Electric and magnetic fields

Lesson D.2 Electric and Magnetic Fields

IB Physics • Standard Level and Higher Level

Electric forces help explain atomic structure, chemical bonding and much of modern technology. This lesson develops the ideas of electric charge, electric force, electric field strength and magnetic field patterns. Higher Level sections connect electric fields to potential, potential energy and work. All derivations use algebra and the IB formulas, without calculus.

Learning goals

Explain charge quantization, charge conservation and charging by friction, contact, induction and grounding.
Apply Coulomb’s law and distinguish attractive from repulsive forces.
Calculate electric field strength and interpret electric field patterns.
Describe magnetic fields around magnets, straight wires, circular coils and solenoids.
Explain how Millikan’s oil-drop experiment supports charge quantization.
HL: Calculate electric potential and potential energy, use equipotentials and interpret potential gradients.

Symbols and useful data

Use $E_p$ for potential energy and $V_e$ for electric potential. $E$ is electric field strength, $B$ is magnetic flux density, $Q$ is a source charge and $q$ is a test charge. In the parallel-plate formula, $V$ denotes the magnitude of the potential difference between the plates.

$e=1.60\times10^{-19}\,\mathrm{C},\qquad k=8.99\times10^9\,\mathrm{N\,m^2\,C^{-2}}$

$\varepsilon_0=8.85\times10^{-12}\,\mathrm{C^2\,N^{-1}\,m^{-2}},\qquad g=9.81\,\mathrm{m\,s^{-2}}$

Unless stated otherwise, use the vacuum value of $k$ for air. Keep extra figures during calculations and give final numerical answers to 2–3 significant figures.

1. Electric charge: quantization and conservation (SL and HL)

Electric charge can be positive or negative. Like charges repel and unlike charges attract. An electron has charge $-e$ and a proton has charge $+e$ . An object is neutral when its total positive and negative charges balance.

Quantization of charge

The net charge on an ordinary isolated object occurs in integer multiples of the elementary charge:

$Q=ne$

Here $n$ is an integer, including zero and negative integers. A negatively charged object has excess electrons; a positively charged object has an electron deficit.

Conservation of charge

The total electric charge of an isolated system remains constant. When objects become charged, charge is transferred or separated. If two initially neutral objects exchange electrons, their final net charges are equal in magnitude and opposite in sign, provided no charge enters or leaves the system.

How objects become charged

Friction: rubbing suitable materials transfers electrons between them.
Contact or conduction: electrons move between objects that touch, if a conducting path is available.
Induction: a nearby charged object causes charge separation without touching the object being charged.
Grounding: a conducting connection allows electrons to flow to or from Earth.

In ordinary charging of solid objects, electrons are transferred; protons remain within atomic nuclei. Conductors allow charge to redistribute readily. Insulators restrict charge movement, although they can still become charged or polarized.

Charging a conductor by induction

To leave an initially neutral conducting sphere positively charged using a negatively charged rod:

Bring the rod close to the sphere without touching it. Electrons move to the far side of the sphere.
Ground the sphere while the rod remains nearby. Some electrons leave the sphere for Earth.
Disconnect the grounding connection first.
Remove the rod. The remaining positive net charge redistributes over the sphere.

Important: charge separation alone does not change the sphere’s net charge. Removing the rod before disconnecting Earth can allow the sphere to become neutral again.

Worked example 1 — Counting transferred electrons

A sphere acquires a charge of $-4.80\times10^{-9}\,\mathrm{C}$ . Determine how many electrons it gained.

Show answer and working

$N=\frac{|Q|}{e}=\frac{4.80\times10^{-9}}{1.60\times10^{-19}}=3.00\times10^{10}$

The sphere gained $3.00\times10^{10}$ electrons. Its negative charge indicates an excess of electrons.

2. Coulomb’s law (SL and HL)

The electrostatic force between two point charges acts along the line joining them. In vacuum, its magnitude is:

$F=k\frac{|q_1q_2|}{r^2},\qquad k=\frac{1}{4\pi\varepsilon_0}$

The IB expression is commonly written $F=k\frac{q_1q_2}{r^2}$ . When calculating a force magnitude, use the magnitudes of the charges and state the direction separately: like charges repel and unlike charges attract. $r$ is the separation between the charges.

