D.2 Electric and magnetic fields

Syllabus
First assessment 2025
Topic
Level
SL

Model Electric Charge Forces

Use the charge signs

There are two types of electric charge. Like charges repel: positive–positive and negative–negative. Unlike charges attract: positive–negative. The force on each charge acts along the line joining the two charges, with equal magnitude and opposite direction.

Draw the interaction

For two like point charges, draw arrows away from each other. For two unlike point charges, draw arrows toward each other. The direction is determined by the sign combination; the force magnitude also depends on charge magnitudes and separation, which Coulomb’s law quantifies.

Extend to a third charge

If a third charge is present, find the force from each other charge separately and add the force vectors. Do not decide the net direction by charge sign alone: compare the individual vectors and their magnitudes.

Common trap

Do not say that a negative charge always repels or that a positive charge always attracts. Attraction and repulsion depend on the pair of charges, and Newton’s third-law pair acts on different charges.

D.2.1 Exam Analysis

Assessment in practice

1 marks
How it is assessed

Questions predict the motion of a displaced charge or ask for the direction of an electric force.

Command terms

Explain / Which list

What earns marks

Identify like-charge repulsion or unlike-charge attraction, then draw each force along the joining line with equal and opposite directions.

Watch for

Assigning attraction or repulsion to one charge in isolation, or drawing the force on both charges in the same direction.

Representative question

Question 1

[Maximum number: 1]

N is just displaced along L , closer to q, and released.

Explain the subsequent motion of N .

Apply Coulomb’s Law

Use the inverse-square law

For two point charges, rr is their centre-to-centre separation. In a medium of permittivity ε\varepsilon, k=1/(4πε)k=1/(4\pi\varepsilon); in vacuum, k=8.99×109Nm2C2k=8.99\times10^9\,\mathrm{N\,m^2\,C^{-2}}. Calculate magnitude, then use charge signs to state attraction or repulsion.

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

Worked example — unlike charges in air

For q1=4.5×108Cq_1=4.5\times10^{-8}\,\mathrm{C}, q2=1.3×107Cq_2=-1.3\times10^{-7}\,\mathrm{C} and r=3.2×102mr=3.2\times10^{-2}\,\mathrm{m}, F=(8.99×109)q1q2/r2=5.1×102NF=(8.99\times10^9)|q_1q_2|/r^2=5.1\times10^{-2}\,\mathrm{N}. The force is attractive because the charges have opposite signs.

Read the scaling

Doubling either charge doubles the force. Doubling the separation reduces the force to one quarter. If the medium has permittivity ε\varepsilon rather than ε0\varepsilon_0, use k=1/(4πε)k=1/(4\pi\varepsilon); greater permittivity reduces the force for the same charges and separation.

Choose the point-charge model

Spherical charged bodies can be treated as point charges at their centres when the geometry permits. Use centre-to-centre separation and convert charge units, such as microcoulombs, before substitution. For several charges, calculate each force vector and add them.

Common trap

Do not use diameter or a single radius as r, and do not forget that a change in separation is squared. Keep the force magnitude positive in the calculation, then state attraction or repulsion separately.

D.2.2 Exam Analysis

Assessment in practice

1–2 marks
How it is assessed

Questions compare forces after changing separation or permittivity, or compare electric fields at two distances from one charge.

Command terms

What is

What earns marks

Use F=k|q1q2|/r², select k for the medium, convert units, and state attraction or repulsion from the charge signs.

Watch for

Forgetting the square on separation, using vacuum k in a dielectric without adjustment, or confusing force magnitude with force direction.

Representative question

Question 1

[Maximum number: 1]

An isolated point charge q is located at point X. Two other points Y and Z are such that Y Z=2 X Y.

What is  electric field at Y electric field at Z?\frac{\text { electric field at } Y}{\text { electric field at } Z} ?

A

19\frac{1}{9}

B

13\frac{1}{3}

C

3

D

9

Apply Charge Conservation

Core idea

Electric charge is conserved: in an isolated system, the total charge before an interaction equals the total charge after it. Charge can move between objects, but it is not created or destroyed in the transfer.

Use it at a junction

In a steady circuit, charge does not accumulate at a junction. The current entering equals the current leaving, for example I1=I2+I3I_1=I_2+I_3. This is a consequence of charge conservation, not a separate rule that overrides it.

