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8.02: Electricity and Magnetism

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!

Preview of Maxwell's Equations

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.

ΦE=SEdA=qencϵ0\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.

CBds=μ0Ienc+μ0ϵ0ddtEdA\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.

BdA=0\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.

ϵ=emf=dΦBdt\epsilon = \text{emf} = -\frac{d \Phi_B}{dt}

No idea what all this means? Read on!

Outline (WIP!)

The course is divided into 15 weeks, which cover:

  1. (Weeks 1-5, exam 1) Electrostatics: What are charges? What's an electric field, and what electric field does an object create? What forces and torques does another object experience inside an electric field? What if there are multiple objects?
  2. (Week 6) Applying electrostatics(?) What are circuits and currents? How do currents continuously flow in a circuit? How can we simplify circuits? How does this stuff relate to potential energy? What's a capacitor?
  3. (Weeks 9-10, quiz 1) Magnetostatics: What's a magnetic field and how does it compare/contrast with an electric field? What magnetic field does an object create? What forces and torques does another object experience due to a magnetic field?
  4. (Weeks 11-15) Moving stuff, more circuits, EM waves, energy flow:

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.

Basic Concepts of Electrostatics

Charges (unit: Coulomb)

  • There are two types of electric charges, termed positive and negative. Like charges repel, and opposites attract.
  • The elementary charge is the charge of an electron (negative) or proton (positive): ±e=1.602×1019\pm e = 1.602 \times 10^{-19}
  • Charge is quantized—all charges are integer multiples of ee.
  • Charge is conserved—it cannot be created or destroyed.

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.

F12=kq1q2r122r^12\mathbf{\vec{F}_{12}} = k \frac{q_1 q_2}{r_{12}^2} \mathbf{\hat{r}_{12}}

where r^12\mathbf{\hat{r}_{12}} is the unit vector from q1q_1 to q2q_2 and k=1/4πϵ0k = 1/{4 \pi \epsilon_0}

Electric fields (unit: Newton/Coulomb)

  • Charged objects produce electric fields, which act on other charged objects.
  • An electric field is a vector field, which can be represented by a set of arrows, field lines, or grass seeds.

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:

ES(P)=keqsrSP2r^SP(P)\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 qq experiences due an electric field E\mathbf{\vec{E}} is:

FE=qE\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:

pi=1Nqiri\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:

  • Field lines point away from positive charges to negative and terminate on charges.
  • The direction of the field at any point is tangent to the field line at that point.
  • Field lines never cross each other.

Superposition principle

  • The net force on any charge is the vector sum of forces from other individual charges:
Fj=i=1,ijNFij\mathbf{\vec{F}}_j = \sum_{i=1, i \neq j}^N \mathbf{\vec{F}}_{ij}
  • The electric field due to a collection of N point charges is the vector sum of the individual electric fields due to each charge:
E=i=1NE\mathbf{\vec{E}} = \sum_{i=1}^N \mathbf{\vec{E}}

Problem Solving: Finding Electric Fields

Continuous charge distributions:

Dimension Charge density if uniform Differential charge element
1D (e.g. wire) λ=Q/L\lambda = Q/L dQ=λdLdQ = \lambda dL
2D (e.g. plane) σ=Q/A\sigma = Q/A dQ=σdAdQ = \sigma dA
3D (e.g. cylinder) ρ=Q/V\rho = Q/V dQ=ρdVdQ = \rho dV

To find the electric field of a continuous charge distribution:

  • Break the distribution into small elements with charge dqdq
  • The E field at PP due to the small charged element is:
dE=kdqrdq,P2r^dq,Pd \mathbf{\vec{E}} = k \frac{dq}{r_{dq,P}^2} \mathbf{\hat{r}}_{dq,P}
  • The superposition of all the dqdq’s becomes an integration :0

Common problems: Finding the electric field of a finite line of charge, charged rod, ring, disk

Gauss’s law:

  • First of Maxwell’s equations
  • Shortcut for finding electric field for certain highly symmetrical objects
  • Tells us that electric flux through a closed surface is proportional to the charge enclosed
ΦE=SEdA=qencϵ0\mathbf{\Phi}_E = \oiint\limits_S \mathbf{\vec{E}} \cdot d\mathbf{\vec{A}} = \frac{q_{enc}}{\epsilon_0}

But first, what is flux? I still don’t have a great intuition for it, but mathematically, flux is:

ΦE=surfaceEdA=surfaceEn^dA\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, dAd \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 EAEA. If it’s parallel, then it equals 0.

Source Symmetry Gaussian surface
Spherical Concentric sphere
Cylindrical Coaxial cylinder
Planar Gaussian “pillbox”

Electric Potential Energy & Difference, Capacitors

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.

