IB Physics HL Electric & Magnetic Fields — Complete Cheatsheet
Every formula, field-line rule, trick and trap for IB Physics HL Topic D.2 — Coulomb's law, electric and magnetic fields, Millikan's experiment, electric potential, and the gravitational-vs-electric comparison. Hand-built by an IBO-certified Singapore tutor.
Topic: D.2 Electric & Magnetic FieldsSyllabus: SL + AHL Topic D.2Read time: ~14 minutesLast updated: Apr 2026
Electric and magnetic fields is the central topic of IB Physics HL Theme D — the bridge between gravitational fields (D.1) and the more demanding D.3 motion-in-EM-fields and D.4 induction. The mathematical structure is identical to gravitation (inverse-square force, $1/r$ potential), but the sign conventions create the most common arithmetic mistakes in the entire HL course. Every IB session pairs a long Paper 2 question with a tricky Paper 1 multiple choice on field lines, conductors or potential.
This cheatsheet condenses every formula, field-line rule, and exam-mark trap from D.2 (SL core and AHL extension) into one revision page. Scroll to the bottom for the printable PDF, the full Notes, the Tutorial booklet, and the marked-up solutions in the gated full library.
§1 — Electric Charge & Coulomb's Law D.2
Key formulas
Coulomb's law:$\displaystyle F = k\frac{q_1 q_2}{r^2} = \frac{q_1 q_2}{4\pi\varepsilon_0 r^2}$
TrickCoulomb's law has the same form as Newton's law of gravitation. Replace $G \to k$, $m \to q$. But gravity is always attractive; the electric force can attract or repel.
TrapAlways use magnitudes of the charges in Coulomb's law and decide direction (attract / repel) by inspection. A negative result does NOT mean "attractive" in IB mark schemes — write the direction in words.
NoteCharge is conserved (cannot be created or destroyed) and quantised (always $q = ne$ for some integer $n$).
Coulomb's law. Like charges repel and unlike charges attract with equal, opposite forces of size $F=kq_1q_2/r^2$ — the same inverse-square form as gravity.
From the Photon question bank
What examiners actually test on this concept:
F = kq₁q₂/r²Use magnitudes + directionCharge quantised q = neSame form as gravity
Coulomb's law mirrors Newton's gravity ($G\to k$, $m\to q$) but can attract OR repel — use magnitudes and state the direction in words. frequent
A negative result doesn't mean "attractive" in IB markschemes — describe the direction explicitly.
Charge is conserved and quantised: $q=ne$, with $e=1.60\times10^{-19}$ C.
§2 — Electric Fields D.2
Key formulas
Electric field:$\displaystyle E = \frac{F}{q}$ [$\mathrm{N\,C^{-1}} = \mathrm{V\,m^{-1}}$]
Point charge:$\displaystyle E = \frac{kQ}{r^2}$
Parallel plates:$\displaystyle E = \frac{V}{d}$ (uniform field only)
Field line rules
Lines run away from $+$, toward $-$.
Lines never cross.
Density of lines $\propto$ field strength.
Field lines are perpendicular to a conductor's surface.
Inside a conductor in equilibrium: $E = 0$.
TrickThe electric field is a vector. To find the net field due to several charges, resolve each contribution into $x$ and $y$ components and add as vectors. Potential is a scalar — much easier to add.
Trap$E = V/d$ applies only to uniform fields (parallel plates). Do NOT use this formula for the field around a point charge or charged sphere.
Electric fields. A point charge has radial field lines ($E=kQ/r^2$); between parallel plates the field is uniform ($E=V/d$).
Result: Millikan found $q = ne$ for every drop, proving the quantisation of charge.
TrickTo find the integer $n$: calculate $q$ from the balance condition, then $n = q/e$. Always round to the nearest integer.
