Spring 2020 · "eight oh two" · Instructor: C. Paus (shoutout to my TA and technical instructor too!)
In physics, there are four known fundamental interactions, which cannot be reduced to more basic interactions. One of them is gravitation, and another is electromagnetism, the focus of 8.02. Maxwell's equations, named after James Clerk Maxwell (who was born about 100 years after Isaac Newton died, and died the year Albert Einstein was born), form the foundation of classical electromagnetism.
I generally don't know wtf is going on in physics, but this is my best attempt to make sense of it. If you see any errors, feel free to let me know!
8.02 didn't introduce all these equations at once (and it's much more than simply memorizing equations), but I find it helpful to have a high-level overview of what this field (like, academic subject, not electric/magnetic field) is all about.
Edit: As I’m finishing up this summary to review for my final, I realized that the first class’s slides actually do include all of Maxwell’s equations. I must’ve not remembered because I didn’t understand what any of it meant.
(1) Gauss's Law helps us find the electric field produced by an object if we know the enclosed charge (or can derive it from charge density). The surface integral is over a closed surface called the Gaussian surface, and charge enclosed refers to the charge that passes through that surface. It's applicable for objects with a high enough degree of symmetry that we can infer the direction of the field: spheres, very long (i.e. infinite) cylinders/rods, and infinite slabs/sheets.
$\mathbf{\Phi}_E = \oiint\limits_S \mathbf{\vec{E}} \cdot d\mathbf{\vec{A}} = \frac{q_{enc}}{\epsilon_0}$(2) Ampere-Maxwell Law helps us find the magnetic field produced by an object if we know the enclosed current (or can derive it from current density). The line integral is over a closed path called the Amperian loop, and current enclosed refers to the current that passes through the surface enclosed by the loop. It’s applicable for solenoids, infinite current slabs, and very long (i.e. infinite) cylindrical wires/shells.
$\oint\limits_C \mathbf{\vec{B}} \cdot d \mathbf{\vec{s}} = \mu_0 I_{enc} + \mu_0 \epsilon_0 \frac{d}{dt} \iint \mathbf{\vec{E}} \cdot d \mathbf{\vec{A}}$The second term was added by Maxwell and represents the contribution from the displacement current.
(3) Gauss's Magnetism Law, like the other Gauss's Law, is a statement about fields generated by point charges. Indeed, it says that magnetic point charges (i.e. monopoles—an isolated north pole or south pole) do not exist.
$\oint \mathbf{\vec{B}} \cdot d \mathbf{\vec{A}} = 0$(4) Faraday's Law says that when the magnetic flux through a coil of wire changes, an electromotive force will be induced. This emf tries to get the coil back to the previous magnetic flux state—that negative sign is so important that it itself forms Lenz's law.
$\epsilon = \text{emf} = -\frac{d \Phi_B}{dt}$No idea what all this means? Read on!
The course is divided into 15 weeks, which cover:
We didn’t cover anything during weeks 7 and 8 of the term due to COVID-19, so my summary doesn’t include what what normally be covered in the last 2 weeks of 8.02.
Charges (unit: Coulomb)
Electric forces (unit: Newton) and Coulomb’s law: Point charges or charged objects exert forces on each other, and those forces are given by Coulomb’s law.
$\mathbf{\vec{F}_{12}} = k \frac{q_1 q_2}{r_{12}^2} \mathbf{\hat{r}_{12}}$where $\mathbf{\hat{r}_{12}}$ is the unit vector from $q_1$ to $q_2$ and $k = 1/{4 \pi \epsilon_0}$
Electric fields (unit: Newton/Coulomb)
By defining electric fields, we now separate source (what electric field does this object create?) from effect (what force does this electric field have some object?).
(Source) The field that a point charge creates can be found from Coulomb’s law:
$\mathbf{\vec{E}_S(P)} = k_e \frac{q_s}{r_{SP}^2} \mathbf{\hat{r}}_{SP}(P)$(Effect) The force that a point charge $q$ experiences due an electric field $\mathbf{\vec{E}}$ is:
$\mathbf{\vec{F}_E} = q \mathbf{\vec{E}}$Dipoles and dipole moments: A dipole is a distribution of charges that is overall neutral. The dipole moment is defined as:
$\mathbf{\vec{p}} \equiv \sum_{i=1}^N q_i \mathbf{\vec{r}}_i$We looked at dipole behavior in external fields. When a given dipole is placed in a given external electric field, it feels a force if the field is non-uniform, and it feels a torque if it’s not aligned with the external field. The torque tries to align the dipole with the external field.
