Electromagnetism

Electric Field

Define the electric field as force per unit charge, build it from point charges, and explore superposition and field lines.

Gauss’ Law

Understand electric flux, relate closed surfaces to enclosed charge, and use symmetry to find the field.

Electric Potential

Move from force to energy: potential energy landscapes, the electric potential, and its relationship to the field.

Electric Current

Here

Follow charges into motion: electron drift, the surface charge that steers the field, Ohm’s law, and circuits.

Electric Current

The microscopic motion of charges and the macroscopic behavior of circuits are two views of the same physics. Inside a metal, a sea of conduction electrons collides with the lattice while being nudged along by an electric field. In a circuit, components shape that field to control current and energy transfer. Bridging the two pictures is the essence of Ohm’s law.

In the classical Drude model, a metal hosts a dense gas of conduction electrons that ricochet off the ionic lattice. Each collision resets an electron’s random thermal velocity, but an applied field biases the motion into a slow net drift.

The drift velocity can be estimated from the average time between collisions $\tau$ and the field $E$ :

v_d = -\frac{eE\tau}{m_e},

where $e$ and $m_e$ are the electron charge and mass. The negative sign means electrons drift opposite the field. The drift speed is tiny compared with the random thermal speed (about $10^5 \text{m/s}$ at room temperature; the visualization above is very exaggerated). In copper, with $\tau \approx 2.5\times10^{-14}\,\text{s}$ and a field of $1.0\,\text{V/m}$ , the drift speed is only about $0.4\,\text{mm/s}$ .

How does this connect to Ohm's law?

Define the conductivity $\sigma$ and the current density $J = I/A$ . Higher conductivity means more current for a given field, and current density is current per cross-sectional area. They are related by

J = \sigma E.

For a wire of length $L$ in a constant field, the potential difference is $\Delta V = EL$ , so

J = \sigma E = \sigma\,\frac{\Delta V}{L} \quad\Longrightarrow\quad \Delta V = \frac{L}{\sigma A}\,I.

This is Ohm’s law with resistance $R = \dfrac{L}{\sigma A}$ .

To connect back to drift, a geometric argument gives the current density as $J = -n e v_d$ , where $n$ is the electron number density. Combining with the drift velocity,

v_d = -\frac{eE\tau}{m_e} \quad\Longrightarrow\quad J = \frac{n e^2 \tau}{m_e}\,E,

so the conductivity is

\sigma = \frac{n e^2 \tau}{m_e}.

Materials with more conduction electrons or longer times between collisions conduct better.

The Drude model is a useful first picture, but it has limits: it treats electrons as classical particles and ignores quantum effects such as the Pauli exclusion principle. Explaining superconductivity or the temperature dependence of resistivity needs more.

Polarization Dynamics

Before wiring anything up, let’s consider the simpler case of a neutral conducting block placed in a uniform electric field. Its conduction electrons are free to move, so they drift until the net field within the conductor is zero.

As these electrons move, the sides of the block acquire a net positive or net negative charge. We say a nonzero surface charge distribution has formed. Note that electrons did not move from one side to the other; rather, the entire population of conduction electrons simply shifted ever so slightly. Furthermore, this shift isn’t really caused by the electrons pushing each other like a packed crowd of people. The dynamics are determined by the electric field.

Time (t / τ)0.00

E inside / E₀ = 1.000face charge = 0.000ε₀E₀

Electrons drift against E₀ until the two charged faces produce exactly the field needed to cancel it, after which nothing moves.

How quickly does a conductor screen a field?

Charge arriving at a face is just the conduction current that got there, so the surface charge density $\sigma$ grows at the rate

\frac{d\sigma}{dt} = J = \sigma_c E_{\text{in}},

writing $\sigma_c$ for the conductivity to keep it apart from the surface charge. Meanwhile the two charged faces act as a parallel-plate pair, pulling the interior field down from its applied value:

E_{\text{in}} = E_0 - \frac{\sigma}{\varepsilon_0}.

Eliminating $E_{\text{in}}$ leaves a single relaxation equation,

\frac{d\sigma}{dt} = \frac{\sigma_c}{\varepsilon_0}\left(\varepsilon_0 E_0 - \sigma\right) \quad\Longrightarrow\quad \sigma(t) = \varepsilon_0 E_0\left(1 - e^{-t/\tau}\right), \qquad \tau = \frac{\varepsilon_0}{\sigma_c}.

The interior field decays as $E_0 e^{-t/\tau}$ . This dielectric relaxation time is startlingly short: copper’s $\sigma_c = 5.8\times10^{7}\,\text{S/m}$ gives $\tau = 1.5\times10^{-19}\,\text{s}$ .

Note this number is far below the collision time $\tau_{\text{coll}} \approx 2.5\times10^{-14}\,\text{s}$ that the Drude model itself depends on, so electrons cannot possibly redistribute that fast. The true settling in a good metal is governed by the collision time and by plasma oscillations of the electron gas, and takes something closer to $10^{-14}\,\text{s}$ . In any case, these surface charges form on a very fast time scale.