TrapIf the question requires you to include buoyancy: $qE = (\rho_{\text{oil}} - \rho_{\text{air}})V g$, where $V$ here is the volume of the droplet, not the voltage. Don't confuse volume $V$ with voltage $V$ in the same equation.
Millikan's oil-drop experiment. A charged drop hangs still when the upward electric force balances gravity: $qE=mg$, so $q=mgd/V$.
From the Photon question bank
What examiners actually test on this concept:
qE = mgq = mgd/Vn = q/eBuoyancy variant
Balanced drop: $qE=mg$ with $E=V/d$ gives $q=mgd/V$; then $n=q/e$ (rounded) proves charge is quantised. frequent
If buoyancy is included: $qE=(\rho_{\text{oil}}-\rho_{\text{air}})Vg$ — here $V$ is the drop VOLUME, not the voltage.
Millikan's $q=ne$ result established the elementary charge $e$.
§4 — Magnetic Force & Fields D.2
Key facts
$\vec{B}$ field lines form closed loops (N $\to$ S outside the magnet, S $\to$ N inside).
No magnetic monopoles exist.
The magnetic force acts only on moving charges.
The magnetic force is always $\perp$ to the velocity, so it does no work.
Field patterns to memorise
Configuration
Field pattern
Long straight wire
Concentric circles around the wire — right-hand grip rule
Flat coil
Looks like a short bar magnet
Solenoid
Uniform inside; bar-magnet shape outside
Bar magnet
N to S outside; closed loops inside
TrickRight-hand grip rule. Wire: thumb = current direction, fingers curl in the direction of $\vec{B}$. Solenoid: fingers curl with the current, thumb = N pole.
Magnetic field patterns. Concentric circles around a straight wire (right-hand grip), and a uniform field inside a solenoid with bar-magnet poles outside.
From the Photon question bank
What examiners actually test on this concept:
B lines are closed loopsForce on moving charges onlyNo work (⊥ to v)Right-hand grip rule
The magnetic force acts only on MOVING charges, is ⊥ to $v$, and does NO work. $B$ field lines are closed loops (no monopoles). frequent
Right-hand grip — wire: thumb along $I$, fingers curl with $B$; solenoid: fingers with the current, thumb = N pole.
Solenoid: uniform field inside, bar-magnet field outside.
Electric vs magnetic fields — at a glance
Electric
Magnetic
Source
Charges
Moving charges / currents
Acts on
Any charge
Moving charges only
Force direction
Parallel to $\vec{E}$
$\perp$ to both $\vec{v}$ and $\vec{B}$
Does work?
Yes
No
Monopoles?
Yes
No
§5 — Electric Potential Energy & Potential D.2 AHL
Key formulas
Electric potential:$\displaystyle V = \frac{W}{q} = \frac{kQ}{r}$ [V]
Sign of $V$:$V > 0$ near $+Q$, $V < 0$ near $-Q$, $V = 0$ at $\infty$
Superposition:$\displaystyle V_{\text{total}} = \sum \frac{kQ_i}{r_i}$ (scalar sum — no vectors!)
Field from $V$:$\displaystyle E = -\frac{\Delta V}{\Delta r}$ (negative gradient)
Conducting sphere of radius $R$ carrying charge $Q$
Outside ($r > R$): $V = kQ/r$
Surface ($r = R$): $V = kQ/R$
Inside ($r < R$): $V = kQ/R$ (constant — same as the surface)
TrickPotential is a scalar; field is a vector. If a question asks for the potential at a point, just add $kQ_i/r_i$ with signs. No component resolution needed.
TrapInside a conducting sphere, $E = 0$ but $V \neq 0$. The potential equals the surface potential. Students often write $V = 0$ inside — this is wrong and loses marks.
NoteEquipotentials are perpendicular to field lines. Closely spaced equipotentials = strong field.
Electric potential. $V=kQ/r$ for a point charge; for a charged conducting sphere $V$ is constant inside (equal to the surface value) even though $E=0$ there.