This model applies to point charges, or to suitable spherically symmetric charge distributions when considering exterior points. For extended objects, check whether their sizes and charge distributions allow a point-charge approximation.

Inverse-square behaviour and the medium

For fixed charges, doubling the separation reduces the force to one-quarter; tripling the separation reduces it to one-ninth. The spherical symmetry of a point-charge field is consistent with its spreading over surfaces of area $4\pi r^2$ .

The medium also matters. In a homogeneous insulating medium, the permittivity replaces $\varepsilon_0$ . A greater permittivity gives a smaller force between the same fixed charges at the same separation.

Superposition

If several charges are present, calculate the force from each source and add the forces as vectors. Add forces in the same direction and subtract forces in opposite directions. For perpendicular components, use a vector diagram and Pythagoras.

Worked example 2 — Force between unlike charges

Two small charged spheres carry $+2.0\,\mu\mathrm{C}$ and $-3.0\,\mu\mathrm{C}$ . Their centres are $0.15\,\mathrm{m}$ apart. Treat them as point charges and calculate the force on each.

Show answer and working

$F=\frac{(8.99\times10^9)(2.0\times10^{-6})(3.0\times10^{-6})}{(0.15)^2}=2.40\,\mathrm{N}$

Each sphere experiences a force of approximately $2.4\,\mathrm{N}$ towards the other sphere. The forces are equal in magnitude and opposite in direction.

3. Electric field strength (SL and HL)

An electric field is a region in which a charge experiences an electric force. Its direction is the direction of the force on a small positive test charge. Electric field strength is force per unit positive test charge:

$E=\frac{F}{q}$

The vector relationship is $\vec{F}=q\vec{E}$ . A positive charge experiences force in the field direction; a negative charge experiences force in the opposite direction. For magnitudes, use $F=|q|E$ .

$1\,\mathrm{N\,C^{-1}}=1\,\mathrm{V\,m^{-1}}$

Deriving the field of a point charge

For a positive test charge $q$ at distance $r$ from source charge $Q$ , combine Coulomb’s law with the field definition:

$F=k\frac{|Q|q}{r^2},\qquad E=\frac{F}{q}=k\frac{|Q|}{r^2}$

The field magnitude is independent of the test charge. It points away from a positive source and towards a negative source. With outward radial direction defined as positive, the signed radial component is $E_r=k\frac{Q}{r^2}$ .

Electric field lines

Field lines point away from positive charges and towards negative charges; they may also extend to or from infinity.
The tangent to a field line gives the field direction at that point.
Closer line spacing represents a stronger field when the diagram uses a consistent drawing convention.
Field lines never cross, because the field has a single resultant direction at any point.
Two unlike charges form a dipole pattern, with lines directed from the positive charge to the negative charge.
For two like charges, lines bend away from the region between them. Equal like charges have zero resultant field at the midpoint.

The resultant electric field is the vector sum of the fields from all source charges. The field at a point exists whether or not a test charge is placed there.

Uniform fields between parallel plates

Between two large, oppositely charged parallel plates, the field is approximately uniform away from the edges. It points from the positive plate to the negative plate. If the potential difference has magnitude $V$ and the plate separation is $d$ :

$E=\frac{V}{d}$

A positive charge moved slowly against the field across the gap gains potential energy $qV$ . The external work is also $Fd=qEd$ . Equating these magnitudes gives:

$qV=qEd\quad\Rightarrow\quad E=\frac{V}{d}$

The approximation becomes less accurate near the edges, where the field lines curve outwards. These are fringing fields.

Worked example 3 — Parallel plates and an electron

Two plates have a potential difference of $600\,\mathrm{V}$ and a separation of $0.020\,\mathrm{m}$ . Calculate the field strength and the force magnitude on an electron between them. State the force direction.

Show answer and working

$E=\frac{600}{0.020}=3.0\times10^4\,\mathrm{N\,C^{-1}}$

$F=eE=(1.60\times10^{-19})(3.0\times10^4)=4.8\times10^{-15}\,\mathrm{N}$

The electric field points from the positive plate to the negative plate. The electron’s force points towards the positive plate, opposite to the field.

