Deriving equation of stored electrical energy in a charge distribution.

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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
 
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Answers and Replies

  • #2
Born2bwire
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V is the potential on the surface of your integrating volume right? So what's the potential of a point charge at infinity? Not just the potential, but also the electric flux density. The electric field falls off as 1/r^2 (faster than the potential) so what is the electric flux density of a charge at an infinite distance?
 
  • #3
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V is the potential on the surface of your integrating volume right? So what's the potential of a point charge at infinity? Not just the potential, but also the electric flux density. The electric field falls off as 1/r^2 (faster than the potential) so what is the electric flux density of a charge at an infinite distance?

Thanks for the answer.

I can see that V goes to zero when the surface getting to infinity because the charge is far away from the surface and the surface integral goes to zero......ONLY due to the V term alone, not the D.

I never quite understand why they even use the electric flux density. D term is also in the volume intergal. If one drag the D term in, the volume integral term should fade away too with the same logic....... Which to me absolutely don't make sense.

Bottom line, if the book only give the reason V goes to zero as surface keep expanding to infinity, I can accept it. But two books keep hopping on the D proportional to 1/r^2 so VD is proportional to 1/r^3 yada yada. Once you drag in the D then the argument should apply to both integrals and both should fade to zero as the size goes to infinity!!!
 
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  • #4
Born2bwire
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When you are doing the volume integration, the fact that the fields drop off by 1/r^4 doesn't come into it. It's an integral over the VOLUME, so you are still including the contributions from the near-fields. You can argue that the integral will converge to a constant value as you increase the integrating volume because the contributions for large distances will diminish quickly. The surface integral is different because it only includes values at the large distances. So while the surface integral evaluates its arguments at infinite distances, the volume integral evaluates its arguments at distances from 0 to infinity.

You still have to take into account the electric flux density in the treatment of the surface integral. If V falls off as 1/r but D asymptotically behaves as r^n where n>0, then the integral would be non-zero or infinite. So you have to take into account that both V and D together behave as 1/r^3 asymptotically to reason that the integral will go to zero.
 
  • #5
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When you are doing the volume integration, the fact that the fields drop off by 1/r^4 doesn't come into it. It's an integral over the VOLUME, so you are still including the contributions from the near-fields. You can argue that the integral will converge to a constant value as you increase the integrating volume because the contributions for large distances will diminish quickly. The surface integral is different because it only includes values at the large distances. So while the surface integral evaluates its arguments at infinite distances, the volume integral evaluates its arguments at distances from 0 to infinity.

You still have to take into account the electric flux density in the treatment of the surface integral. If V falls off as 1/r but D asymptotically behaves as r^n where n>0, then the integral would be non-zero or infinite. So you have to take into account that both V and D together behave as 1/r^3 asymptotically to reason that the integral will go to zero.

Maybe this is exactly what I am missing. My understand of the surface integral without V is "the total electric flux out of the surface is equal to the charge inside." This mean the surface integral should be constant even if the surface keep expanding to infinity just like the volume integral. Surface integral do not say anything about the surface flux density. It is only the V term that fades in great distance that cause the surface integral in this particular case goes to zero. The D is still irrelavent as in the volume integral.

Or do you mean that D is only relavent in this because of V fading at 1/r, then we have to start justifying that D is fading at 1/r^2 and surface is growning at r^2. So the conclusion is the "surface integral is fading at 1/r". If that is the way you put it, I can accept. BUt the real cause is still the V term. Can you tell me whether I am correct?
 
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  • #6
Born2bwire
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By itself, the surface integral of the electric flux density will be equal to the contained charge. Assuming a finite system, then yes this integral will become a constant as we take the limit to infinite volume. However, this physical interpretation goes out the window once we scale the flux density by the potential. You shouldn't think of it as being a constant contribution times a 1/r contribution because we can't separate the integral of a product of functions as the product of the integral of the individual functions. So without this physical insight we can see that the product falls off as 1/r^3 which means that a surface integral, over a sphere, will have an integral of 1/r (since the surface integration element scales as r^2). Of course, by the same token we know that if fell off by 1/r^2 then the integrand for the surface integral would be O(1) which shows why we can consider the surface integral of the flux density as being a constant. So we can see that the integrand approaches 0 in the limit as r->\infty.
 
  • #7
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I think that is what I was asking in the very last part of my last post. Because V is part of the inside of the surface integral, then we need to check the rest and make sure that together the integral will fall off. Which in this case the integral will fall off at 1/r. So we conclude the integral tends to zero as the surface approach infinity.

Am I correct?

thanks
 
  • #8
Born2bwire
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I think that is what I was asking in the very last part of my last post. Because V is part of the inside of the surface integral, then we need to check the rest and make sure that together the integral will fall off. Which in this case the integral will fall off at 1/r. So we conclude the integral tends to zero as the surface approach infinity.

Am I correct?

thanks

Ja, that's it.
 
  • #9
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Thanks for your help.
 

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