From the Photon question bank
What examiners actually test on this concept:
V = kQ/rScalar sum with signsE = −ΔV/ΔrInside conductor V ≠ 0
Potential is a SCALAR ($V=kQ/r$) — add algebraically with signs, no components; the sign follows $Q$. frequent (AHL)
Inside a conductor $E=0$ but $V\ne0$ (it equals the surface potential) — a common lost mark.
$E$ is the negative gradient of $V$; equipotentials are ⊥ to field lines.
§6 — Gravitational vs Electric Potentials D.2 AHL
Parallel equations
Gravitational
Electric
Force
$F = \dfrac{GMm}{r^2}$
$F = \dfrac{kQq}{r^2}$
Field
$g = \dfrac{GM}{r^2}$
$E = \dfrac{kQ}{r^2}$
Potential
$V_g = -\dfrac{GM}{r}$
$V_e = \dfrac{kQ}{r}$
Potential energy
$E_p = -\dfrac{GMm}{r}$
$E_p = \dfrac{kQq}{r}$
TrickGravitational potential is always negative (mass attracts). Electric potential can be $+$ or $-$ depending on the sign of $Q$.
From the Photon question bank
What examiners actually test on this concept:
Same 1/r² structureV_g always −, V_e ±ShieldingeV conversion
Identical maths ($G\leftrightarrow k$, $m\leftrightarrow q$): inverse-square force, $1/r$ potential, $E=-\mathrm{d}V/\mathrm{d}r$ — only the SIGN differs. frequent
Gravitational potential is always negative; electric can be $+$ or $-$. Electric fields can be shielded (Faraday cage); gravity can't.
$1\,\text{eV}=1.60\times10^{-19}$ J; to convert J → eV DIVIDE by $e$ (never multiply).
Electronvolt
$1\;\mathrm{eV} = 1.60 \times 10^{-19}\;\mathrm{J}$ — the kinetic energy gained by an electron accelerated through $1\;\mathrm{V}$. Quick conversion: $E_K = qV$ gives the energy in eV directly when $q$ is expressed in units of $e$ and $V$ in volts.
TrapWhen converting joules to eV, divide by $1.60 \times 10^{-19}$. Students often multiply instead — that error makes a tiny number unmanageably large.
§7 — Three Key Differences D.2 AHL
Sign of force: Gravity always attracts; the electric force can attract or repel.
Sign of potential: $V_g$ is always negative; $V_e$ can be $+$ or $-$.
Shielding: Electric fields can be shielded (Faraday cage); gravitational fields cannot.
NoteThe mathematical structure is identical — inverse-square force, $1/r$ potential, field $= -\mathrm{d}V/\mathrm{d}r$. Only the sign conventions differ, because mass is always positive while charge can be $+$ or $-$.
§8 — Exam Attack Plan All sections
When you see this in the question — reach for that:
Question trigger
Reach for
Two point charges, find force
Coulomb: use magnitudes; state direction in words
"Field at point P" — multiple charges
Vector sum: resolve into $x,y$ components
"Potential at point P" — multiple charges
Scalar sum $\sum kQ_i/r_i$ with signs; no components
Charged droplet stationary between plates
Millikan: $qE = mg$, then $q = mgd/V$
"Sketch field lines" or "Sketch equipotentials"
Apply field-line rules; equipotentials $\perp$ to field lines
Field inside a conductor
$E = 0$, $V$ = surface potential (constant)
Convert energy to eV
Divide by $e = 1.60 \times 10^{-19}$ — never multiply
Compare gravitational and electric
Same structure, $G \leftrightarrow k$, $m \leftrightarrow q$, sign differs
Long straight wire — direction of $\vec{B}$
Right-hand grip: thumb along $I$, fingers curl with $\vec{B}$
Worked Example — IB-Style Two-Charge Field & Potential
Question (HL Paper 2 style — 6 marks)
Two point charges are placed on the $x$-axis: $Q_1 = +3.0\;\mathrm{nC}$ at $x = 0$ and $Q_2 = -3.0\;\mathrm{nC}$ at $x = 0.40\;\mathrm{m}$. Take $k = 8.99 \times 10^9\;\mathrm{N\,m^2\,C^{-2}}$.
