Electromagnetism Domain Guide

Concise answer

Electromagnetism is four statements and their consequences: electric field lines begin and end on charge (Gauss), magnetic field lines never end (no monopoles), a changing magnetic flux drives an electric field around a loop (Faraday), and currents plus changing electric flux drive a magnetic field around a loop (Ampere-Maxwell). Electrostatics, circuits, magnetostatics, induction, and light are not five topics but five regimes of that one system, and the exam tests whether you can identify which regime you are in and which symmetry makes the integral trivial.

Definitions

Electric flux: $\Phi_E = \int \vec{E}\cdot d\vec{A}$ , the field-weighted area threaded by the field. Gauss's law says the flux through a closed surface depends only on the charge inside it, so a charge outside contributes exactly zero net flux however close it sits.
Electric potential: $V = U/q$ , the potential energy per unit charge, with $\vec{E} = -\nabla V$ . It is a scalar, so potentials from several charges add arithmetically while fields must be added as vectors — which is why the potential route is almost always faster.
Capacitance: $C = Q/V$ , a purely geometric property: $C = \varepsilon_0 A/d$ for parallel plates, multiplied by the dielectric constant $\kappa$ when the gap is filled. The stored energy is $U = \tfrac{1}{2}CV^{2} = Q^{2}/2C$ .
Electromotive force: The work per unit charge a source supplies around a circuit. It is not a force and not necessarily the terminal voltage: a real source with internal resistance $r$ delivers $V = \mathcal{E} - Ir$ .
Magnetic flux and Lenz's law: $\Phi_B = \int \vec{B}\cdot d\vec{A}$ ; Faraday's law gives $\mathcal{E} = -d\Phi_B/dt$ . The minus sign is Lenz's law: the induced current opposes the change in flux, not the flux itself, so a growing flux and a shrinking flux drive currents in opposite directions.
Poynting vector: $\vec{S} = \dfrac{1}{\mu_0}\vec{E}\times\vec{B}$ , the instantaneous energy flow per unit area of an electromagnetic field. Its time average is the intensity, $I = \tfrac{1}{2}\varepsilon_0 c E_0^{2}$ .

Intuition

Gauss's law is not a way of computing fields; it is a statement that flux counts enclosed charge. It becomes a computational tool only in the three symmetries where $\vec{E}$ is constant in magnitude and perpendicular to a surface you can draw — sphere, cylinder, plane. If the charge distribution has no such symmetry, Gauss's law is still true and still useless, and recognising that instantly is what separates a thirty-second item from a three-minute one.

Circuits are conservation laws in disguise. The junction rule is charge conservation and the loop rule is energy conservation, so a circuit question is never really about circuits — it is about whether you can account for every joule and every coulomb. That framing also explains the capacitor and inductor time constants: $RC$ and $L/R$ are simply how long the stored energy takes to move, and after roughly five time constants the transient is over.

Induction is Newton's third law for fields. Lenz's minus sign is not a convention to memorise; it is energy conservation. If the induced current reinforced the change instead of opposing it, a small disturbance would grow without limit and you would have a free energy source. Any time you are unsure of the direction, compute the mechanical power an external agent must supply and check that it matches $I^{2}R$ — the two must agree exactly, and they never agree if the sign is wrong.

Concept walkthrough

ETS lists electromagnetism as the second-largest content area, and our learning model groups electrostatics, circuits, magnetism, induction, and electromagnetic waves beneath it — again our own reading of the outline, not an official sub-division. Treat it as a single subject with a shared opening question: is the field static, is it carried by a steady current, or is something changing in time? The answer selects which of Maxwell's equations survives, and the surviving equation is usually enough.

