DP Physics SL D.2 Electric and Magnetic Fields

DP Physics SL D.2 Electric and Magnetic Fields - Full notes

RevisionPrep

DP Physics SL | D.2 Electric and Magnetic Fields

DP PHYSICS · THEME D: FIELDS

D.2

Electric and Magnetic Fields

Full notes for SL · IB-style

Contents

1	Electric charge
2	Coulomb's law
3	Electric fields
4	Magnetic fields
5	Fields in the natural world and technology
6	Exam technique and further practice
7	Common misconceptions and links
8	Exam-style question with markscheme
9	Check your understanding
10	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
describe charging and use conservation and quantisation of charge
use Coulomb's law, including for several charges
define and calculate electric field strength, including E = V/d
draw electric field patterns and describe Millikan's experiment
draw magnetic field patterns for magnets, wires and solenoids
compare gravitational, electric and magnetic fields

This booklet covers the SL content of this subtopic. Additional Higher Level material is not included.

Guiding questions

How do charged particles interact through fields? How are electric and magnetic fields represented and quantified? What are the similarities and differences between gravitational, electric and magnetic fields?

How this topic is examined

D.2 appears in Papers 1 and 2. Expect Coulomb's law, field strength calculations, field patterns, uniform fields and (HL) potential and equipotentials.

Treat forces and field strengths as vectors; potentials (HL) as scalars.

Command term	What the examiner wants
State	a brief answer with no explanation
Outline	a brief account or summary
Describe	a detailed account without reasons
Explain	a detailed account including causes, reasons or mechanisms
Calculate	a numerical answer with working and correct units
Determine	obtain the only possible answer, often from a graph or data
Show that	obtain the given result, showing every step (give one more significant figure)
Estimate	an approximate value using reasonable assumptions
Sketch	a graph or diagram showing the general shape and key features
Deduce / Suggest	reach a conclusion from given information; propose a reasonable explanation

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the physics shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Electric charge

The source of electric forces.

Think about it

Why does a balloon rubbed on your hair stick to a wall, even though the wall is uncharged?

Electric forces hold atoms and molecules together and govern all of chemistry and electronics. This section begins with charge itself, then builds the field model used for both electric and magnetic effects.

1.1 Properties of charge

Charge comes in two types: positive and negative. Like charges repel; unlike charges attract. Charge is measured in coulombs (C), is conserved and is quantised in multiples of the elementary charge e = 1.60 × 10⁻¹⁹ C. Electrons carry −e; protons +e.

WORKED EXAMPLE 1 Electrons (basic)

A rod gains a charge of −4.8 × 10⁻⁹ C. How many electrons has it gained?

Solution

4.8 × 10⁻⁹/1.6 × 10⁻¹⁹M1

= 3.0 × 10¹⁰A1

Quick check

1. Can an object have a charge of 2.4 × 10⁻¹⁹ C?

Answers: 1. no, it is not a whole multiple of e

1.2 Conductors and insulators

In conductors such as metals, some electrons are free to move through the material. In insulators, electrons are bound to their atoms. Semiconductors lie in between. Charge placed on a conductor spreads over its outer surface.

Type	Examples	Charge carriers
conductors	metals, graphite, salt solutions	free electrons or ions
semiconductors	silicon, germanium	few free electrons and holes
insulators	plastic, glass, rubber, dry air	almost none

1.3 Charging by friction and contact

Rubbing two insulators together transfers electrons from one to the other: one becomes negative and the other equally positive. Touching a charged object to a conductor shares charge by conduction.

WORKED EXAMPLE 2 Friction (standard)

A polythene rod rubbed with wool gains 2.0 × 10¹¹ electrons. Find the charge on the rod and on the wool.

Solution

Rod: −2.0 × 10¹¹ × 1.6 × 10⁻¹⁹ = −3.2 × 10⁻⁸ CM1

Wool: +3.2 × 10⁻⁸ C (equal and opposite)A1

1.4 Charging by induction

A charged rod brought near a neutral conductor attracts opposite charge to the near side and repels like charge to the far side. Earthing the conductor briefly and then removing the earth before the rod leaves the conductor with a net charge opposite to the rod's.

•Bring the negatively charged rod near the sphere.
•Touch the far side to earth: electrons flow away to earth.
•Remove the earth connection, then the rod.
•The sphere is left positively charged.