4. Millikan’s oil-drop experiment (SL and HL)

Millikan studied charged oil drops between horizontal plates. By adjusting the electric field, a drop could be held stationary. Repeated measurements showed that drop charges were consistent with integer multiples of the elementary charge.

In the simplified model, neglect air buoyancy and consider a stationary drop. Its upward electric force balances its downward weight:

$|q|E=mg\quad\Rightarrow\quad |q|=\frac{mg}{E}$

The equation gives the charge magnitude. The field direction is chosen so that the electric force acts upwards. For a negatively charged drop, the electric field must point downwards. A more precise analysis accounts for buoyancy and uses drag when the drop is moving.

Worked example 4 — A suspended oil drop

An oil drop of mass $6.40\times10^{-15}\,\mathrm{kg}$ is stationary in an electric field of magnitude $1.31\times10^5\,\mathrm{N\,C^{-1}}$ . Neglect buoyancy. Find its charge magnitude and the likely number of elementary charges.

Show answer and working

$|q|=\frac{(6.40\times10^{-15})(9.81)}{1.31\times10^5}=4.79\times10^{-19}\,\mathrm{C}$

$\frac{|q|}{e}=\frac{4.79\times10^{-19}}{1.60\times10^{-19}}=2.99\approx3$

The charge magnitude is approximately $3e$ . Its sign cannot be determined from the magnitudes alone. If the field points downwards, the suspended drop has charge approximately $-3e$ .

5. Electric potential energy and electric potential (HL)

Electric potential energy

Taking the potential energy to be zero when two point charges are infinitely far apart, their electric potential energy at separation $r$ is:

$E_p=k\frac{q_1q_2}{r}$

Keep the signs of both charges in this equation. Like charges give positive potential energy; unlike charges give negative potential energy. Bringing like charges closer requires external work, while bringing unlike charges closer decreases their electric potential energy.

This expression describes the interaction energy of the pair. If more than two charges are present, add the energy of each distinct pair once.

Electric potential

Electric potential at a point is electric potential energy per unit charge. Equivalently, with zero potential at infinity, it is the external work per unit positive charge required to bring a small test charge slowly from infinity to that point.

$V_e=\frac{E_p}{q}$

For source charge $Q$ and test charge $q$ , substitute $E_p=k\frac{Qq}{r}$ and cancel $q$ :

$V_e=\frac{kQq/r}{q}=k\frac{Q}{r}$

Potential is a scalar, measured in volts: $1\,\mathrm{V}=1\,\mathrm{J\,C^{-1}}$ . A positive source produces positive potential and a negative source produces negative potential, using the zero-at-infinity convention. The sign of the test charge does not change the potential created by the source.

Adding potentials

For several point charges, add the individual potentials algebraically, retaining the source-charge signs:

$V_e=k\frac{Q_1}{r_1}+k\frac{Q_2}{r_2}+\cdots$

At the midpoint between equal unlike charges, the total potential is zero but the electric field is not zero. Between equal like charges, the field is zero at the midpoint but the potential is not zero. Zero potential and zero field are different statements.

Worked example 5 — Potential and potential energy

A source charge is $Q=-3.0\times10^{-9}\,\mathrm{C}$ . Calculate the electric potential $0.20\,\mathrm{m}$ from it and the potential energy of a $+2.0\times10^{-9}\,\mathrm{C}$ test charge placed there.

Show answer and working

$V_e=\frac{(8.99\times10^9)(-3.0\times10^{-9})}{0.20}=-135\,\mathrm{V}$

$E_p=qV_e=(2.0\times10^{-9})(-135)=-2.70\times10^{-7}\,\mathrm{J}$

The potential is negative because the source charge is negative. The potential energy is negative because the two charges have opposite signs. To 2 significant figures, the answers are $-1.3\times10^2\,\mathrm{V}$ and $-2.7\times10^{-7}\,\mathrm{J}$ .

