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The total work done to build up the free charge from zero to the final configuration in the presence of a dielectric material is: $$W = \frac{1}{2}\int \vec{D} \cdot \vec{E} d\tau.$$ We got this by supposing the dielectric material is fixed in position and then we bring in the free charge $\Delta \rho_f$ a bit at a time. Thus the work done on the incremental free charge is given by: $$\Delta W = \int (\Delta \rho_f)V d \tau.$$

Since $\nabla \cdot \vec{D} = \rho_f$, $\Delta \rho_f = \nabla \cdot (\Delta \vec{D})$, where $\Delta \vec{D}$ is the resulting change in $\vec{D}$, so $$\Delta W = \int [\nabla \cdot (\Delta \vec{D})]V d \tau.$$ and hence by integration by parts $$ \Delta W = \int \nabla \cdot [(\Delta \vec{D})V]d \tau + \int(\Delta \vec{D}) \cdot \vec{E} d \tau.$$ The divergence theorem turns the first term into a surface integral, which vanishes if we integrate over all space. Therefore , the work done is equal to $$\Delta W = \int (\Delta \vec{D}) \cdot \vec{E} d \tau.$$

Questions: Why does $\Delta \vec{D}V$ have to go to zero faster that $\frac{1}{r^2}$ for the surface integral (from the divergence theorem) $\int \nabla \cdot [(\Delta \vec{D})V]d \tau = \oint V \Delta \vec{D} \cdot da = 0$?

In this derivation (a full explanation is here) are we accounting for the fact that as we bring in free charge the dielectric material becomes polarised and hence the electric field (and therefore the potential at any point $V(r)$) changes due to the addition of $\Delta \rho_f$? If so, where in the derivation?

Thanks.

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Read the demonstration you posted carefully. You have to bring the charges from infinity, so the initial integral is applied to all the space (in the link this is represented as $d^3r$).

The term $\Delta \vec{D}V$ should go as $1/r^2$ because, as we said, we need to evaluate the integral in all the space. We change the $d^3r \equiv dV$ to $dA$ in the divergence theorem, but after integrating we still have to do the limit $A\rightarrow+\infty$ in order to integrate over the space. If we think in this differential as a growing sphere, then we can write $dA=r^2\sin\theta d\theta d\phi$. We have to integrate this, and then take the limit becomes $r\rightarrow +\infty$. This gives an infinite result if we don't eliminate the term $r^2$ inside of the integral. To avoid this result, $\Delta \vec{D}V$ should go as $1/r^2$ to cancel the $r^2$ term. Moreover, the integral has to be zero, because if the surface goes to infinite and we suppose that the potential is zero at this point, we are not going to have any contribution from this integral.

About the polarization of the dielectric, I'm pretty sure it is incluided in the derivation, and I think that it is incluided in the fact that you're using $\vec{D}$. Remember that $\vec{D} = \varepsilon \vec{E}+\vec{P}$ where $\vec{P}$ is the polarization density. If the medium is polarized, you have to include the polarization density vector in your calculus.

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  • $\begingroup$ Oh I think I understand. If the integrand doesn't go to zero fast enough (faster than $\frac{1}{r^2}$), we would have a non-zero integrand $V \Delta \vec{D}$ at the surface of the sphere which is going to infinity (so $\text{limit}A \to \infty$) hence the surface integral would diverge and $\Delta W$ would diverge as well which we don't want. What if we consider $V \vec{D}$ goes to zero exactly as fast as $\frac{1}{r^2}$, say $V\vec{D} = \frac{1}{r^2}\hat{r}$ so the surface integral doesn't have to go to zero and $\Delta W$ can still converge? So why does the surface integral have to be zero? $\endgroup$
    – user100411
    Commented Sep 24, 2016 at 18:31
  • $\begingroup$ @V_Programmer If you have a chance, check out my post on this topic. $\endgroup$
    – Alex
    Commented Sep 26, 2016 at 13:22
  • $\begingroup$ @JohnDoe Not 100% sure but I think it has to go to zero because you set the potential zero at the infinite (the document you linked to said clearly this election of potential).@Alex can't see the post, wrong link I think ^^" $\endgroup$ Commented Sep 30, 2016 at 16:39

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