Quick check

1. Why does a charged balloon attract a neutral wall?

Answers: 1. it induces opposite charge on the near surface of the wall, which is attracted more strongly than the repelled charge further away

1.5 Detecting charge: the electroscope

A gold-leaf electroscope detects charge: when the cap is charged, charge spreads to the stem and leaf, which repel so the leaf rises. A larger charge makes the leaf rise further.

Figure 1. A gold-leaf electroscope.

Quick check

1. What happens to the leaf if the cap is earthed?

Answers: 1. charge flows away and the leaf falls

1.6 Generating large charges: the Van de Graaff

A Van de Graaff generator uses a moving belt to carry charge to a hollow metal dome, where it spreads over the outer surface. Potentials of hundreds of kilovolts can build up until the air breaks down and a spark jumps.

Figure 2. A Van de Graaff generator.

WORKED EXAMPLE 3 Breakdown (exam level)

Air breaks down at about 3 × 10⁶ V m⁻¹. Estimate the maximum potential of a dome of radius 0.15 m (surface field V/r).

Solution

V = Er = 3 × 10⁶ × 0.15M1

≈ 450 kVA1

1.7 Electrostatics in technology

Electrostatic precipitators charge smoke particles, which are then attracted to earthed plates and removed from chimney gases. Electrostatic spray painting charges paint droplets so they spread evenly and wrap around the object. Photocopiers and laser printers use charged drums to attract toner.

Figure 3. An electrostatic precipitator removing smoke particles.

Quick check

1. Why does electrostatic spray painting waste less paint?

Answers: 1. charged droplets are attracted to the (earthed) object, including its back surfaces

2 Coulomb's law

The force between point charges.

2.1 The law

Two point charges exert forces on each other given by F = kq₁q₂/r², where k = 1/4πε₀ = 8.99 × 10⁹ N m² C⁻² and ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻² is the permittivity of free space. The force acts along the line joining the charges.

Figure 4. Force between two point charges against separation.

WORKED EXAMPLE 4 Two charges (standard)

Find the force between charges of +3.0 μC and −5.0 μC 0.20 m apart.

Solution

F = 8.99 × 10⁹ × 3.0 × 10⁻⁶ × 5.0 × 10⁻⁶/0.040M1

= 3.4 N (attractive)A1

Quick check

1. How does the force change if both charges double and the distance doubles?

Answers: 1. unchanged

2.2 Testing Coulomb's law

Small charged spheres on an insulating balance can measure the force at different separations. A graph of F against 1/r² is a straight line through the origin, confirming the inverse-square law; the gradient is kq₁q₂.

Figure 5. Force against 1/r² for two charged spheres.

WORKED EXAMPLE 5 Charge from gradient (exam level)

The gradient is 2.24 × 10⁻⁵ N m² and the spheres carry equal charges. Find the charge on each.

Solution

kq² = 2.24 × 10⁻⁵M1

q = 5.0 × 10⁻⁸ CA1

2.3 Electric and gravitational forces compared

Electric forces between particles are vastly stronger than gravitational forces. Gravity is always attractive; electric forces can attract or repel, so large bodies are usually neutral and gravity dominates on astronomical scales.

WORKED EXAMPLE 6 Hydrogen atom (exam level)

Find the ratio of electric to gravitational force between a proton and an electron.

Solution

Fe/Fg = ke²/GmempM1

≈ 2.3 × 10³⁹A1

2.4 Forces from several charges

When several charges act on one, find each force by Coulomb's law and add them as vectors, using components if they are not in a line.

WORKED EXAMPLE 7 Three in a line (exam level)

+2.0 μC at x = 0 and +4.0 μC at x = 0.30 m. Find the force on −1.0 μC at x = 0.10 m.

Solution

From +2.0 μC: 1.80 N towards x = 0; from +4.0 μC: 0.90 N towards x = 0.30 mM1 A1

Resultant = 0.90 N towards x = 0A1

WORKED EXAMPLE 8 Right angle (exam level)

Charges of +5.0 nC sit at (0.10 m, 0) and (0, 0.10 m). Find the resultant force on a +5.0 nC charge at the origin.

Solution

Each force = k(5.0 × 10⁻⁹)²/0.010 = 2.25 × 10⁻⁵ N, at right anglesM1

Resultant = √2 × 2.25 × 10⁻⁵ = 3.2 × 10⁻⁵ N at 45° below the negative x-axis (away from both)A1

2.5 Balancing forces

A charged object can hang in equilibrium when the electric force balances other forces. Resolve forces as in mechanics.