Electrostatics begins with Coulomb's law, $F = kq_1q_2/r^{2}$ with $k = 1/4\pi\varepsilon_0 \approx 8.99\times10^{9}\ \mathrm{N\,m^{2}/C^{2}}$ , but the exam rewards the three Gauss's-law results far more often than the integral. For a spherically symmetric charge, the field outside is exactly that of a point charge at the centre and the field inside a uniformly charged insulating sphere grows linearly, $E = kQr/R^{3}$ . For an infinite line of charge, $E = \lambda/2\pi\varepsilon_0 r$ . For an infinite non-conducting sheet, $E = \sigma/2\varepsilon_0$ , independent of distance — and note the contrast that generates wrong answers: just outside the surface of a conductor the field is $\sigma/\varepsilon_0$ , twice as large, because the conductor's charge sits on one face with zero field behind it.

Potential is the labour-saving device. Because $V = kq/r$ is a scalar, the potential of a charge assembly is a sum with no components, and the field then follows from $\vec{E} = -\nabla V$ . Two consequences are worth having ready: the interior of a conductor in electrostatic equilibrium is an equipotential with $\vec{E} = 0$ and all excess charge on the surface, and the work to move a charge between two points is $W = q\,\Delta V$ regardless of path. A cavity inside a conductor with no enclosed charge has zero field — electrostatic shielding — which is a Gauss's-law consequence, not a separate rule.

Circuits are where most of the domain's items live because they are quick to state. Series resistances add, parallel resistances add reciprocally, and capacitors do exactly the reverse: capacitors in parallel add, capacitors in series add reciprocally. Power is $P = IV = I^{2}R = V^{2}/R$ — pick the form whose two quantities you already know. In transient problems the time constants are $\tau = RC$ for a capacitor and $\tau = L/R$ for an inductor, with the universal behaviour that a capacitor behaves as a wire the instant it starts charging and as a break once fully charged, while an inductor does the opposite: a break at the first instant, a wire in the steady state. Those two sentences answer most 'immediately after the switch closes' questions without any exponential at all.

Magnetostatics is dominated by three field results and one force law. A long straight wire gives $B = \mu_0 I/2\pi r$ ; a solenoid gives $B = \mu_0 n I$ with $n$ turns per unit length, not total turns; a circular loop gives $B = \mu_0 I/2R$ at its centre. The force is $\vec{F} = q\vec{v}\times\vec{B}$ , always perpendicular to the velocity, so a magnetic field can change a charge's direction but never its speed, and never its kinetic energy. That gives circular motion of radius $r = mv/qB$ at cyclotron angular frequency $\omega_c = qB/m$ — a frequency independent of speed and radius in the non-relativistic regime, which is the fact that makes a cyclotron work.

Induction and waves close the loop. Faraday's law, $\mathcal{E} = -d\Phi_B/dt$ , has three ways to make the flux change — change $B$ , change the area, or rotate the loop — and the third gives $\mathcal{E} = NBA\omega\sin(\omega t)$ , the generator. A bar sliding on rails at speed $v$ produces the motional EMF $\mathcal{E} = BLv$ . Adding Maxwell's displacement-current term to Ampere's law completes the set and predicts self-propagating waves at $c = 1/\sqrt{\mu_0\varepsilon_0} \approx 3.00\times10^{8}\ \mathrm{m/s}$ , with $\vec{E}$ , $\vec{B}$ , and the propagation direction mutually perpendicular, oscillating in phase, and related by $E_0 = cB_0$ . The time-averaged intensity is $I = \tfrac{1}{2}\varepsilon_0 cE_0^{2}$ , and the radiation pressure on a perfect absorber is $I/c$ , doubling to $2I/c$ on a perfect reflector.