WAB=ABF21dsW_{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=keqsq1(1rB1rA)<0W = -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 qtq_t) from A to B.

ΔVABΔUABqt=AB(Fst/qt)ds=ABEsds\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:

  • When a charged particle (charge qq) is moved across an electric potential difference, then the change in potential energy is:
ΔU=qΔV=q(VfVi)\Delta U = q \Delta V = q (V_f - V_i)
  • The change in mechanical energy of the charged particle is:
ΔEmech=ΔK+qΔV=12mΔv2+qΔV2\Delta E_{mech} = \Delta K + q \Delta V = \frac{1}{2} m \Delta v^2 + q \Delta V^2

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:

VBVA=ABEdsV_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: E=V\mathbf{\vec{E}} = - \vec{\nabla} V

Electric potential function:

Vs(P)=14πϵ0qsrV_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


  • All points on an equipotential curve are at the same potential. Each curve is represented by V(x,y)=constantV(x,y) = \text{constant}
  • E field points from high to low potential
  • E field lines are perpendicular to equipotentials
  • E field has no component along equipotential
  • Electrostatic force does zero work to move a charged particular along equipotential

Conductors are equipotential surfaces. When a conductor is placed in an external electric field, charges in the conductor will move until

  1. E = 0 inside conductor
  2. E is perpendicular to the surface
  3. A charge distribution (generally non-uniform) is induced on the surface of the conductor
  4. All points in conductor are at the same potential

Capacitors and capacitance

  • Capacitors are the first of 3 standard electronics devices (capacitors, resistors, and inductors)
  • For an isolated conductor with total charge QQ and equipotential surface with potential VV (with V=0V=0 at infinity), the charge on the conductor QQ is linearly proportional to the potential:

    Q=CVQ = CV

    where CC is capacitance (unit: 1 Coulomb/Volt = 1 farad) and depends only on the size and shape of the capacitor.

  • A capacitor consists of two isolated conductors with equal and opposite charges ±Q\pm Q. ΔV\Delta V is the potential difference between the two conductors, and the capacitance of the two conducting surfaces is C=Q/ΔVC = Q/|\Delta V|
  • The energy stored in a capacitor is given by:
U=Q22C=12QΔV=12CΔV2U = \frac{Q^2}{2C} = \frac{1}{2} Q |\Delta V| = \frac{1}{2} C |\Delta V|^2

Conductors as shields


Faraday Ice Pail


Currents, Ohm’s Law, DC Circuits

Current (unit: Coulomb/second = Ampere)

  • Flow of charge
  • I=dQ/dtI = dQ/dt
  • Positive charge moves in the direction of current (in reality, negative charges, i.e. electrons, are the ones that move, but people were working on circuit theory before it was known that electrons flow)
  • Current flows because if an electric field is set up inside a conductor, charge will move, making a current in the direction of the electric field. Note that when a current is flowing, the conductor is not an equipotential surface and Einside0E_{\text{inside}} \neq 0

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:

Jnqvq=iniqivqi\mathbf{\vec{J}} \equiv n q \mathbf{\vec{v}}_q = \sum_i n_i q_i \mathbf{\vec{v}}_{q_i}

where nn is the number of charged objects per unit volume.

Current density and current are related by:

I=SJn^dA=SJdAI = \int\limits_S \mathbf{\vec{J}} \cdot \mathbf{\hat{n}} dA = \int\limits_S \mathbf{\vec{J}} \cdot d \mathbf{\vec{A}}

So if J\mathbf{\vec{J}} is uniform and perpendicular to the surface, then I=JAI = JA.

Ohm’s Law, Resistance, Resistivity

Instead of E field, think of the potential difference across a conductor. For Ohmic materials:

ΔV=IR\Delta V = IR

where RR, the constant of proportionality, is resistance (unit: Volts/Ampere = Ohm).

Potential difference is also equal to:

ΔV=EL\Delta V = EL

where LL is the length of the conductor.

Resistivity ρr\rho_r, the inverse of conductivity, depends only on the microscopic properties of the conductor:

ρr1σc\rho_r \equiv \frac{1}{\sigma_c}

Resistivity and conductivity are related to electric field and current density by:

E=ρrJ\mathbf{\vec{E}} = \rho_r \mathbf{\vec{J}} J=σcE\mathbf{\vec{J}} = \sigma_c \mathbf{\vec{E}}

From the microscopic Ohm’s law, we can find resistance:

R=ρrLAR = \frac{\rho_r L}{A}

EMF & Batteries

Electromotive force (emf) is the work per unit charge around a closed path. It’s not a force :/

E=closed pathfsds\mathcal{E} = \oint\limits_{\text{closed path}} \mathbf{\vec{f}}_s \cdot d \mathbf{\vec{s}}

where fs\mathbf{\vec{f}}_s is force per unit charge. If a closed conducting path is present, then E=IR\mathcal{E} = IR.