WORKED EXAMPLE 9 Charged spheres (exam level)

Two identical 0.10 g spheres with equal charges hang from threads and repel to 4.0 cm apart, each thread at 10° to the vertical. Find the charge.

Solution

Fe = mg tan 10° = 1.73 × 10⁻⁴ NM1

q = √(Fr²/k) = 5.5 × 10⁻⁹ CA1

3 Electric fields

Force per unit charge.

3.1 Electric field strength

An electric field is a region where a charge experiences a force. Electric field strength E = F/q is the force per unit positive test charge (N C⁻¹ or V m⁻¹). For a point charge, E = kQ/r², directed away from positive and towards negative charges.

Figure 6. Field strength against distance from a point charge.

WORKED EXAMPLE 10 Point charge (standard)

Calculate E 0.50 m from a +8.0 nC charge.

Solution

E = 8.99 × 10⁹ × 8.0 × 10⁻⁹/0.25M1

= 288 N C⁻¹A1

WORKED EXAMPLE 11 Force on an electron (standard)

Find the force on an electron in a field of 5000 N C⁻¹.

Solution

F = eE = 1.6 × 10⁻¹⁹ × 5000M1

= 8.0 × 10⁻¹⁶ N opposite to the fieldA1

3.2 Electric field patterns

Field lines start on positive charges and end on negative charges; they never cross; their density shows field strength. Around a point charge the field is radial. Between opposite charges, lines run from one to the other; between like charges there is a neutral point.

Figure 7. Electric field lines for a point charge, a dipole and two like charges.

Quick check

1. What does a neutral point between two equal positive charges mean?

Answers: 1. the resultant field is zero there

3.3 Field of a charged conducting sphere

Outside a charged conducting sphere, the field is the same as if all the charge were at the centre: E = kQ/r². Inside, the field is zero.

Figure 8. Field strength inside and outside a conducting sphere (radius 10 cm, Q = 20 nC).

WORKED EXAMPLE 12 Surface field (standard)

Find the field at the surface of the sphere in the figure.

Solution

E = 8.99 × 10⁹ × 2.0 × 10⁻⁸/0.10²M1

= 1.8 × 10⁴ N C⁻¹A1

3.4 Combining fields along a line

Along the line between two charges, the field from each adds as a vector. For like charges, there is a point where the resultant is zero.

Figure 9. Resultant field between +1.0 μC (at 0) and +4.0 μC (at 30 cm).

Quick check

1. Why is the zero-field point nearer the smaller charge?

Answers: 1. the smaller charge's field must be at a shorter distance to match the larger charge's field

3.5 More complex field patterns

With unequal charges, more field lines leave the larger charge, and some lines from it go off to infinity rather than ending on the smaller charge. With several charges the pattern can be found by adding the fields from each at many points (as a computer does).

Figure 10. Field lines for unequal charges and for three charges.

Quick check

1. Why do some lines from +2q not end on −q?

Answers: 1. the net charge is +q, so lines representing that excess go to infinity

3.6 Uniform fields between parallel plates

Between two parallel plates with potential difference V and separation d, the field is uniform (except near the edges): E = V/d. A charge q there experiences a constant force F = qE.

Figure 11. The uniform field between parallel plates, with edge effects.

WORKED EXAMPLE 13 Plates (standard)

Two plates 5.0 mm apart have a pd of 1500 V. Find E and the force on an electron between them.

Solution

E = 1500/0.0050 = 3.0 × 10⁵ V m⁻¹A1

F = 1.6 × 10⁻¹⁹ × 3.0 × 10⁵ = 4.8 × 10⁻¹⁴ NA1

3.7 Field strength across a uniform field

Between ideal parallel plates, a graph of E against position is a horizontal line: the field has the same value everywhere. Near the edges the field weakens and the lines curve outwards (edge effects).

Figure 12. Field strength between plates 1.0 cm apart with 200 V across them.

Quick check

1. What happens to E if the plate separation doubles at constant V?

Answers: 1. it halves

3.8 Field along the axis of a dipole

Along the line through two opposite charges, the field between them points from + to − and is strong; outside, the fields partly cancel and the resultant falls off rapidly.