After this page, you should be able to

Decide in one glance whether a field problem has spherical, cylindrical, or planar symmetry, and apply Gauss's law only when it does.
Move fluently between $\vec{E}$ , $V$ , and $U$ , using the scalar potential whenever superposition would otherwise require vector addition.
Reduce any resistor network to one equivalent resistance, then walk back through it to recover branch currents and voltages, and apply Kirchhoff's rules when the network is not reducible.
Compute magnetic forces with $\vec{F} = q\vec{v}\times\vec{B}$ and $\vec{F} = I\vec{L}\times\vec{B}$ , and use the fact that magnetic forces do no work as a consistency check.
Apply Faraday's and Lenz's laws to a moving conductor and verify the answer with an independent energy-balance calculation.
State the relationships in an electromagnetic wave — $E \perp B \perp$ propagation, in phase, $E_0 = cB_0$ , $c = 1/\sqrt{\mu_0\varepsilon_0}$ — and compute intensity and radiation pressure from them.

Formulas and assumptions

Gauss's law and its three standard results

\oint \vec{E}\cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}, \quad E_{\text{sphere}} = \frac{kQ}{r^{2}}, \quad E_{\text{line}} = \frac{\lambda}{2\pi\varepsilon_0 r}, \quad E_{\text{sheet}} = \frac{\sigma}{2\varepsilon_0}, \quad E_{\text{conductor}} = \frac{\sigma}{\varepsilon_0}

Variables

Q_enc: charge inside the closed surface only
lambda: charge per unit length
sigma: charge per unit area

Assumptions

The law is always true but only computable when spherical, cylindrical, or planar symmetry makes E constant over the surface.
Charge outside the surface contributes zero net flux, however close it is.

Potential, field, and stored energy

V = \frac{kq}{r}, \quad E_x = -\frac{\partial V}{\partial x}, \quad W = q\Delta V, \quad C = \frac{\kappa\varepsilon_0 A}{d}, \quad U = \tfrac{1}{2}CV^{2} = \frac{Q^{2}}{2C}

Variables

V: potential, a scalar that superposes arithmetically
kappa: dielectric constant, at least 1
d: plate separation

Assumptions

V = kq/r takes the zero of potential at infinity, which fails for an infinite charge distribution.
Whether U rises or falls when a dielectric is inserted depends on whether the battery stays connected (V fixed) or is removed (Q fixed).

Resistor and capacitor combinations, and power

R_{\text{series}} = \sum R_i, \quad \frac{1}{R_{\text{par}}} = \sum \frac{1}{R_i}, \quad C_{\text{par}} = \sum C_i, \quad \frac{1}{C_{\text{series}}} = \sum \frac{1}{C_i}, \quad P = IV = I^{2}R = \frac{V^{2}}{R}

Variables

R: resistance
C: capacitance
P: power dissipated or delivered

Assumptions

Series elements share the same current; parallel elements share the same voltage. Everything else follows from that.
Capacitors combine oppositely to resistors, which is the single most reversed pair of rules in the domain.

Transients: what a capacitor and an inductor do at t = 0 and t = infinity

\tau_{RC} = RC, \quad \tau_{LR} = \frac{L}{R}, \quad q(t) = Q\left(1 - e^{-t/\tau}\right), \quad q(t) = Qe^{-t/\tau}

Variables

tau: time constant, the time to reach 1 - 1/e of the final value
L: inductance
Q: final (or initial) charge

Assumptions

The exponential describes a single-loop RC or LR circuit; a network must first be reduced to its Thevenin equivalent.
After about five time constants the transient is complete to better than 1%.

Magnetic fields of standard current geometries

B_{\text{wire}} = \frac{\mu_0 I}{2\pi r}, \quad B_{\text{solenoid}} = \mu_0 nI, \quad B_{\text{loop centre}} = \frac{\mu_0 I}{2R}, \quad \oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}

Variables

mu0: 4 pi x 10^-7 T m / A
n: turns per unit length, N / length
I_enc: current threading the Amperian loop

Assumptions

The solenoid result is the ideal long-solenoid interior field; it uses n, not the total turn count N.
Ampere's law in this form omits the displacement-current term, which matters whenever the electric flux is changing.