Ideal batteries:

  • Zero force outside
  • Inside: non-zero chemical force fs\mathbf{\vec{f}}_s that moves charges through a region in which a static electric field Estatic=fs\mathbf{\vec{E}}_{\text{static}} = -\mathbf{\vec{f}}_s opposes motion
  • Net force on charges is zero
  • TODO

Real batteries have an internal resistance rr, so ΔV=EIr\Delta V = \mathcal{E} - Ir.

Circuit Analysis

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:

  • Resistors in series: Req=R1+R2+...R_{eq} = R_1 + R_2 + ...
  • Resistors in parallel: 1/Req=1/R1+1/R2+...1/R_{eq} = 1/R_1 + 1/R_2 + ...

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 ΔV=V(after)V(before)\Delta V = V(\text{after}) - V(\text{before}).

Circulating from the positive to negative terminals of a battery increases potential:

ΔV=+E\Delta V = + \mathcal{E}

Circulating across a resistor in the direction of current decreases potential:

ΔV=IR\Delta V = -IR

Kirchhoff’s laws are also helpful:

  1. Loop rule: The sum of potential differences across all elements around any closed circuit loop must be 0:
iΔVi=closed pathEstaticds=0\sum\limits_{i} \Delta V_i = -\oint\limits_{\text{closed path}} \mathbf{\vec{E}}_{\text{static}} \cdot d \mathbf{\vec{s}} = 0
  1. Junction rule (current conservation): The sum of currents entering any junction in a circuit must equal the sum of currents exiting that junction.

Magnetic Fields and Forces

  • Unit for magnetic fields: Tesla = NsCm=kgCs\frac{N \cdot s}{C \cdot m} = \frac{kg}{C \cdot s}
  • Magnetic fields are created by currents (i.e. moving charges)
  • Static (i.e. not changing with time) magnetic fields only exert forces on moving charges. That force is given by:

    FB=qv×B\mathbf{\vec{F}}_B = q \mathbf{\vec{v}} \times \mathbf{\vec{B}}
  • When an electric field is present too, use the Lorentz force equation:

    FB=q(E+v×B)\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 LL and current II is:

    F=IL×B\mathbf{\vec{F}} = I \mathbf{\vec{L}} \times \mathbf{\vec{B}}
  • We can generalize that “wire”

    dF=Ids×Bd \mathbf{\vec{F}} = I d \mathbf{\vec{s}} \times \mathbf{\vec{B}} F=Iabds×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:

    dB=μ04πIds×(rfrs)rfrs3d \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 rf\mathbf{\vec{r}}_f is the position where we’re interested in the magnetic field, and rs\mathbf{\vec{r}}_s is the position of infinitesimal currents.

  • Common Biot-Savart law problems: 1) intuit magnitude/direction of B fields generated by wires in various shapes (semi-circles, rectangles), 2) calculate B field for objects like a ring of current or a straight wire.
  • For a wire:

    B=μ0I2πrθ^\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=mv2RqvB = \frac{mv^2}{R} R=mvqBR = \frac{mv}{qB}
  • Magnetic field lines are directed out of the north pole of a bar magnet and terminate at the south pole. Inside the magnet, they point from the south pole to the north.
  • The North Pole of the Earth is the south pole of Earth’s magnet. The north pole of a compass needly points in the direction of the magnetic field lines.
  • Magnetic dipole moment is given by:

    μIAn^RHR\vec{\mu} \equiv I A \mathbf{\hat{n}}_{\text{RHR}}

    where n^RHR\mathbf{\hat{n}}_{\text{RHR}} is a unit vector perpendicular to the area.

  • I had this video (AP Physics 2: Magnetism 15: Magnetic Dipole Moment and Magnetic Dipole in an External Magnetic Field) labeled with just “big helpful” in my notes.
  • Torque on a magnetic dipole due to external magnetic field:

    τμ×B\vec{\tau} \equiv \vec{\mu} \times \mathbf{\vec{B}}

Ampere’s Law

Sources with enough symmetry:

  1. Infinitely long straight wire
  2. Infinitely long solenoid (tightly wound coil of wires)
  3. Toroid with tightly wrapped turns of wire
  4. Infinite sheet or slab (with a current density)

Faraday’s Law, Lenz’s Law, and Inductance

Back emf

RC, LR, and LC Circuits

  • R: Resistor
  • C: Capacitor
  • L: Inductor

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.

Displacement Current, Waves (Solutions to Maxwell’s Equations), Poynting Vector, Energy Flow

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