Figure 13. Field along the axis of a pair of opposite charges (schematic).

Quick check

1. Between the charges of a dipole, which way does the field point?

Answers: 1. from the positive charge towards the negative charge

3.9 Millikan's oil-drop experiment

Charged oil drops are balanced between horizontal plates when the electric force equals the weight: qV/d = mg. Millikan found every charge was a whole-number multiple of 1.6 × 10⁻¹⁹ C, showing that charge is quantised.

Figure 14. Forces on a charged oil drop held stationary between plates.

WORKED EXAMPLE 14 Balanced drop (exam level)

A 4.9 × 10⁻¹⁵ kg drop is held stationary between plates 8.0 mm apart with 1200 V. Find its charge and the number of excess electrons.

Solution

q = mgd/V = 4.9 × 10⁻¹⁵ × 9.81 × 0.0080/1200 = 3.2 × 10⁻¹⁹ CM1 A1

= 2 electronsA1

3.10 Charges in fields: acceleration

A charge q of mass m in a uniform field accelerates at a = qE/m. Electrons have enormous accelerations because their mass is so small.

WORKED EXAMPLE 15 Electron acceleration (standard)

Find the acceleration of an electron in a field of 2.0 × 10⁴ V m⁻¹.

Solution

a = 1.6 × 10⁻¹⁹ × 2.0 × 10⁴/9.11 × 10⁻³¹M1

= 3.5 × 10¹⁵ m s⁻²A1

3.11 Conductors, shielding and points

Excess charge on a conductor sits on its outer surface and the field inside is zero. Charge concentrates at sharp points, where the field is strongest. A closed metal box (Faraday cage) shields its interior from external fields.

In the real world

Cars and aircraft act as Faraday cages: if struck by lightning, the charge flows over the outer metal skin, protecting the occupants.

Quick check

1. Why do lightning conductors have pointed tips?

Answers: 1. the field is strongest at points, allowing controlled discharge

4 Magnetic fields

Fields from magnets and currents.

Think about it

A compass needle swings when a nearby wire carries current. What is the connection between electricity and magnetism?

4.1 Magnets and magnetic field lines

Magnetic fields are produced by permanent magnets and by moving charges (currents). Field lines go from north to south outside a magnet, form closed loops and never cross; their density shows the strength of the field, measured as magnetic flux density B in teslas (T).

Figure 15. Magnetic field of a bar magnet.

Quick check

1. Where is the field of a bar magnet strongest?

Answers: 1. near the poles

4.2 Field of a long straight wire

A current in a long straight wire produces circular field lines centred on the wire. The right-hand grip rule gives the direction: thumb along the conventional current, fingers curl in the direction of the field. The field weakens with distance from the wire.

Figure 16. Circular field lines around a straight current-carrying wire.

Quick check

1. Using the right-hand grip rule, describe the field around a wire with current flowing out of the page.

Answers: 1. anticlockwise circles

4.3 How B varies near a wire

The magnetic flux density around a long straight wire is proportional to the current and inversely proportional to the distance from the wire: B = μ₀I/2πr, with μ₀ = 4π × 10⁻⁷ T m A⁻¹ (used in D.3).

Figure 17. Magnetic flux density against distance from a wire carrying 10 A.

WORKED EXAMPLE 16 Field near a wire (standard)

Find B 2.0 cm from a wire carrying 10 A.

Solution

B = 4π × 10⁻⁷ × 10/(2π × 0.020)M1

= 1.0 × 10⁻⁴ T (100 μT)A1

4.4 Fields of two parallel wires

The fields of two parallel currents combine. Between wires carrying currents in the same direction the fields partly cancel; with opposite currents they add. (This leads to the forces between wires in D.3.)

Figure 18. Combined field patterns of two parallel wires (• out of page, × into page).

Quick check

1. Where is the field zero for two equal currents in the same direction?

Answers: 1. midway between the wires

4.5 Field of a solenoid

A solenoid (long coil) produces a field like that of a bar magnet outside and a nearly uniform field inside. The field inside increases with the current and with the number of turns per unit length, and greatly with an iron core.

Figure 19. Magnetic field of a solenoid (cross-section).

WORKED EXAMPLE 17 Electromagnet (standard)

Give three ways to increase the strength of an electromagnet.