Track the system boundary

When charge appears to change on one object, include the other object, the conductor or the ground in the system. Electrons may move across the chosen boundary, so the object’s charge changes while the total charge of the larger isolated system remains constant.

Common trap

Do not answer “Kirchhoff’s law” alone when asked for the fundamental law behind current balance. State conservation of electric charge.

D.2.3 Exam Analysis

Assessment in practice

1 marks
How it is assessed

Questions identify the fundamental law behind current balance or explain an apparent charge change during transfer.

Command terms

State

What earns marks

State conservation of electric charge and identify the complete system boundary; at a circuit junction, current entering equals current leaving.

Watch for

Naming Kirchhoff’s law without stating conservation of electric charge, or treating transferred charge as newly created.

Representative question

Question 1

[Maximum number: 1]

The diagram shows a junction in a circuit.

The currents in the three wires are related by I1=I2+I3I_{1}=I_{2}+I_{3}.
State the fundamental law of Physics from which this relation is derived.

Explain Millikan’s Oil-Drop Experiment

Set the force balance

Millikan observed charged oil drops between parallel plates. By adjusting the potential difference, the electric force on a drop can balance its weight so the drop is stationary. With E=V/dE=V/d, the balance is
qE=mgq=mgE=mgdVqE=mg\quad\Rightarrow\quad q=\frac{mg}{E}=\frac{mgd}{V}
for the simplified model in which buoyancy is neglected.

Read the evidence

Repeating the measurement for many drops gives charges that are integer multiples of a smallest value, the elementary charge ee: q=neq=ne, where nn is an integer. This pattern is evidence that electric charge is quantized rather than continuously variable.

Explain the method

The experiment varies the electric field until a drop is held stationary, then uses the known mass and field to infer its charge. It is the repeated integer-multiple pattern—not one isolated drop—that supports the quantization conclusion.

Common trap

Do not say that Millikan directly measured a continuous range of charge or that the drop is uncharged when it is stationary. Stationary means the electric and gravitational forces balance; the charge is non-zero and can be calculated from the balance.

D.2.4 Exam Analysis

Assessment in practice

1 marks
How it is assessed

Questions identify Millikan as the scientist associated with quantized charge or identify a valid electron-charge value.

Command terms

Who was / What is

What earns marks

Describe the electric–weight balance, use q=mg/E when calculation is required, and connect repeated integer multiples of e to charge quantization.

Watch for

Confusing quantization with charge conservation, or treating a stationary drop as evidence of zero charge.

Representative question

Question 1

[Maximum number: 1]

What is a correct value for the charge on an electron?

A

1.60×1012μC1.60 \times 10^{-12} \mu \mathrm{C}

B

1.60×1015mC1.60 \times 10^{-15} \mathrm{mC}

C

1.60×1022kC1.60 \times 10^{-22} \mathrm{kC}

D

1.60×1024MC1.60 \times 10^{-24} \mathrm{MC}

Model Charge Transfer

Transfer by friction

Rubbing two insulating materials can move electrons from one surface to the other. One object becomes negatively charged and the other positively charged; the total charge of the pair is conserved. The material that loses electrons is positive, and the material that gains electrons is negative.

Transfer by induction

Bring a charged object near a conductor without touching it. Charges in the conductor separate by repulsion and attraction. If the conductor is connected to ground while the charged object remains nearby, electrons can flow to or from Earth. Disconnect the ground first, then remove the external charged object, leaving the conductor with a net charge.

Transfer by contact and grounding

Touching a charged conductor to another conductor allows charge to redistribute between them. Grounding connects an object to a very large charge reservoir: electrons can leave an object or enter it, depending on the nearby charge and the object’s potential.

Common trap

Induction does not require contact with the charged rod. In a grounding sequence, remove the ground before removing the inducing charge; reversing the order can leave the conductor neutral.

D.2.5 Exam Analysis

Assessment in practice

1–2 marks
How it is assessed

Questions ask what charge remains after a grounding sequence or distinguish induction from contact charging.

Command terms

What is correct

What earns marks

Identify whether electrons move by friction, contact or induction, track the system boundary, and state the role of grounding as an electron reservoir.

Watch for

Removing the inducing rod before the ground, or treating polarization in a conductor as a net charge transfer without grounding.

Representative question

Question 1

[Maximum number: 1]

A positively charged rod is near a metal plate that is grounded as shown.

