Static Electric and Magnetic Fields in Matter

Abstract

The effects of electric polarization and magnetization are studied. The electric displacement D and the auxiliary field H are defined, and their relationship with the densities of free charge and free current, respectively, is found.

Appendices

Questions

• 1. What is a magnetic dipole? Should this term be interpreted literally? What justifies the standard use of the term?

• 2. Describe the behavior of an electric dipole inside an electric field, as well as the behavior of a magnetic dipole inside a magnetic field.

• 3. In what respect is the polarization mechanism in CO₂ different from that in H₂O?

• 4. What is the difference between free and bound charges, as well as between free and bound currents?

• 5. For what practical purposes do we place a dielectric in the interior of a capacitor?

• 6. A piece of metal and a piece of dielectric are placed inside an electrostatic field. In what respects will the responses of these substances to the field differ?

• 7. Show that the polarization (bound) charges of a uniformly polarized dielectric are confined to the surface of the dielectric. Similarly, show that the magnetization (bound) currents of a uniformly magnetized substance are confined to the surface of the substance.

• 8. (a) Consider a homogeneous linear dielectric in the interior of which there are no free charges. Show that the total electric charge inside the dielectric is zero. (b) Consider a homogeneous linear material in the interior of which there are no free currents. Show that the total current passing through any closed curve inside the material is zero.

• 9. Show that μ > μ₀ for paramagnetic substances while μ < μ₀ for diamagnetic substances.

• 10. The magnetic field in the interior of a linear medium is of the form $\overrightarrow{B}={B}_0\;{\hat{u}}_z\kern0.24em \left({B}_0>0\right)$. What will be the directions of the field $\overrightarrow{H}$ and the magnetization $\overrightarrow{M}$ inside this medium if the medium is (a) paramagnetic, (b) diamagnetic?

Problems

• 1. A charge +q is located at ${\overrightarrow{r}}_{+}\equiv \left(1,-1,1\right)$, while a charge –q is located at ${\overrightarrow{r}}_{-}\equiv \left(1,0,1\right)$. (a) Find the electric dipole moment of the system. (b) Find the torque exerted on the system by a uniform electric field $\overrightarrow{E}={E}_0\;{\hat{u}}_x$.

• Solution: We have:

$$ \overrightarrow{p}=q\;\overrightarrow{s}=q\kern0.22em \left({\overrightarrow{r}}_{+}-{\overrightarrow{r}}_{-}\right)\equiv q\kern0.22em \left(0,-1,0\right)\equiv \left(0,-q,0\right)\equiv -q\;{\hat{u}}_y.\mathrm{Also}, $$

$$ \overrightarrow{E}={E}_0\;{\hat{u}}_x\equiv \left({E}_0,0,0\right)\kern0.24em \mathrm{and}\kern0.24em \overrightarrow{T}=\overrightarrow{p}\times \overrightarrow{E}=\kern0.36em \left|\begin{array}{ccc}{\hat{u}}_x& {\hat{u}}_y& {\hat{u}}_z\\ {}0& -q& 0\\ {}{E}_0& 0& 0\end{array}\right|\kern0.36em =q\kern0.1em {E}_0\;{\hat{u}}_z\equiv \left(0,\kern0.4em 0,\kern0.4em q\kern0.1em {E}_0\right). $$

• 2. Consider a circular current loop of radius R, lying on the xy-plane and carrying a current Ι (Fig. 8.9). The current is flowing clockwise as we look the xy-plane from the side of the positive z-semiaxis. Find the magnetic dipole moment of the loop, as well as the torque exerted on the loop by a uniform magnetic field $\overrightarrow{B}={B}_0\;{\hat{u}}_x$.

• Solution:

Fig. 8.9

A circular current loop on the xy-plane

The area of the loop is a = π R² or, in vector form, $\overrightarrow{a}=-a\;{\hat{u}}_z=-\pi {R}^2\kern0.2em {\hat{u}}_z$. Then,

$$ \overrightarrow{m}=I\;\overrightarrow{a}=- I\pi {R}^2\kern0.2em {\hat{u}}_z\equiv \left(0,0,- I\pi {R}^2\right) $$

Also,

$$ \overrightarrow{B}={B}_0\;{\hat{u}}_x\equiv \left({B}_0,0,0\right) $$

Therefore,

$$ \overrightarrow{T}=\overrightarrow{m}\times \overrightarrow{B}=\kern0.36em \left|\begin{array}{ccc}{\hat{u}}_x& {\hat{u}}_y& {\hat{u}}_z\\ {}0& 0& - I\pi {R}^2\\ {}{B}_0& 0& 0\end{array}\right|\kern0.36em =- I\pi {R}^2{B}_0\;{\hat{u}}_y\equiv \left(0,- I\pi {R}^2{B}_0,0\right) $$

Note that our results do not depend on the location of the center O of the circular loop on the xy-plane.