We also looked at field lines for dipoles:
Superposition principle
Dimension | Charge density if uniform | Differential charge element |
---|---|---|
1D (e.g. wire) | $\lambda = Q/L$ | $dQ = \lambda dL$ |
2D (e.g. plane) | $\sigma = Q/A$ | $dQ = \sigma dA$ |
3D (e.g. cylinder) | $\rho = Q/V$ | $dQ = \rho dV$ |
To find the electric field of a continuous charge distribution:
Common problems: Finding the electric field of a finite line of charge, charged rod, ring, disk
But first, what is flux? I still don’t have a great intuition for it, but mathematically, flux is:
$\Phi_E = \iint\limits_{\text{surface}} \mathbf{\vec{E}} \cdot d \mathbf{\vec{A}} = \iint\limits_{\text{surface}} \mathbf{\vec{E}} \cdot \mathbf{\hat{n}} dA$So if the electric field is parallel to the unit normal, then flux is positive.
For a closed surface, $d \mathbf{\vec{A}}$ is normal to the surface and points outward.
The flux through a closed surface with no charge enclosed is zero. Think about having a point charge outside the surface. Its field lines will enter the surface on one side and simply go out the other, so flux cancels out.
In choosing a Gaussian surface, we want to try to choose ones where the electric field is either perpendicular or parallel to the faces of the surface. If it’s perpendicular and has uniform magnitude on the surface, then it equals $EA$. If it’s parallel, then it equals 0.
Source Symmetry | Gaussian surface |
---|---|
Spherical | Concentric sphere |
Cylindrical | Coaxial cylinder |
Planar | Gaussian “pillbox” |
Electric force is conservative due to the radially symmetric nature of Coulomb’s law. Therefore, the work done by en electrical force moving an object from A to B is path-independent.
$W_{AB} = \int_{A}^{B} \mathbf{\vec{F}}_{21} \cdot d \mathbf{\vec{s}}$It takes negative work to bring a negatively charged object toward a positively charged source.
$W = -k_e q_s q_1 \left(\frac{1}{r_B} - \frac{1}{r_A}\right) < 0$Electric potential difference (unit: Joule/Coulomb = Volt) is the change in potential energy per test charge in moving the test object charge $q_t$) from A to B.
$\Delta V_{AB} \equiv \frac{\Delta U_{AB}}{q_t} = -\int\limits_A^B (\mathbf{\vec{F}}_{st} / q_t) \cdot d \mathbf{\vec{s}} = -\int\limits_A^B \mathbf{\vec{E}}_s \cdot d \mathbf{\vec{s}}$Effects of electric field and potential:
Using Gauss’s law to find electric potential from electric field: If a charge distribution has enough symmetry to use Gauss’s law to calculate the electric field, then you can calculate the electric potential difference between 2 points A and B:
$V_B - V_A = - \int\limits_A^B \mathbf{\vec{E}} \cdot d \mathbf{\vec{s}}$This is a path-independent line integral.
Deriving E from V: $\mathbf{\vec{E}} = - \vec{\nabla} V$
Electric potential function:
$V_s(P) = \frac{1}{4 \pi \epsilon_0} \frac{q_s}{r}$Problems like finding electric potential difference between infinitey and a point along the z-axis of a ring
Equipotentials
Conductors are equipotential surfaces. When a conductor is placed in an external electric field, charges in the conductor will move until
For an isolated conductor with total charge $Q$ and equipotential surface with potential $V$ (with $V=0$ at infinity), the charge on the conductor $Q$ is linearly proportional to the potential:
$Q = CV$where $C$ is capacitance (unit: 1 Coulomb/Volt = 1 farad) and depends only on the size and shape of the capacitor.
Conductors as shields
TODO
Faraday Ice Pail
TODO
Drift velocity: Average velocity forced by applied electric field in the presence of collisions (not completely sure what this means)
Current density is defined as:
$\mathbf{\vec{J}} \equiv n q \mathbf{\vec{v}}_q = \sum_i n_i q_i \mathbf{\vec{v}}_{q_i}$where $n$ is the number of charged objects per unit volume.
Current density and current are related by:
$I = \int\limits_S \mathbf{\vec{J}} \cdot \mathbf{\hat{n}} dA = \int\limits_S \mathbf{\vec{J}} \cdot d \mathbf{\vec{A}}$So if $\mathbf{\vec{J}}$ is uniform and perpendicular to the surface, then $I = JA$.
