## PHYS102 electromagnetism课程简介

The physics of our universe is dominated by four fundamental forces: gravity, electromagnetism, and weak and strong nuclear forces. These forces control how matter, energy, space, and time interact. For example, when someone sits on a chair, gravitational forces balance with the material forces that “push up” to hold the person in place. This upward push results from electromagnetic forces on microscopic length scales. On the larger stage, gravity holds the celestial bodies in their orbits, but without electromagnetic radiation (light), none of these bodies would be visible to us.

Electromagnetism extends our understanding beyond classical mechanics because it introduces the concept of charge – a property we can observe in macroscopic objects and the smallest building blocks of matter. Electromagnetism is the invisible hand that allows charged objects to interact with each other. It also allows you to take this course: the modern world would be impossible without telecommunications and microelectronics, two of the major applications of electromagnetism.

## PREREQUISITES

Scientists began studying electromagnetism in the 18th century. They prepared the groundwork for developments in the 20th century and our modern understanding of atomic structure and the cosmos. In this course, we will learn why electromagnetism is so important for everyday applications and fundamental physics. To put this information into proper context, you should be familiar with the force concept of classical mechanics.

Building on the idea of force, we develop the more abstract concept of fields. This study culminates in Maxwell’s theory which, among other achievements, led to the discovery of radio waves. We begin by discussing waves and oscillations in the more familiar setting of mechanics to review how forces relate to the motion of objects. This preparation will help you understand Maxwell’s insights into the nature of electromagnetic radiation as a wave phenomenon.

The term electromagnetism combines two effects we will study separately: electricity and magnetism. We explore electrical measurements and circuits to learn how to observe, quantify, and apply the laws that govern how charges cause static electricity and magnetism. In Maxwell’s equations, we will finally unify electric and magnetic effects and discover electromagnetic radiation. This will also put radio waves on the same footing as light: they are the same phenomenon, differing only in their wavelength.

## PHYS102 electromagnetism HELP（EXAM HELP， ONLINE TUTOR）

1. A parallel-plate capacitor with plates of area $A$ and separation $d$ has its bottom plate grounded $(V=0)$ and its top plate held at potential $V_0$. (a) Solve Laplace’s equation inside the capacitor. (Note: use Cartesian coordinates and ignore edge effects; then the potential depends on only one variable.) Apply the boundary conditions and find $V$ as a function of position, then take the gradient to get $\vec{E}$. (b) Use your result in part (a) to find the surface charge densities on the two plates. (Note that this solution proves that $\sigma$ is uniform in a parallel plate capacitor when you can neglect edge effects.) (c) Now imagine that the space between the plates of the capacitor is filled with a linear dielectric with electric susceptibility $\chi_e$. Do your results for $V$ and/or $\vec{E}$ in part (a) change? Explain. Then find the polarization in the dielectric and all bound and free charge densities. (d) Find the capacitance of this capacitor using the definition $C=Q / \Delta V$, both without and with the dielectric. Check that the ratio of your results agrees with Griffiths example 4.6 (page 183).

(a) Since we can ignore edge effects, the potential between the plates depends only on the distance $z$ between them. Laplace’s equation $\nabla^2 V = 0$ reduces to $\frac{d^2 V}{dz^2} = 0$. The general solution is $V(z) = Az+B$, where $A$ and $B$ are constants determined by the boundary conditions. At $z=0$, the bottom plate is grounded so $V(0)=0$. At $z=d$, the top plate is held at potential $V_0$, so $V(d)=V_0$. Solving for $A$ and $B$ gives $A=\frac{V_0}{d}$ and $B=0$, so the potential is $$V(z) = \frac{V_0}{d}z$$ Taking the gradient gives the electric field: $\vec{E} = -\nabla V = -\frac{V_0}{d}\hat{z}$.

(b) The surface charge density on the plates is given by $\sigma = \epsilon_0 E_z$. From part (a), we have $E_z = -\frac{V_0}{d}$, so the surface charge densities on the top and bottom plates are $\sigma_1 = -\sigma_2 = \frac{\epsilon_0 V_0}{d}$.

(c) When a dielectric is introduced, the electric field inside it is reduced by a factor of $1/\epsilon_r$, where $\epsilon_r = 1+\chi_e$ is the relative permittivity of the dielectric. Therefore, the potential between the plates becomes $V(z) = \frac{V_0}{d}z/\epsilon_r$, and the electric field becomes $\vec{E} = -\frac{V_0}{d\epsilon_r}\hat{z}$. The polarization in the dielectric is $\vec{P} = \epsilon_0\chi_e\vec{E} = \frac{\epsilon_0\chi_e V_0}{d\epsilon_r}\hat{z}$, and there are no free charges in the dielectric. The bound charge density on each plate is equal and opposite to $\sigma$, so the total charge density on the plates is $\pm\sigma+\sigma_b$, where $\sigma_b = \frac{\epsilon_0\chi_e V_0}{d\epsilon_r}$ is the bound charge density.