6. Work, energy and potential difference (HL)

For a charge moved from an initial to a final position, define $\Delta V_e=V_{e,\mathrm{final}}-V_{e,\mathrm{initial}}$ . Its potential-energy change is:

$\Delta E_p=q\Delta V_e$

If an external agent moves the charge slowly so that its kinetic energy does not change, the external work is:

$W_{\mathrm{external}}=q\Delta V_e$

The electric field does the opposite amount of work. If only the electric force does work, the kinetic-energy change is:

$W_{\mathrm{field}}=-q\Delta V_e,\qquad \Delta E_k=-q\Delta V_e$

A positive charge released from rest accelerates towards lower potential. A negative charge accelerates towards higher potential. In both cases, the electric potential energy decreases and the kinetic energy increases.

The electronvolt

One electronvolt is the energy gained by a charge of magnitude $e$ moving through a potential difference of magnitude $1\,\mathrm{V}$ in the accelerating direction:

$1\,\mathrm{eV}=1.60\times10^{-19}\,\mathrm{J}$

Worked example 6 — An electron accelerated through a voltage

An electron moves through a potential increase of $200\,\mathrm{V}$ . Assuming only the electric force does work, determine its kinetic-energy gain in joules and electronvolts.

Show answer and working

$\Delta E_k=-q\Delta V_e=-(-1.60\times10^{-19})(200)=3.20\times10^{-17}\,\mathrm{J}$

$\Delta E_k=\frac{3.20\times10^{-17}}{1.60\times10^{-19}}=200\,\mathrm{eV}$

The electron gains $200\,\mathrm{eV}$ of kinetic energy. Its negative charge means that a potential increase corresponds to a potential-energy decrease.

7. Equipotentials and potential gradients (HL)

Equipotential surfaces

An equipotential surface joins points with the same electric potential. Moving a charge along it changes neither its electric potential energy nor the work done by the electric field, because $\Delta V_e=0$ .

Electric field lines are perpendicular to equipotential surfaces.
The electric field points towards decreasing potential.
For a point charge, equipotential surfaces are concentric spheres; a two-dimensional section shows concentric circles.
Between large parallel plates, equipotential surfaces are approximately parallel planes.
For equal potential steps, closer equipotential spacing indicates a stronger field.

The IB potential-gradient relationship

Along a chosen coordinate direction, the signed electric field component is the negative potential gradient:

$E=-\frac{\Delta V_e}{\Delta r}$

In a uniform field, this gives the exact component over a finite interval. In a non-uniform field, the finite ratio gives the average component over that interval; the local field is represented by the negative slope of the tangent to a potential–distance graph. The negative sign means that the field points towards decreasing potential.

For a positive point charge, the potential decreases with increasing distance, so the field points radially outwards. For a negative point charge, the potential increases towards zero with increasing distance, so the field points radially inwards.

Worked example 7 — Reading a potential gradient

Along a straight line in a uniform field, potential decreases from $120\,\mathrm{V}$ to $40\,\mathrm{V}$ as the coordinate increases by $0.040\,\mathrm{m}$ . Determine the field component and its direction.

Show answer and working

$\Delta V_e=40-120=-80\,\mathrm{V}$

$E=-\frac{-80}{0.040}=+2.0\times10^3\,\mathrm{V\,m^{-1}}$

The field has magnitude $2.0\times10^3\,\mathrm{N\,C^{-1}}$ and points in the positive coordinate direction, from the higher-potential position towards the lower-potential position.

8. Charged conductors in electrostatic equilibrium (HL)

In electrostatic equilibrium, the electric field within the conducting material is zero. Otherwise, free charges would continue to move. The conductor has the same potential throughout, and any excess net charge resides on its surface.

Immediately outside a conductor, the electric field is perpendicular to its surface. A component along the surface would make surface charges move. An empty cavity completely enclosed by a conductor also has zero electric field when no charge is enclosed in the cavity.

An isolated charged conducting sphere

For an isolated conducting sphere of radius $R$ and charge $Q$ , with no external fields, the exterior field and potential are the same as those of a point charge at its centre:

$E=k\frac{|Q|}{r^2},\qquad V_e=k\frac{Q}{r}\qquad (r>R)$

Inside the conducting sphere, the field is zero and the potential is constant, equal to its surface value:

$E=0,\qquad V_e=k\frac{Q}{R}\qquad (R>r)$

The same interior result applies to the empty interior of an isolated hollow conducting sphere when there is no charge inside. Zero field does not require zero potential.