Magnetic force and circular motion

\vec{F} = q\vec{v}\times\vec{B}, \quad \vec{F} = I\vec{L}\times\vec{B}, \quad r = \frac{mv}{qB}, \quad \omega_c = \frac{qB}{m}

Variables

theta: angle between the velocity (or wire) and the field
r: radius of the circular path
omega_c: cyclotron angular frequency

Assumptions

The magnetic force is always perpendicular to the velocity, so it does no work and cannot change the speed.
The cyclotron frequency is independent of speed and radius only in the non-relativistic limit.

Faraday's law, Lenz's law, and motional EMF

\mathcal{E} = -N\frac{d\Phi_B}{dt}, \quad \Phi_B = BA\cos\theta, \quad \mathcal{E} = BLv, \quad \mathcal{E} = NBA\omega\sin(\omega t), \quad U = \tfrac{1}{2}LI^{2}

Variables

N: number of turns
theta: angle between B and the surface normal
L: self-inductance

Assumptions

The minus sign means the induced current opposes the change in flux, not the flux itself.
Flux can change through B, through area, or through orientation; identify which before differentiating.

Electromagnetic waves

c = \frac{1}{\sqrt{\mu_0\varepsilon_0}}, \quad E_0 = cB_0, \quad I = \tfrac{1}{2}\varepsilon_0 cE_0^{2}, \quad P_{\text{rad}} = \frac{I}{c}\ \text{or}\ \frac{2I}{c}

Variables

E0, B0: peak field amplitudes
I: time-averaged intensity in W/m^2
P_rad: radiation pressure

Assumptions

These are vacuum relations; in a medium of refractive index n the speed is c/n and the wavelength shortens accordingly.
B0 = E0/c makes the magnetic amplitude numerically tiny in SI units, which does not mean the magnetic field is unimportant.

Worked example

A sliding bar, checked twice: Faraday first, then energy

A conducting bar of length $L = 0.50\ \mathrm{m}$ slides at constant speed $v = 4.0\ \mathrm{m/s}$ along frictionless rails closed by a resistor $R = 2.0\ \Omega$ . A uniform field $B = 0.30\ \mathrm{T}$ is perpendicular to the plane of the circuit. Find the induced EMF, the current, the force needed to keep the bar moving, and the power the external agent supplies — then verify the last number independently.

1Identify what makes the flux change. Neither $B$ nor the orientation is changing; the enclosed area is, at rate $dA/dt = Lv$ . So $|\mathcal{E}| = B\,dA/dt = BLv$ , which is the motional-EMF result rather than a separate formula.
2Compute the EMF: $\mathcal{E} = (0.30\ \mathrm{T})(0.50\ \mathrm{m})(4.0\ \mathrm{m/s}) = 0.60\ \mathrm{V}$ . Check the units: $\mathrm{T\cdot m\cdot m/s} = \mathrm{Wb/s} = \mathrm{V}$ .
3Compute the current from Ohm's law for the loop: $I = \mathcal{E}/R = 0.60/2.0 = 0.30\ \mathrm{A}$ . Lenz's law fixes its direction: the flux is growing, so the induced current circulates to oppose the increase.
4Find the force on the bar. The induced current sits in the same field, so it feels $F = BIL = (0.30)(0.30)(0.50) = 0.045\ \mathrm{N}$ , directed opposite to the motion — again Lenz's law, now as a mechanical statement. To keep the speed constant the external agent must supply exactly $0.045\ \mathrm{N}$ .
5Compute the mechanical power supplied: $P = Fv = (0.045\ \mathrm{N})(4.0\ \mathrm{m/s}) = 0.18\ \mathrm{W}$ .
6Verify with an independent route. The resistor dissipates $P = I^{2}R = (0.30)^{2}(2.0) = 0.18\ \mathrm{W}$ , and equivalently $\mathcal{E}^{2}/R = 0.36/2.0 = 0.18\ \mathrm{W}$ . The two agree exactly, which is the content of the minus sign in Faraday's law: every joule of electrical energy came from the agent pushing the bar. Had Lenz's sign been reversed, the force would have pushed the bar along and the circuit would have produced energy from nothing.