Surface Charge Around a Circuit

A wire carrying a steady current is not in electrostatic equilibrium. The electric field inside it is small, but it is not zero. That field cannot be coming directly from the battery. A battery’s own field falls off with distance and has no way to suddenly turn a corner as a wire bends. It instead comes from the surface charge distribution of the circuit.

Start with the switch open. Every conductor is in static equilibrium, but that does not mean the surfaces are bare. Each branch is still wired to a battery terminal, so each branch sits at that terminal’s potential, and its surface carries some charge.

Close the switch, and relaxation dynamics similar to the polarized block occur. However, this time it cannot finish: the battery keeps re-supplying charge to its terminals as electrons begin drifting through the circuit. Instead of settling into a zero field equilibrium, the surfaces acquire a gradient of charge, one that leaves a small electric field pointing along the wire at every point on the loop.

Resistor R (Ω)6.0Drift electrons

I through R = 0.00AΔV across R = 0.00VΔV across switch = 6.00V

Charges indicate the relative surface charge density along the circuit elements and the arrows are the resulting electric field. The small blue dots are conduction electrons, drifting the opposite way to the conventional current.

Surface charge density tracks the local potential. It is strongly positive near the positive terminal, passing through zero somewhere around the loop, and strongly negative near the negative terminal.
Its gradient sets the interior field, so the charge piles up most steeply where the field has to be strongest. A resistor needs a far larger field than the wire to carry the same current, so a significant change in surface charge happens across it.

Macroscopic Circuits

The previous section gives a taste of what it means for a circuit to guide the electric field. However, once the circuit has settled, following the charge in that much detail is rarely worth the trouble. It is simpler to track the potential differences across components, since a charge carrier’s potential energy changes as it moves with or against the field.

A battery provides a potential difference that drives the current. Resistors, bulbs, and other components limit and direct that current, each with its own voltage drop. Applying Ohm’s law to each component, together with how the components are wired, predicts the current in every branch.

Build your own circuit below. Drag components from the palette onto the grid, wire them together, and run the simulation. Beyond resistors and batteries, you can add capacitors, inductors, and switches.

Circuit Builder

Drag components onto the grid, wire them up, and run a transient simulation with a live voltage/current scope. Use Fullscreen for more room.

How does energy flow in a circuit?

Power is the rate of change of potential energy. For a charge $q$ crossing a potential difference $\Delta V$ , the energy change is $\Delta U = q\,\Delta V$ . If that charge crosses in time $\Delta t$ ,

P = \frac{\Delta U}{\Delta t} = \frac{q\,\Delta V}{\Delta t} = I\,\Delta V,

using $I = \dfrac{q}{\Delta t}$ . Components with a larger voltage drop or more current dissipate more power. For a resistor, $\Delta V = IR$ , so

P = I\,\Delta V = I^2 R,

which is converted into thermal energy, heating the resistor.

Problem Solving

Drift Speed in Copper

Problem

A copper wire $0.10\,\text{m}$ long with cross-section $1.0\,\text{mm}^2$ sits in a field of $2.0\,\text{V/m}$ . Estimate the drift speed if $\tau = 2.5\times10^{-14}\,\text{s}$ .

Solve

With $e = 1.60\times10^{-19}\,\text{C}$ and $m_e = 9.11\times10^{-31}\,\text{kg}$ ,

v_d = \frac{eE\tau}{m_e} = \frac{(1.60\times10^{-19})(2.0)(2.5\times10^{-14})}{9.11\times10^{-31}} \approx 8.8\times10^{-3}\,\text{m/s}.

Resistance and Current

Problem

Copper has conductivity $5.8\times10^7\,\text{S/m}$ . Find the resistance of the wire above, then estimate the current from the field using Ohm’s law.

Solve

R = \frac{L}{\sigma A} = \frac{0.10}{(5.8\times10^7)(1.0\times10^{-6})} \approx 1.7\times10^{-3}\,\Omega.

The potential difference is $\Delta V = EL = (2.0)(0.10) = 0.20\,\text{V}$ , so

I = \frac{\Delta V}{R} = \frac{0.20}{1.7\times10^{-3}} \approx 120\,\text{A}.

Series vs Parallel

Problem

Two resistors, $30\,\Omega$ and $60\,\Omega$ , are connected across a $12\,\text{V}$ battery. Find the total current if they are (a) in series and (b) in parallel.

Solve

(a) Series: $R_{eq} = 30 + 60 = 90\,\Omega$ , so $I = \dfrac{12}{90} \approx 0.13\,\text{A}$ .

(b) Parallel: $\dfrac{1}{R_{eq}} = \dfrac{1}{30} + \dfrac{1}{60} = \dfrac{1}{20}$ , so $R_{eq} = 20\,\Omega$ and $I = \dfrac{12}{20} = 0.60\,\text{A}$ .

Electric Current Checkpoint

In the Drude model, which way do conduction electrons drift relative to the applied electric field?