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Suppose in a given situation, a dielectric extends in all space till infinity. Now there is a cavity of radius $R$ centered at origin. At the origin, there is a point charge $Q$.

The question asks us to find the net bound surface charge. While finding this, should I consider the charge that would supposedly appear at infinity too (is it defined at all?), which would make the answer zero, or should it be the non-zero value that would appear at the inner surface of the cavity in the dielectric?

(The question was worded specifically to include "net", so my real question is whether any bound charge can be defined in this situation except that at the inner surface)

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  • $\begingroup$ Polarization in dielectrics — if you understand how polarization works, then you'll get your answer easily. $\endgroup$ – Yashas Apr 3 '17 at 14:19
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What's the electric field at infinity?

$$\vec{E}=\frac{1}{4\pi\epsilon}\frac{Q}{r^2}\hat{r}|_{\substack{r=\infty}}=0$$ Polarization=$$\vec{P}=\epsilon_0\chi_e\vec{E}=0$$

Bound surface charge=

$$\sigma_b=\vec{P}\cdot{\hat{n}}=0$$

So no bound surface charge at infinity.

Consider only the charge at the surface of the cavity.

UPDATE:

As suggested by @Radial Apps I'm putting this (useful) link that explains dielectrics. This is the link from internet archive.

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    $\begingroup$ But actually keeping everything as limits, you would get sigma b tending to zero. Now when you try to integrate it over area to find the total charge, your area is tending to infinity, so you could possibly get a finite charge. $\endgroup$ – PulseJet Apr 1 '17 at 20:33
  • $\begingroup$ That's like saying, okay my current is zero, but if I take it over an infinitesimal time interval, I should get some charge. You can't evaluate a limit here because $\sigma_b$ isn't a function over all space, it's only defined at a boundary. So it doesn't tend to zero it is zero. You can't get any charge no matter what the area you multiply zero with. $\endgroup$ – GeeJay Apr 2 '17 at 3:27
  • $\begingroup$ @RadialApps is right. Even if the surface charge density is $0$, the area is infinite so you can't conclude that the bound surface charge is $0$. That's why you first consider a finite dielectric then take the limit to make it extend in all space. $\endgroup$ – honey.mustard Apr 2 '17 at 5:23
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    $\begingroup$ Okay, yes if you were to do that you would get a bound charge opposite in sign and equal in value to the bound charge at the surface of the cavity. But the thing is, this charge is at infinity. It doesn't count. You will find this interesting, especially the second line from the top on page 8. $\endgroup$ – GeeJay Apr 2 '17 at 7:24
  • $\begingroup$ @GeeJay, +1 for a very nice find! I would suggest putting the link in the answer along with a link to web archive (in case it ever goes down). $\endgroup$ – PulseJet Apr 3 '17 at 10:29
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The problem is probably asking for only the induced charge on the surface of the cavity. After all, what would be the point in detailing the configuration if the answer is going to be $0$ regardless? In addition, it's also difficult to talk about bound charges at infinity. The dielectric extends in all space so there's no identifiable outer boundary at which charges can be induced.

The problem itself can be solved by integrating $\nabla\cdot\bf{D}=\rho_{\mathrm{free}}$ over a sphere that includes the entire cavity to obtain an expression for $\bf{D}$. Assuming the permittivity of the dielectric is a scalar $\epsilon$, you can then write $\bf{E}=\bf{D}/\epsilon$. By noting that $\bf{E}$ must be equal to the electric field produced by both the charge at the center and the charges induced on the surface of the cavity, you should be able to find an expression for the total induced charge.

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  • $\begingroup$ What I really want to know is whether the bound charge is meaningful at all at infinity going strictly by its definition (exactly what you said was "difficult to talk about" ;) ) $\endgroup$ – PulseJet Apr 1 '17 at 17:10
  • $\begingroup$ Btw, I didn't downvote, in case you're wondering. $\endgroup$ – PulseJet Apr 1 '17 at 20:24
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There is nothing special about the bound charge except that it is derived only from the boundary of the dielectric in this particular case (volume charge density of bound charges, that is bound charge inside the dielectric is zero here). As long as a boundary is defined so is the bound charge. Therefore the net charge is zero.

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  • $\begingroup$ That's the whole point; there is no proper boundary defined $\endgroup$ – PulseJet Apr 1 '17 at 20:37

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