The grounding wire and then the rod are removed. What is correct about the overall charge on the plate before and after grounding is removed?

Charge on plate before

grounding is removed

Charge on plate after

grounding is removed

neutral

neutral

neutral

negative

negative

neutral

negative

negative

Calculate Electric Field Strength

Define the field

Electric field strength is force per unit positive test charge. For a point source QQ, it is directed away from positive QQ and toward negative QQ. Its SI unit is NC1\mathrm{N\,C^{-1}}.

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

Worked example — point-charge field

At r=1.0mr=1.0\,\mathrm{m} from Q=+2.9×108CQ=+2.9\times10^{-8}\,\mathrm{C}, E=(8.99×109)(2.9×108)/(1.0)2=2.6×102NC1E=(8.99\times10^9)(2.9\times10^{-8})/(1.0)^2=2.6\times10^2\,\mathrm{N\,C^{-1}}. Because the source is positive, the field points radially outward.

Add fields as vectors

For more than one source, calculate each electric-field vector at the point and add them. Do not add magnitudes unless all field vectors point in the same direction. The force on a particular charge is then F=qEF=qE, with its direction reversed from the field if the charge itself is negative.

Common trap

The field direction is defined using a positive test charge, not the sign of the test charge in the question. Also distinguish field strength EE from force FF: changing the test charge changes F but not the source field E.

D.2.6 Exam Analysis

Assessment in practice

1–2 marks
How it is assessed

Questions compare field strength at different distances or find the resultant field direction from two charges.

Command terms

What is / What is the direction

What earns marks

Use E=F/q or E=k|Q|/r², keep field direction defined by a positive test charge, and add multiple source fields as vectors.

Watch for

Using the sign of the test charge to define field direction, or comparing field strengths linearly rather than with inverse-square scaling.

Representative question

Question 1

[Maximum number: 1]

Two point charges, -Q and +Q, are placed as shown. Point P is at the same distance from both charges.

What is the direction of the electric field strength at P ?

Q+Q\begin{array}{cc} \circ & \circ \\ -Q & +Q \end{array}

Read Electric Field Lines

Interpret a field line

Electric field lines show the direction of the force on a small positive test charge. Their arrows point away from positive charges and toward negative charges. The tangent to a line gives the local field direction.

Required geometry Electric-field pattern
Single point charge radial; outward for positive, inward for negative
Two point charges resultant curves; from positive toward negative; lines never cross
Charged spherical conductor radial outside and normal to surface; no field lines in conducting material or an empty shielded cavity
Opposite parallel plates straight, parallel central lines from positive to negative; curved edge lines show fringing

Read qualitative strength

Where field lines are closer together, the field is stronger; where they spread out, it is weaker. This is a qualitative representation unless the diagram specifies equal field-line intervals or a scale.

Common trap

Do not point electric field lines from negative to positive, make them cross, or draw them tangent to equipotential surfaces. Field lines follow the positive-test-charge convention.

D.2.7 Exam Analysis

Assessment in practice

1–2 marks
How it is assessed

Questions judge correct field-line statements or ask you to draw lines between charged plates.

Command terms

Which / Draw

What earns marks

Point arrows in the positive-test-charge direction, keep lines non-crossing, and use density to compare qualitative field strength.

Watch for

Drawing arrows from negative to positive or allowing lines to cross; also confusing line density with the number of charges.

Representative question

Question 1

[Maximum number: 1]

The diagram shows the electric field pattern due to two point charges X and Y . Y is a negative charge.

Which of the following correctly identifies the charge X and the direction of the electric field?

Sign of charge X

Direction of electric field

positive

Y to X

positive

X to Y

negative

X to Y

negative

Y to X

Read Field-Line Density

Core idea

In a field-line diagram, greater line density represents a stronger electric field. Compare density over equal areas or equal widths of the same diagram; the visual spacing is a qualitative encoding of E|E|, not a new physical force.

Connect density to distance

Around an isolated point charge, field lines spread as distance increases, so the field becomes weaker. A denser pattern near the charge is consistent with the inverse-square dependence of field strength. In a uniform field, equal spacing indicates constant field strength.

Check the representation

Density comparisons are meaningful only when the diagram uses the same line convention and potential/field intervals. Do not infer exact numerical values from arbitrary artwork; use labels or a scale if a calculation is required.

