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I have problem understanding why the surface integral disappear when taking the volume to very large in this equation:

[tex]W_e = \frac 1 2 \int_{v'} \rho_v V dv' = \frac 1 2 \int_{v'} (\nabla \cdot \vec D)V dv' = \frac 1 2 \int_{s'} V \vec D

\cdot \hat n ds' \;+\; \frac 1 2 \int_{v'} \vec E \cdot \vec D dv' [/tex]

I understand as long as [itex]v'[/itex] enclose all the charge distribution, the volume integral will be constant even when you extend the [itex]v'[/itex] to infinite size.

But I thought the surface integral behave the same. The definition of surface integral of charges is that the total fields coming out of the complete surface equal to the charges enclosed inside also. This mean if the closed surface contain all the charge distribution, the surface integral will be a constant even when further increase the size.

The book keep talking about potential decrease proportion to 1/r and E proportion to 1/(r^2) and the surface integral goes to zero as r goes to infinity.

The book then conclude that when volume goes to infinity:

[tex]W_e = \frac 1 2 \int_{v'} \vec E \cdot \vec D dv' [/tex]

I don't get why if the surface integral should equal to constant as long as the closed surface enclosed all charges at the same time will goes to zero as the surface expand. I understand that the "surface field density" goes to zero as the surface keep expanding, but not the total field emmitted from the surface no matter how big it gets!!!

Can someone explain this to me?

Thanks

Alan

[tex]W_e = \frac 1 2 \int_{v'} \rho_v V dv' = \frac 1 2 \int_{v'} (\nabla \cdot \vec D)V dv' = \frac 1 2 \int_{s'} V \vec D

\cdot \hat n ds' \;+\; \frac 1 2 \int_{v'} \vec E \cdot \vec D dv' [/tex]

I understand as long as [itex]v'[/itex] enclose all the charge distribution, the volume integral will be constant even when you extend the [itex]v'[/itex] to infinite size.

But I thought the surface integral behave the same. The definition of surface integral of charges is that the total fields coming out of the complete surface equal to the charges enclosed inside also. This mean if the closed surface contain all the charge distribution, the surface integral will be a constant even when further increase the size.

The book keep talking about potential decrease proportion to 1/r and E proportion to 1/(r^2) and the surface integral goes to zero as r goes to infinity.

The book then conclude that when volume goes to infinity:

[tex]W_e = \frac 1 2 \int_{v'} \vec E \cdot \vec D dv' [/tex]

I don't get why if the surface integral should equal to constant as long as the closed surface enclosed all charges at the same time will goes to zero as the surface expand. I understand that the "surface field density" goes to zero as the surface keep expanding, but not the total field emmitted from the surface no matter how big it gets!!!

Can someone explain this to me?

Thanks

Alan

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