$\mathcal{E} = 0.60\ \mathrm{V}$ , $I = 0.30\ \mathrm{A}$ , $F = 0.045\ \mathrm{N}$ opposing the motion, and $P = 0.18\ \mathrm{W}$ supplied mechanically and dissipated resistively. Doubling $v$ doubles the EMF and the current, so it quadruples the power — the drag force is proportional to $v$ , exactly like a viscous damper.

Common traps

Using Gauss's law where the symmetry does not exist. It remains true for a dipole or a finite rod, and remains useless: the flux integral cannot be factored unless $E$ is constant over the surface.
Confusing $\sigma/2\varepsilon_0$ (infinite sheet of charge) with $\sigma/\varepsilon_0$ (just outside a conductor's surface). The factor of two is the whole distinction between the two configurations.
Assuming the field inside any charged object is zero. It is zero inside a conductor in electrostatic equilibrium, but inside a uniformly charged insulating sphere it grows linearly with $r$ .
Adding parallel resistors directly, or forgetting to invert the reciprocal sum at the end. A parallel combination is always smaller than the smallest branch — use that as an instant sanity check.
Combining capacitors with the resistor rules. Capacitors in parallel add; capacitors in series add reciprocally, which is the reverse of resistors.
Reading Lenz's law as 'the induced current opposes the flux'. It opposes the change in flux, so a decreasing flux drives a current that reinforces the existing field.
Putting the magnetic force into a work-energy equation. $\vec{F} = q\vec{v}\times\vec{B}$ is always perpendicular to $\vec{v}$ , so it does no work and never changes the speed.
Using the total number of turns $N$ in $B = \mu_0 nI$ for a solenoid. The formula takes turns per unit length; using $N$ inflates the field by the solenoid's length in metres.
Treating the cyclotron radius and frequency as varying together. The radius $r = mv/qB$ grows with speed while $\omega_c = qB/m$ does not — that separation is what a cyclotron exploits.
Assuming the terminal voltage of a battery equals its EMF. With internal resistance $r$ it is $\mathcal{E} - Ir$ , which is why a large current makes a real battery's terminal voltage sag.

Gauss's Law and Sphere Radius Series-Parallel Branch Current DC Circuits Gauss's Law and Electric Fields GRE Physics overview Optics and Waves Domain Guide GRE Physics reference sheet Sign in for current practice

Question depth and domain coverage vary by exam. Practice answers are checked after submission.

Sources

GRE Subject Test Content and Structure — ETS. Accessed 2026-07-06. Use as a cited source for exam facts; do not imply affiliation or reproduce protected test material.
University Physics Volume 2, Section 6.2: Explaining Gauss's Law — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 2, Section 6.3: Applying Gauss's Law — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 2, Section 6.4: Conductors in Electrostatic Equilibrium — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 2, Section 7.4: Determining Field from Potential — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 2, Section 9.4: Ohm's Law — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 2, Section 10.2: Resistors in Series and Parallel — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 2, Section 10.3: Kirchhoff's Rules — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 2, Section 10.5: RC Circuits — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.

Sources and corrections

Sources last checked 2026-08-15

Every source cited on this page was checked on the date shown, and we update the page when a source changes. If something looks wrong, tell us and we'll recheck it.

Report an issue with this page

Electromagnetism Domain Guide

Concise answer

Definitions

Intuition

Concept walkthrough

After this page, you should be able to

Formulas and assumptions

Gauss's law and its three standard results

Potential, field, and stored energy

Resistor and capacitor combinations, and power

Transients: what a capacitor and an inductor do at t = 0 and t = infinity

Magnetic fields of standard current geometries

Magnetic force and circular motion

Faraday's law, Lenz's law, and motional EMF

Electromagnetic waves

Worked example

A sliding bar, checked twice: Faraday first, then energy

Common traps

Related pages and practice

Sources

Sources and corrections