar is calculated in comsol at 1[m] so I think it is constant with the distance..it only differs with the angle as shown in the attached figure.. so why my result differ when I integrate over closed surfaces with different radius?
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Posted:
6 years ago
2 mai 2019, 18:50 UTC−4
Energy conservation ensures one can calculate the power in the near field. Or maybe you are interested in more than the final result?
-------------------
ZHANG, Pu
School of Physics,
Huazhong University of Science and Technology
Energy conservation ensures one can calculate the power in the near field. Or maybe you are interested in more than the final result?