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feynman1

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When a dielectric has an infinite dielectric constant, the dielectric degenerates into a conductor. But can the bound charges also be degenerated into induced charges?

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- Thread starter feynman1
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- #1

feynman1

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When a dielectric has an infinite dielectric constant, the dielectric degenerates into a conductor. But can the bound charges also be degenerated into induced charges?

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- #3

feynman1

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I don't get it. The Maxwell's eqs or Gauss' law on the boundary will show that the induced charges on conductors should appear as free charges, otherwise these eqs give wrong answers.But usually the induced surface charge for conductors are not treated as free charges but free charges is extra charge you put in addition to the charges of the matter constituents.

- #4

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$$\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{D}=0, \quad \vec{D}=\epsilon_0 \vec{E}.$$

Then you have the condition ##\vec{E}(\vec{r})=0## for ##|\vec{r}|<a##, ##\vec{E}(\vec{r}) \rightarrow \vec{E}_0## for ##|\vec{r}| \rightarrow \infty##. Another condition is of course that the total charge on the surface is 0 (for an insulated sphere). Here are no "free charges" involved. All that happens is that charge already present in the uncharged sphere is shifted.

The solution is found by introducing the potential

$$\vec{E}=-\vec{\nabla} \Phi.$$

Let ##\vec{E}_0=E_0 \vec{e}_z##. By symmetry the part of the potential describing the induced field can only be characterized by a vector. It should vanish at infinity and not lead to additional net charge. So it can only be a dipole field with ##\vec{p}=p \vec{e}_z##. So the ansatz for ##r>a## is

$$\Phi=-z \left (E_0-\frac{p}{r^3} \right).$$

For ##r \leq a## you have ##\Phi=0##. With this choice of the arbitrary additive constant the potential should vanish along the sphere, we have

$$E_0-\frac{p}/a^3=0 \; \Rightarrow \; p=E_0 a^3$$

and thus

$$\Phi=-E_0 z \left (1-\frac{a^3}{r^3} \right).$$

The surface charge distribution is most easily calculated in spherical coordinates,

$$\Phi=-E_0 \cos \vartheta \left (r-\frac{a^3}{r^2} \right),$$

leading to

$$\sigma=D_r=-\epsilon_0 \partial_r \Phi|_{r=a} = 3 \epsilon_0 E_0 \cos \vartheta.$$

From Gauss's Law it's clear that the total charge on the sphere is 0, but of course you can verify it easily directly

$$Q_{\text{ind}}=3 \epsilon_0 E_0 \int_{0}^{\vartheta} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi a^2 \sin \vartheta \cos \vartheta =0.$$

- #5

feynman1

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Of course the total free charge is 0 on the boundary for an initially neutral sphere. But there's free charges on 1 side and opposite free ones on the other side, where the D field is discontinuous.

$$\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{D}=0, \quad \vec{D}=\epsilon_0 \vec{E}.$$

Then you have the condition ##\vec{E}(\vec{r})=0## for ##|\vec{r}|<a##, ##\vec{E}(\vec{r}) \rightarrow \vec{E}_0## for ##|\vec{r}| \rightarrow \infty##. Another condition is of course that the total charge on the surface is 0 (for an insulated sphere). Here are no "free charges" involved. All that happens is that charge already present in the uncharged sphere is shifted.

The solution is found by introducing the potential

$$\vec{E}=-\vec{\nabla} \Phi.$$

Let ##\vec{E}_0=E_0 \vec{e}_z##. By symmetry the part of the potential describing the induced field can only be characterized by a vector. It should vanish at infinity and not lead to additional net charge. So it can only be a dipole field with ##\vec{p}=p \vec{e}_z##. So the ansatz for ##r>a## is

$$\Phi=-z \left (E_0-\frac{p}{r^3} \right).$$

For ##r \leq a## you have ##\Phi=0##. With this choice of the arbitrary additive constant the potential should vanish along the sphere, we have

$$E_0-\frac{p}/a^3=0 \; \Rightarrow \; p=E_0 a^3$$

and thus

$$\Phi=-E_0 z \left (1-\frac{a^3}{r^3} \right).$$

The surface charge distribution is most easily calculated in spherical coordinates,

$$\Phi=-E_0 \cos \vartheta \left (r-\frac{a^3}{r^2} \right),$$

leading to

$$\sigma=D_r=-\epsilon_0 \partial_r \Phi|_{r=a} = 3 \epsilon_0 E_0 \cos \vartheta.$$

From Gauss's Law it's clear that the total charge on the sphere is 0, but of course you can verify it easily directly

$$Q_{\text{ind}}=3 \epsilon_0 E_0 \int_{0}^{\vartheta} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi a^2 \sin \vartheta \cos \vartheta =0.$$

- #6

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In my example you cannot so easily provide the induced surface charge as free charge, because you don't know how it looks beforehand. You only can calculate it after solving the boundary-value problem for the electrostatic fields. That's why in this case one chooses that ##\rho_{\text{f}}=0## everywhere and then calculates the induced surface-charge density after the boundary-value problem is solved.

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