Common trap

Do not count field lines as individual objects or compare the total number of lines in two drawings with different scales. It is the local density that represents relative field strength.

Model the Parallel-Plate Field

Use the uniform-field model

Between two large opposite parallel plates, away from the edges, field strength equals potential difference VV divided by perpendicular plate separation dd. The field points from the positive plate to the negative plate; edge regions are not uniform.

E=\frac{V}{d}

Worked example — required potential difference

For E=1.0×106Vm1E=1.0\times10^6\,\mathrm{V\,m^{-1}} and d=0.50cm=5.0×103md=0.50\,\mathrm{cm}=5.0\times10^{-3}\,\mathrm{m}, V=Ed=(1.0×106)(5.0×103)=5.0×103VV=Ed=(1.0\times10^6)(5.0\times10^{-3})=5.0\times10^3\,\mathrm{V}. The result applies to the uniform central region.

Read the direction

Electric field lines point from the positive plate to the negative plate, so a positive charge accelerates in that direction and a negative charge accelerates oppositely. The field strength can be expressed in NC1\mathrm{N\,C^{-1}} or equivalently Vm1\mathrm{V\,m^{-1}}.

Check the boundary

The formula assumes a uniform region and neglects edge effects. Use the perpendicular plate separation in metres; do not use the diagonal distance or the plate length.

Common trap

Do not reverse the field direction because the test charge is negative. Field direction is defined by a positive test charge; the force on a negative charge is opposite.

D.2.9 Exam Analysis

Assessment in practice

1 marks
How it is assessed

Questions calculate the field between plates from voltage and spacing.

Command terms

Calculate

What earns marks

Use E=V/d with perpendicular separation in SI units, report N C⁻¹ or V m⁻¹, and state the direction from positive to negative plate.

Watch for

Using plate length instead of separation, forgetting to convert centimetres to metres, or reversing the field direction for a negative test charge.

Representative question

Question 1

[Maximum number: 2]

The plastic film begins to conduct when the electric field strength in it exceeds 1.5MNC11.5 \mathrm{MNC}^{-1}. Calculate the maximum charge that can be stored on the capacitor.

Read Magnetic Field Lines

Interpret a magnetic field line

Magnetic field lines show the local direction of the magnetic field; a compass north pole or a suitable test direction follows the arrow convention. Unlike isolated electric field lines, magnetic field lines form continuous closed loops.

Use the right-hand rule

Around a long straight current-carrying wire, the field lines are concentric circles centred on the wire. Point the right thumb in the conventional current direction; curled fingers give the magnetic-field direction. If electrons move into the page, conventional current is out of the page, so reverse the electron-motion direction before applying the rule.

Source Magnetic-field pattern Direction rule
Bar magnet closed loops; outside from north to south arrows return through the magnet
Straight wire concentric circles around the wire right thumb = conventional current; curled fingers = field
Circular coil loops combine into a field through the coil centre along its axis curl fingers with current; thumb gives axial field
Air-core solenoid nearly parallel, uniform lines inside; bar-magnet-like return field outside curl fingers with coil current; thumb gives the solenoid’s north end

Common trap

Do not use electron motion as though it were conventional current, and do not draw magnetic field lines starting or ending on an isolated magnetic pole.

D.2.10 Exam Analysis

Assessment in practice

1 marks
How it is assessed

Questions determine field direction at a point near one or more current-carrying wires.

Command terms

What is / What is the direction

What earns marks

Convert electron motion to conventional current when needed, apply the right-hand rule, and identify the magnetic-field direction from the local circular or closed-loop pattern.

Watch for

Applying the right-hand rule directly to electron motion instead of conventional current, or reversing the field direction around the wire.

Representative question

Question 1

[Maximum number: 1]

Two parallel wires carry equal currents in the same direction out of the paper. Which diagram shows the magnetic field surrounding the wires?

A
B
C
D

Retrieve the Core D.2 Electric and Magnetic Fields Model

D.2 core fields is secure when you can move between charge, force and field representations.

  • Like charges repel and unlike charges attract
  • Coulomb’s law gives inverse-square force
  • Charge is conserved, quantized and transferable
  • Millikan’s experiment supports q=ne
  • E=F/q and field lines show direction and relative density
  • Parallel plates give E=V/d
  • Magnetic field lines are closed and follow current direction

Objective notes

10 learning objectives