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:$\displaystyle F = k\frac{q_1 q_2}{r^2} = \frac{q_1 q_2}{4\pi\varepsilon_0 r^2}$
$k$:$k = 8.99 \times 10^{9}\;\mathrm{N\,m^2\,C^{-2}}$
$\varepsilon_0$:$\varepsilon_0 = 8.85 \times 10^{-12}\;\mathrm{C^2\,N^{-1}\,m^{-2}}$
Elementary charge:$e = 1.60 \times 10^{-19}\;\mathrm{C}$ (charge is quantised: $q = ne$)
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 $n$).
§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.
§3 — Millikan's Oil-Drop Experiment D.2
Balance condition
For a stationary droplet (ignoring buoyancy), the upward electric force balances gravity:
$$qE = mg \quad \Longrightarrow \quad q = \frac{mgd}{V}$$
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.
§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.
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.
§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 $Q$.
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)
- Potential — scalar sum with signs:
$V_P = \dfrac{kQ_1}{r} + \dfrac{kQ_2}{r} = \dfrac{(8.99\times10^9)(+3.0\times10^{-9})}{0.20} + \dfrac{(8.99\times10^9)(-3.0\times10^{-9})}{0.20}$
$= +135 - 135 = 0\;\mathrm{V}$ (M1)(A1) - 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.
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