(a) Calculate the electric potential at the midpoint $P$ ($x = 0.20\;\mathrm{m}$). (b) Calculate the magnitude of the electric field at $P$.
Solution
Identify: Both distances from $P$ are $r = 0.20\;\mathrm{m}$. (R1)
Field — magnitude from each charge (use $|Q|$): $E_1 = \dfrac{k|Q_1|}{r^2} = \dfrac{(8.99\times10^9)(3.0\times10^{-9})}{(0.20)^2} = 6.74 \times 10^{2}\;\mathrm{N\,C^{-1}}$ (M1)
Direction at $P$: $\vec{E}_1$ points away from $Q_1$ (i.e. in $+x$). $\vec{E}_2$ points toward $Q_2$ (also in $+x$, since $Q_2$ is to the right and negative). Both fields are co-linear and add. (R1)
Net field magnitude: $E_P = E_1 + E_2 = 2 \times 6.74 \times 10^2 = 1.35 \times 10^3\;\mathrm{N\,C^{-1}}$, directed in $+x$. (A1)
Examiner's note: The classic trap here is to assume that "potential = 0" implies "field = 0". They are independent: $V$ can be zero while $E$ is non-zero, and vice versa. Students who skip the explicit direction analysis and just add field magnitudes ignoring signs frequently get $E_P = 0$ and lose all the field marks.
Common Student Questions
When can I use $E = V/d$ in IB Physics?
Only inside a uniform electric field — for example, between two parallel plates. Do not use $E = V/d$ for the field around a point charge or a charged sphere; for those use $E = kQ/r^2$. Mark schemes routinely strip marks for using the parallel-plate formula on a non-uniform field.
Why is electric potential a scalar but electric field a vector?
Potential $V = W/q$ is energy per unit charge — energy has no direction. Field $E = F/q$ is force per unit charge — force has direction. The practical consequence is huge: when many charges contribute, potentials add as simple algebraic sums (with signs), but fields must be added as vectors using component resolution. Always answer the potential part first if both are asked.
Is the electric field zero inside a charged conducting sphere?
Yes — $E = 0$ inside any conductor in electrostatic equilibrium. But the potential $V$ is not zero inside; it is constant and equal to the surface value, $V = kQ/R$. Many students wrongly write "$V = 0$ inside the sphere" — this is wrong and loses marks every year.
Does the magnetic force ever do work on a moving charge?
No. The magnetic force is always perpendicular to the velocity, so it can change the direction of motion (curve the path) but cannot change the speed or kinetic energy. This is why a charged particle in a uniform $\vec{B}$ field travels in a circle at constant speed. Use this fact in circular-motion-in-B-field questions: KE in = KE out.
How do I avoid sign errors in Coulomb's law?
Use magnitudes of the charges in $F = k|q_1||q_2|/r^2$ and decide the direction (attractive or repulsive) by inspection: opposite signs attract, like signs repel. The IB mark scheme does not accept a negative number as a substitute for "attractive". Compute magnitudes, then state the direction in words.
What's NOT in this cheatsheet
This page gives you the formulas and the traps. The full Photon Academy Electric & Magnetic Fields library (only available to enrolled students or via the resource library subscription) adds:
Notes PDF — every concept worked through in full, with derivations and intuition.
Tutorial booklet — IB-style questions sequenced from foundation to AHL difficulty.
Tutorial Solutions — full mark-scheme-style worked solutions with M1/A1/R1 annotations.
Practice Solutions — extra past-paper-style problems with detailed walk-throughs.
Cheatsheet PDF — print-ready, brand-formatted, the same one our students take into mock exams.