(d) The capacitance is $C = Q/\Delta V$, where $Q$ is the charge on each plate and $\Delta V$ is the potential difference between them. Without the dielectric, we have $Q=\pm\sigma A$ and $\Delta V = V_0$, so $C_0 = \frac{\pm\sigma A}{V_0} = \frac{\epsilon_0 A}{d}$. With the dielectric, the charge on each plate is the sum of the free and bound charges, $Q = \pm \sigma A + \sigma_b A$, and the potential difference is $\Delta V = V_0/\epsilon_r$. Therefore, $C = \frac{\pm\sigma A+\sigma_b A}{V_0/\epsilon_r} = \frac{\epsilon_0\epsilon_r A}{d}$. The ratio of the two capacitances is $C/C_0 = \epsilon_r$, which agrees with Griffiths example 4.6.

A very thin spherical metal shell of radius $a$ is centred inside a thick spherical metal shell with inner radius $b$ and outer radius $c$, such that $a<b<c$ and both spheres have their centres at the origin. The outer sphere is left uncharged and a charge $Q$ is put on the inner sphere. Then the space between the two spheres (i.e., between radii $a$ and $b$ ) is filled with a linear dielectric (insulating) oil with electric susceptibility $\chi_e$. Find all free and bound charge densities (surface and volume), the polarization in the dielectric, and the electric field everywhere.

Since the outer sphere is uncharged, it does not affect the electric field inside the inner sphere. Therefore, we can treat the problem as a spherical capacitor with a dielectric between the plates.

Inside the inner sphere ($r<a$), the electric field is zero since there are no charges. Between the spheres ($a<r<b$), we can use Gauss’s law to find the electric field. Choose a spherical Gaussian surface of radius $r$ with $a<r<b$, so that the charge enclosed is $Q$. The electric field is radial, so the flux through the Gaussian surface is $4\pi r^2 E$, and Gauss’s law gives:

$$4\pi r^2 E = \frac{Q}{\epsilon_0}$$

or

$$E(r) = \frac{Q}{4\pi\epsilon_0 r^2}$$

Inside the dielectric ($b<r<c$), the electric field is reduced by a factor of $1/\epsilon_r$, where $\epsilon_r=1+\chi_e$ is the relative permittivity of the dielectric. Therefore, the electric field is

$$E(r) = \frac{Q}{4\pi\epsilon_0 r^2 \epsilon_r}$$

The polarization in the dielectric is $\vec{P} = \epsilon_0 \chi_e \vec{E}$. Inside the dielectric, this gives

$$\vec{P}(r) = \frac{Q\chi_e}{4\pi r^2}\hat{r}$$

Since the dielectric is insulating, there are no free charges inside it. However, the dielectric induces bound charges on the surfaces of the inner and outer spheres. These bound charges are equal and opposite to each other, and their densities are given by

$$\sigma_b = -\frac{\vec{P}\cdot\hat{r}}{\epsilon_0}$$

where $\hat{r}$ is the radial unit vector. On the inner sphere, the bound charge density is

$$\sigma_{b,1} = -\frac{Q\chi_e}{4\pi a^2\epsilon_0\epsilon_r}$$

and on the outer sphere it is

$$\sigma_{b,2} = \frac{Q\chi_e}{4\pi c^2\epsilon_0\epsilon_r}$$

Since the outer sphere is uncharged, there is no free charge density on it. On the inner sphere, the free charge density is

$$\rho_{f,1} = \frac{Q}{4\pi a^2}$$

Finally, we can find the electric field everywhere by summing the contributions from the free and bound charges:

$$\vec{E}(r) = \begin{cases} \vec{0} & r<a \ \frac{Q}{4\pi\epsilon_0 r^2}\hat{r} & a<r<b \ \frac{Q}{4\pi\epsilon_0 r^2 \epsilon_r}\hat{r} – \frac{Q\chi_e}{4\pi\epsilon_0 r^2}\hat{r} & b<r<c \ \frac{Q}{4\pi\epsilon_0 r^2}\hat{r} – \frac{Q\chi_e}{4\pi\epsilon_0 r^2}\hat{r} & r>c \end{cases}$$

## Textbooks

• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.