9. Magnetic field patterns (SL and HL)

Magnetic fields are produced by electric currents and magnetic materials. Magnetic flux density is represented by $B$ and measured in tesla, $\mathrm{T}$ . A compass can indicate the local magnetic field direction. Magnetic field lines form continuous loops and do not start or end at isolated poles.

Bar magnet

Outside a bar magnet, field lines point from the north pole to the south pole. Inside the magnet, they continue from south to north, forming closed loops. The field is stronger near the poles, where the lines are closer together.

Straight current-carrying wire

The field lines are concentric circles centred on the wire. Use the right-hand grip rule: point your right thumb in the direction of conventional current, and your curled fingers show the magnetic field direction.

A dot represents current coming out of the page; the field is anticlockwise. A cross represents current going into the page; the field is clockwise. Reversing the current reverses the field direction.

Circular coil

A circular current loop produces a field through its centre and a return field around the outside. Curl your right-hand fingers in the current direction; your thumb gives the field direction through the centre. Viewed from one face, anticlockwise current makes that face a north pole; clockwise current makes it a south pole.

Air-core solenoid

A solenoid is a coil with many turns. Inside a long solenoid, away from its ends, field lines are approximately straight, parallel and evenly spaced, so the field is approximately uniform. Outside, the pattern resembles that of a bar magnet. Curl your right-hand fingers in the conventional current direction; your thumb points along the internal field towards the north end.

A larger current strengthens the field. For otherwise similar solenoids, more turns per unit length also strengthens the field. Reversing the current exchanges the north and south ends. Electron flow in a metal is opposite to conventional current, so use conventional current when applying the grip rule.

Electric and magnetic field lines

Electrostatic field lines begin on positive charges and end on negative charges, or extend to infinity. Magnetic field lines form closed loops. In both types of diagram, the tangent gives the direction and closer spacing represents greater field strength when a consistent scale is used.

10. Comparison: gravitational and electric fields

Feature	Gravitational field	Electric field
Source	Mass $M$	Charge $Q$
Force magnitude	$F=G\frac{Mm}{r^2}$	$F=k\frac{\|Qq\|}{r^2}$
Interaction	Attractive between ordinary masses.	Attractive for unlike charges; repulsive for like charges.
Field definition	$g=\frac{F}{m}$	$E=\frac{F}{q}$
Point-source field magnitude	$g=\frac{GM}{r^2}$	$E=k\frac{\|Q\|}{r^2}$
Field direction	Towards the source mass.	Away from a positive source; towards a negative source.
Potential, zero at infinity (HL)	$V_g=-\frac{GM}{r}$	$V_e=k\frac{Q}{r}$
Potential energy (HL)	$E_p=-\frac{GMm}{r}$	$E_p=k\frac{Qq}{r}$
Potential-energy sign	Negative for two positive masses at finite separation. Whether a system is bound depends on its total energy.	Positive for like charges; negative for unlike charges, with zero at infinite separation.
Uniform-field approximation	Near Earth’s surface over a small height range.	Between large parallel plates, away from edges: $E=\frac{V}{d}$ .
Signed potential gradient (HL)	$g=-\frac{\Delta V_g}{\Delta r}$	$E=-\frac{\Delta V_e}{\Delta r}$

Both point-source fields follow an inverse-square law, while their potentials vary as the inverse of distance. Field quantities add as vectors; potentials add as scalars. In the gradient row, $g$ and $E$ denote signed components along the chosen coordinate.

11. Practice questions with expandable answers

Question 1 — Induction and conservation (SL and HL)

A negatively charged rod is brought near a neutral conducting sphere. Explain why the sphere can be attracted before it has any net charge. Describe how to leave it positively charged without touching it with the rod.

Show answer and working

The rod repels electrons to the far side of the sphere. The near side has an electron deficit and is attracted to the rod. Because the attractive charge is nearer than the repelled charge, attraction can exceed repulsion. The sphere is still neutral overall until charge enters or leaves it.

Ground the sphere while the rod is nearby, allowing electrons to leave. Disconnect Earth first, then remove the rod. The sphere is left positively charged. Charge is conserved when the sphere, rod and Earth are considered together.

Question 2 — Changing force (SL and HL)

Two point charges exert a force of $0.80\,\mathrm{N}$ on each other. One charge is doubled and their separation is tripled. Determine the new force magnitude.

