Induced/bound charges in conductors and dielectrics

In summary, when placed in an electric field, a conductor has induced charges and a dielectric has bound charges. When there's no net bound charge density in the bulk of the dielectric, bound charges stay on the surface only, like induced charges in conductors.
  • #1
feynman1
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When placed in an electric field, a conductor has induced charges and a dielectric has bound charges. When there's no net bound charge density in the bulk of the dielectric, bound charges stay on the surface only, like induced charges in conductors. In Maxwell's eqs, the induced charges are categorized as free charges but bound charges aren't.
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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  • #2
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. The induced surface charge is usually lumped into ##\vec{D}## and can be calcualted as the jump of the normal component of ##\vec{D}## along the surface after the boundary-value problem is solved. Of course you can always shuffle between charges and fields. All that counts at the end is the total electromagnetic field and the total charge-current distribution.
 
  • #3
vanhees71 said:
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.
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.
 
  • #4
What I had in mind is a problem like a conducting sphere of radius ##a## around the origin in an asymptotically homogeneous electric field, i.e.,
$$\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{D}=0, \quad \vec{D}=\epsilon_0 \vec{E}.$$
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
$$Q_{\text{ind}}=3 \epsilon_0 E_0 \int_{0}^{\vartheta} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi a^2 \sin \vartheta \cos \vartheta =0.$$
 
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  • #5
vanhees71 said:
What I had in mind is a problem like a conducting sphere of radius ##a## around the origin in an asymptotically homogeneous electric field, i.e.,
$$\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{D}=0, \quad \vec{D}=\epsilon_0 \vec{E}.$$
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
$$Q_{\text{ind}}=3 \epsilon_0 E_0 \int_{0}^{\vartheta} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi a^2 \sin \vartheta \cos \vartheta =0.$$
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.
 
  • #6
No, by definition these are not free charges but charges belonging to the medium and thus lumped into the fields. You can always shuffle contributions between "matter contributions" (i.e., charges and currents) and "field contributions" (polarization and magnetization). It depends on the problem how to best do it.

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.
 

1. What are induced charges?

Induced charges are charges that are created in a material due to the presence of an external electric field. These charges are temporary and are induced by the movement of free electrons in the material.

2. How are induced charges different from bound charges?

Induced charges are created in response to an external electric field, while bound charges are inherent in the material itself. Bound charges are fixed and cannot move, whereas induced charges can move within the material.

3. What is the difference between conductors and dielectrics in terms of induced and bound charges?

Conductors have free electrons that can easily move in response to an external electric field, resulting in the formation of induced charges. In dielectrics, the electrons are tightly bound to their atoms and cannot move as easily, so the induced charges are smaller. However, dielectrics can also have bound charges due to the alignment of their molecules in an electric field.

4. How does the presence of induced and bound charges affect the behavior of conductors and dielectrics?

In conductors, the movement of induced charges creates an opposing electric field, resulting in shielding of the external field. In dielectrics, the presence of bound charges can enhance or reduce the overall electric field, depending on their orientation.

5. Can induced and bound charges be separated from each other?

No, induced and bound charges are always present together and cannot be separated. They are both necessary for the overall behavior of a material in an electric field.

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