• 3. Consider a system of charges q₁, q₂,… Let $\overrightarrow{p}$ be the electric dipole moment of the system with respect to some reference point Ο. Show that $\overrightarrow{p}$ is independent of the choice of Ο if and only if the total charge of the system is zero.

• Solution: We consider two reference points Ο and Ο΄ (Fig. 8.10) and we call $\overrightarrow{p}$ and ${\overrightarrow{p}}^{\prime }$ the corresponding electric dipole moments of the system relative to these points. We have:

Fig. 8.10

A charge q_i viewed from two reference points O and O΄

$$ \overrightarrow{p}=\sum \limits_i{q}_i\;{\overrightarrow{r}}_i,\kern0.72em {\overrightarrow{p}}^{\prime }=\sum \limits_i{q}_i\;{{\overrightarrow{r}}_i}^{\prime }=\sum \limits_i{q}_i\;\left({\overrightarrow{r}}_i-\overrightarrow{a}\right)\kern0.24em \mathrm{where}\kern0.24em \overrightarrow{a}=\overrightarrow{OO\prime }.\mathrm{Then}, $$

$$ \overrightarrow{p}={\overrightarrow{p}}^{\prime}\kern0.36em \iff \kern0.36em \sum \limits_i{q}_i\;{\overrightarrow{r}}_i=\sum \limits_i{q}_i\;\left({\overrightarrow{r}}_i-\overrightarrow{a}\right)\kern0.36em \iff \kern0.36em \sum \limits_i{q}_i\;{\overrightarrow{r}}_i=\sum \limits_i{q}_i\;{\overrightarrow{r}}_i-\left(\sum \limits_i{q}_i\right)\kern0.22em \overrightarrow{a}\kern0.36em \iff \kern0.36em \sum \limits_i{q}_i=0 $$

This condition is satisfied, in particular, in the case of an electric dipole.

• 4. Show that in the interior of a homogeneous linear dielectric the free-charge density ρ_f and the polarization-charge density ρ_b are related by

$$ {\rho}_b=\left(\frac{\varepsilon_0}{\varepsilon }-1\right)\kern0.22em {\rho}_f=-\frac{\chi_e}{1+{\chi}_e}\kern0.24em {\rho}_f $$

What physical conclusion do you draw for the case where ρ_f = 0?

• Solution: In general, $\overrightarrow{\nabla}\cdot \overrightarrow{D}={\rho}_f$. Since the medium is homogeneous linear, $\overrightarrow{D}=\varepsilon \kern0.1em \overrightarrow{E}$ with ε = const. Hence, $\overrightarrow{\nabla}\cdot \left(\varepsilon \kern0.1em \overrightarrow{E}\right)=\varepsilon \kern0.22em \left(\overrightarrow{\nabla}\cdot \overrightarrow{E}\right)={\rho}_f\kern0.36em \Rightarrow$

$$ \boxed{\kern0.22em \overrightarrow{\nabla}\cdot \overrightarrow{E}=\frac{\rho_f}{\varepsilon}\kern0.22em } $$

(1)

But, by Gauss’ law, $\overrightarrow{\nabla}\cdot \overrightarrow{E}=\rho /{\varepsilon}_0$ where ρ = ρ_f + ρ_b is the total charge density. By comparing with (1), we have:

$$ \frac{\rho_f}{\varepsilon }=\frac{\rho }{\varepsilon_0}=\frac{\rho_f+{\rho}_b}{\varepsilon_0}\kern0.6em \Rightarrow \kern0.6em {\rho}_b=\left(\frac{\varepsilon_0}{\varepsilon }-1\right)\kern0.24em {\rho}_f $$

Moreover, by taking into account that ε = ε₀ (1 + χ_e), we find: $\frac{\varepsilon_0}{\varepsilon }-1=-\frac{\chi_e}{1+{\chi}_e}$.

We notice that ρ_b = 0 when ρ_f = 0. That is, if there is no free charge in the interior of the dielectric, the bound charge distributes itself on the surface of the material. It also follows from Eq. (8.7) that the polarization of the dielectric is uniform in this case.