Show answer and working

$F_{\mathrm{new}}=F_{\mathrm{old}}\frac{2}{3^2}=0.80\times\frac{2}{9}=0.178\,\mathrm{N}\approx0.18\,\mathrm{N}$

The type of interaction is unchanged if the charge signs are unchanged.

Question 3 — A point-charge field (SL and HL)

A point charge of $+5.0\,\mathrm{nC}$ is in air. Calculate the field strength $0.30\,\mathrm{m}$ from it and state the force direction on an electron placed there.

Show answer and working

$E=\frac{(8.99\times10^9)(5.0\times10^{-9})}{(0.30)^2}=4.99\times10^2\,\mathrm{N\,C^{-1}}$

The field is approximately $5.0\times10^2\,\mathrm{N\,C^{-1}}$ radially outwards. An electron experiences force radially inwards, towards the positive source.

Question 4 — Oil-drop data (SL and HL)

A series of oil drops has measured charge magnitudes $3.2,\ 4.8,\ 8.0,\ 11.2$ in units of $10^{-19}\,\mathrm{C}$ . Explain how these results support charge quantization and estimate the elementary charge.

Show answer and working

All four values are integer multiples of $1.6\times10^{-19}\,\mathrm{C}$ : the drops carry magnitudes $2e$ , $3e$ , $5e$ and $7e$ . Together with many more measurements, this common charge step supports the existence of an elementary charge rather than arbitrary continuous charge values.

Question 5 — Field and potential at a midpoint (HL)

Two charges $+Q$ and $-Q$ are separated by distance $2a$ . Find the electric potential and field magnitude at the midpoint. State the field direction.

Show answer and working

$V_e=\frac{kQ}{a}+\frac{k(-Q)}{a}=0$

Both field contributions point from the positive charge towards the negative charge, so their magnitudes add:

$E=\frac{kQ}{a^2}+\frac{kQ}{a^2}=\frac{2kQ}{a^2}$

The potential is zero, but the field is non-zero and points towards the negative charge.

Question 6 — Work done on a positive charge (HL)

An external agent slowly moves a $+3.0\,\mu\mathrm{C}$ charge from a point at $20\,\mathrm{V}$ to a point at $80\,\mathrm{V}$ . Calculate the potential-energy change, the external work and the work done by the electric field.

Show answer and working

$\Delta V_e=80-20=60\,\mathrm{V}$

$\Delta E_p=q\Delta V_e=(3.0\times10^{-6})(60)=1.8\times10^{-4}\,\mathrm{J}$

The external work is $+1.8\times10^{-4}\,\mathrm{J}$ . The electric field does $-1.8\times10^{-4}\,\mathrm{J}$ of work. Slow movement implies negligible change in kinetic energy.

Question 7 — A charged conducting sphere (HL)

An isolated conducting sphere has radius $0.10\,\mathrm{m}$ and charge $+2.0\,\mathrm{nC}$ . Determine the field strength and electric potential at its centre. Take zero potential at infinity.

Show answer and working

The electric field inside the conductor is zero. The potential is constant throughout the sphere and equals the surface potential:

$V_e=\frac{kQ}{R}=\frac{(8.99\times10^9)(2.0\times10^{-9})}{0.10}=180\,\mathrm{V}$

At the centre, $E=0$ and $V_e\approx180\,\mathrm{V}$ . The zero field corresponds to zero potential gradient, rather than zero potential.

Question 8 — Magnetic directions (SL and HL)

(a) A straight wire carries conventional current into the page. State the magnetic field direction around it. (b) Viewed from one end, the current around a solenoid is anticlockwise. Identify that end’s magnetic pole. (c) Describe the effect of reversing the current.

Show answer and working

(a) The field is clockwise around the wire. (b) The viewed end is a north pole; the internal field points towards that end. (c) Reversing current reverses the magnetic field direction and exchanges the solenoid’s north and south ends.

12. Big-picture summary and exam readiness

Core relationships

$Q=ne,\qquad F=k\frac{|q_1q_2|}{r^2}$

$E=\frac{F}{q},\qquad E=k\frac{|Q|}{r^2},\qquad E=\frac{V}{d}$

$|q|E=mg\qquad\text{(stationary oil drop; buoyancy neglected)}$

Higher Level: retain charge signs in potential and energy calculations.