Solution

Increase the current; increase the turns per unit lengthA1

Insert a soft iron coreA1

In the real world

MRI scanners use superconducting solenoids producing uniform fields of 1.5–3 T, about 50 000 times Earth's magnetic field.

4.6 Uses of electromagnets

Electromagnets can be switched on and off and their strength varied. They are used in relays (a small current switches a larger one), scrapyard cranes, door locks, MRI scanners and particle accelerators.

Figure 20. A relay: current in the coil attracts the armature and closes the contacts.

Quick check

1. Why is soft iron used for the core of an electromagnet?

Answers: 1. it is easily magnetised and loses its magnetism when the current stops

4.7 Earth's magnetic field

Earth's field resembles that of a bar magnet, generated by currents in the liquid outer core. Its geographic north pole is near a magnetic south pole, which is why a compass's north pole points north. The field deflects charged particles from the solar wind, which spiral along field lines near the poles, causing aurorae.

Quick check

1. Why do aurorae occur near the poles?

Answers: 1. charged particles follow field lines that converge near the poles

4.8 Plotting field patterns

Iron filings or plotting compasses reveal magnetic field patterns; semolina in oil between charged electrodes shows electric field lines; a probe connected to a voltmeter on conducting paper maps equipotentials.

Try this

Sprinkle iron filings on paper over a bar magnet and over a solenoid. Compare the patterns, then use compasses to add the direction.

Quick check

1. What do iron filings fail to show about a field?

Answers: 1. its direction

5 Fields in the natural world and technology

Applications and contexts.

5.1 Lightning

Collisions between ice particles in thunderclouds separate charge: the cloud base usually becomes negative, inducing positive charge on the ground. When the field exceeds the breakdown strength of air (about 3 × 10⁶ V m⁻¹), air ionises and a conducting path forms, discharging huge currents.

Figure 21. Charge separation in a thundercloud and the lightning discharge.

WORKED EXAMPLE 18 Cloud potential (exam level)

A cloud base 1.5 km above the ground has an average field of 1.0 × 10⁵ V m⁻¹ beneath it. Estimate the potential difference between cloud and ground.

Solution

V = Ed = 1.0 × 10⁵ × 1500M1

= 1.5 × 10⁸ VA1

5.2 Magnetic materials

In ferromagnetic materials (iron, nickel, cobalt), atoms form small regions called domains in which their magnetic fields line up. In an unmagnetised sample the domains point in random directions; magnetising aligns them. Soft iron magnetises and demagnetises easily (electromagnets); steel keeps its magnetism (permanent magnets).

Figure 22. Magnetic domains in an unmagnetised and a magnetised material.

Quick check

1. How can a permanent magnet be demagnetised?

Answers: 1. heating it, hammering it, or placing it in a decreasing alternating field

5.3 Electron beams

In older TVs, oscilloscopes and X-ray tubes, electrons released from a hot filament are accelerated by an electric field between electrodes and then steered by electric or magnetic fields. The motion of charges in fields is studied in D.3.

WORKED EXAMPLE 19 Beam energy (standard)

Electrons are accelerated through 5.0 kV. Find their kinetic energy in eV and J.

Solution

E = 5.0 keVA1

= 5000 × 1.6 × 10⁻¹⁹ = 8.0 × 10⁻¹⁶ JA1

5.4 Shielding sensitive equipment

Sensitive electronic instruments are enclosed in metal cases or wire mesh (Faraday cages) to block external electric fields and radio interference. Mu-metal shields block slowly varying magnetic fields.

Quick check

1. Why do mobile phones lose signal inside lifts?

Answers: 1. the metal lift acts as a Faraday cage, blocking radio waves

6 Exam technique and further practice

Applying field ideas.

6.1 Comparing gravitational, electric and magnetic fields

Feature	Gravitational	Electric	Magnetic
source	mass	charge	moving charge, magnets
force	always attractive	attract or repel	on moving charges and currents
field strength	g = GM/r²	E = kQ/r²	B (tesla)
field lines	towards masses	+ to −	N to S, closed loops

6.2 Method checklist

•Convert μC, nC and cm to SI before substituting.
•Add forces and field strengths as vectors; (HL) add potentials as scalars.
•Use E = V/d only for uniform fields between plates.
•Draw field lines that start on + and end on −, never crossing.
•Use the right-hand grip rule for wires and solenoids.

6.3 Graded practice

WORKED EXAMPLE 20 Zero field point (exam level)

Find where the electric field is zero between +1.0 μC and +4.0 μC charges 0.30 m apart.