Instead of E field, think of the potential difference across a conductor. For Ohmic materials:
$\Delta V = IR$where $R$, the constant of proportionality, is resistance (unit: Volts/Ampere = Ohm).
Potential difference is also equal to:
$\Delta V = EL$where $L$ is the length of the conductor.
Resistivity $\rho_r$, the inverse of conductivity, depends only on the microscopic properties of the conductor:
$\rho_r \equiv \frac{1}{\sigma_c}$Resistivity and conductivity are related to electric field and current density by:
$\mathbf{\vec{E}} = \rho_r \mathbf{\vec{J}}$ $\mathbf{\vec{J}} = \sigma_c \mathbf{\vec{E}}$From the microscopic Ohm’s law, we can find resistance:
$R = \frac{\rho_r L}{A}$Electromotive force (emf) is the work per unit charge around a closed path. It’s not a force :/
$\mathcal{E} = \oint\limits_{\text{closed path}} \mathbf{\vec{f}}_s \cdot d \mathbf{\vec{s}}$where $\mathbf{\vec{f}}_s$ is force per unit charge. If a closed conducting path is present, then $\mathcal{E} = IR$.
Ideal batteries:
Real batteries have an internal resistance $r$, so $\Delta V = \mathcal{E} - Ir$.
We can simplify circuits by replacing multiple resistors with a resistor of equivalent resistance. The way we calculate equivalent resistance depends on the arrangement of the original resistors:
Then we assign a current and current direction for each branch of the circuit. Next, we assign a circulation direction for each loop, which defines a “before” and “after” for each circuit element. We can determine electric potential difference by $\Delta V = V(\text{after}) - V(\text{before})$.
Circulating from the positive to negative terminals of a battery increases potential:
$\Delta V = + \mathcal{E}$Circulating across a resistor in the direction of current decreases potential:
$\Delta V = -IR$Kirchhoff’s laws are also helpful:
Static (i.e. not changing with time) magnetic fields only exert forces on moving charges. That force is given by:
$\mathbf{\vec{F}}_B = q \mathbf{\vec{v}} \times \mathbf{\vec{B}}$When an electric field is present too, use the Lorentz force equation:
$\mathbf{\vec{F}}_B = q (\mathbf{\vec{E}} + \mathbf{\vec{v}} \times \mathbf{\vec{B}})$The force due to a magnetic field on a wire with length $L$ and current $I$ is:
$\mathbf{\vec{F}} = I \mathbf{\vec{L}} \times \mathbf{\vec{B}}$We can generalize that “wire”
$d \mathbf{\vec{F}} = I d \mathbf{\vec{s}} \times \mathbf{\vec{B}}$ $\mathbf{\vec{F}} = I \int\limits_a^b d \mathbf{\vec{s}} \times \mathbf{\vec{B}}$The Biot-Savart law gives the general formula for computing magnetic field:
$d \mathbf{\vec{B}} = \frac{\mu_0}{4 \pi} \frac{I d \mathbf{\vec{s}} \times (\mathbf{\vec{r}}_f - \mathbf{\vec{r}}_s)}{|\mathbf{\vec{r}}_f - \mathbf{\vec{r}}_s|^3}$where $\mathbf{\vec{r}}_f$ is the position where we’re interested in the magnetic field, and $\mathbf{\vec{r}}_s$ is the position of infinitesimal currents.
For a wire:
$\mathbf{\vec{B}} = \frac{\mu_0 I}{2 \pi r} \hat{\theta}$Circular motion problems sometimes come up, and it’s helpful to use Newton’s second law to derive:
$qvB = \frac{mv^2}{R}$ $R = \frac{mv}{qB}$Magnetic dipole moment is given by:
$\vec{\mu} \equiv I A \mathbf{\hat{n}}_{\text{RHR}}$where $\mathbf{\hat{n}}_{\text{RHR}}$ is a unit vector perpendicular to the area.
Torque on a magnetic dipole due to external magnetic field:
$\vec{\tau} \equiv \vec{\mu} \times \mathbf{\vec{B}}$Sources with enough symmetry:
Back emf
An RC circuit contains a resistor and capacitor.
An LR circuit contains an inductor and resistor.
An LC circuit contains both an inductor and capacitor and is modeled as a simple harmonic oscillator.
How can current flow across a capacitor / through the part of a capacitor that's an insulator (air between two plates)?
Current density equals the rate of change of the electric flux vector with time