$E_p=k\frac{q_1q_2}{r},\qquad V_e=\frac{E_p}{q}=k\frac{Q}{r}$

$\Delta E_p=q\Delta V_e,\qquad W_{\mathrm{field}}=-q\Delta V_e$

$E=-\frac{\Delta V_e}{\Delta r},\qquad 1\,\mathrm{eV}=1.60\times10^{-19}\,\mathrm{J}$

Convert microcoulombs and nanocoulombs to coulombs before substituting.
State force and field directions, especially for negative charges.
Distinguish an inverse-square field from an inverse-distance potential.
Add fields as vectors and potentials as scalars.
State the approximation: point charges, uniform plates, isolated conductor or negligible buoyancy.
Distinguish external work from work done by the electric field.
Use conventional current for magnetic right-hand rules.
Draw field lines with arrows and show equipotentials perpendicular to electric field lines.

13. Faith connection — Studying creation with care

Scripture and reflection

“The works of the LORD are great, sought out of all them that have pleasure therein.” — Psalm 111:2 (KJV)

Creation: Christians understand the order and richness of the physical world as part of God’s good creation. Studying electric and magnetic fields invites careful attention to patterns that support living things and make many useful technologies possible.

The Fall: human knowledge and technology can be used selfishly or dishonestly. Scientific work calls for humility: report measurements honestly, acknowledge uncertainty and consider how discoveries affect other people.

Redemption: the Christian message centres on Jesus Christ, who died and rose again to reconcile people to God. Salvation is God’s gift, received through faith in Christ, rather than an achievement earned by knowledge or performance.

Restoration: Christians look forward to God’s renewal of creation. In the present, we can use learning to serve others, care for the world and develop technology responsibly.

Personal response: consider placing your trust in Jesus Christ. Read the Bible to learn who he is, speak to God in prayer and seek a church or community of believers who can help you grow in faith.

Open reflection questions

How can careful measurement and honest reporting honour God and serve your neighbour? Where could knowledge of electric or magnetic fields help improve someone’s life? What would responsible use of that knowledge look like?

IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

Curriculum

Electric and magnetic fields

Lesson D.2 Electric and Magnetic Fields

Learning goals

Symbols and useful data

1. Electric charge: quantization and conservation (SL and HL)

Quantization of charge

Conservation of charge

How objects become charged

Charging a conductor by induction

Worked example 1 — Counting transferred electrons

2. Coulomb’s law (SL and HL)

Inverse-square behaviour and the medium

Superposition

Worked example 2 — Force between unlike charges

3. Electric field strength (SL and HL)

Deriving the field of a point charge

Electric field lines

Uniform fields between parallel plates

Worked example 3 — Parallel plates and an electron

4. Millikan’s oil-drop experiment (SL and HL)

Worked example 4 — A suspended oil drop

5. Electric potential energy and electric potential (HL)

Electric potential energy

Electric potential

Adding potentials

Worked example 5 — Potential and potential energy

6. Work, energy and potential difference (HL)

The electronvolt

Worked example 6 — An electron accelerated through a voltage

7. Equipotentials and potential gradients (HL)

Equipotential surfaces

The IB potential-gradient relationship

Worked example 7 — Reading a potential gradient

8. Charged conductors in electrostatic equilibrium (HL)

An isolated charged conducting sphere

9. Magnetic field patterns (SL and HL)

Bar magnet

Straight current-carrying wire

Circular coil

Air-core solenoid

Electric and magnetic field lines

10. Comparison: gravitational and electric fields

11. Practice questions with expandable answers

Question 1 — Induction and conservation (SL and HL)

Question 2 — Changing force (SL and HL)

Question 3 — A point-charge field (SL and HL)

Question 4 — Oil-drop data (SL and HL)

Question 5 — Field and potential at a midpoint (HL)

Question 6 — Work done on a positive charge (HL)

Question 7 — A charged conducting sphere (HL)

Question 8 — Magnetic directions (SL and HL)

12. Big-picture summary and exam readiness

Core relationships

13. Faith connection — Studying creation with care

Scripture and reflection

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