Solution

1/x² = 4/(0.30 − x)²M1

x = 0.10 m from the 1.0 μC chargeA1

WORKED EXAMPLE 21 Field from force (standard)

A 2.0 nC charge experiences 6.0 × 10⁻⁵ N. Find E.

Solution

E = F/qM1

= 3.0 × 10⁴ N C⁻¹A1

WORKED EXAMPLE 22 Plate separation (standard)

Find the plate separation that gives 4.0 × 10⁴ V m⁻¹ with 200 V.

Solution

d = V/EM1

= 5.0 mmA1

6.4 Multi-step problem: charged drop and field

WORKED EXAMPLE 23 Accelerating drop (exam level)

A 3.0 × 10⁻¹⁵ kg drop carrying 4.8 × 10⁻¹⁹ C is between plates 6.0 mm apart. Find (a) the pd that holds it stationary; (b) its acceleration if the pd is doubled.

Solution

(a) V = mgd/q = 370 VM1 A1

(b) Electric force doubles to 2mg: net upward force mg, a = 9.81 m s⁻² upwardsM1 A1

6.5 Self-test

Try these without notes, then check the answers.

Quick check

1. State the value of the elementary charge.

2. How does F change if one charge triples and r doubles?

3. Find E between plates 2.0 cm apart with 300 V.

4. Which way do field lines point near a negative charge?

5. What rule gives the direction of the field around a wire?

6. Why is the field inside a charged conducting sphere zero?

Answers: 1. 1.60 × 10⁻¹⁹ C 2. × 3/4 3. 1.5 × 10⁴ V m⁻¹ 4. towards the charge 5. the right-hand grip rule 6. charges arrange themselves on the outer surface so their fields cancel inside

6.6 Graded practice: Coulomb and fields

WORKED EXAMPLE 24 Equilateral triangle (exam level)

Three +2.0 μC charges sit at the corners of an equilateral triangle of side 0.10 m. Find the resultant force on one of them.

Solution

Each force = k(2.0 × 10⁻⁶)²/0.010 = 3.6 N at 60° to each otherM1

Resultant = 2 × 3.6 cos 30° = 6.2 NA1

WORKED EXAMPLE 25 Field at a point (exam level)

Find E at the midpoint between +3.0 nC and −3.0 nC charges 0.20 m apart.

Solution

Each gives 8.99 × 10⁹ × 3.0 × 10⁻⁹/0.010 = 2700 N C⁻¹, same directionM1

E = 5400 N C⁻¹ towards the negative chargeA1

WORKED EXAMPLE 26 Proton in a field (standard)

Find the acceleration of a proton between plates 2.0 cm apart with 400 V.

Solution

E = 2.0 × 10⁴ V m⁻¹; F = 3.2 × 10⁻¹⁵ NM1

a = F/m = 1.9 × 10¹² m s⁻²A1

6.7 Graded practice: uniform fields and Millikan

WORKED EXAMPLE 27 Drop data (exam level)

In a Millikan-type experiment, drops are balanced with charges of 3.2, 4.8, 8.0 and 6.4 × 10⁻¹⁹ C. What do these results suggest?

Solution

All are whole multiples of 1.6 × 10⁻¹⁹ C (2, 3, 5 and 4 times)M1

Charge is quantised in units of eR1

WORKED EXAMPLE 28 Changing separation (standard)

Plates 1.0 cm apart at 500 V are moved to 2.5 cm apart, still at 500 V. Find the old and new field strengths.

Solution

Old: 5.0 × 10⁴ V m⁻¹A1

New: 2.0 × 10⁴ V m⁻¹A1

WORKED EXAMPLE 29 Balancing a heavier drop (exam level)

A drop twice as heavy but with the same charge must be balanced between the same plates. By what factor must the pd change?

Solution

qV/d = mg: V ∝ mM1

pd must doubleA1

6.8 Graded practice: magnetic fields

WORKED EXAMPLE 30 Compass near a wire (standard)

A compass is placed directly above a wire carrying a current northwards. Which way does the needle point?

Solution

Right-hand grip rule: field above the wire points westM1

Needle points west (if the wire's field dominates)A1

WORKED EXAMPLE 31 Doubling current (standard)

The current in a solenoid doubles and its turns per metre halve. What happens to the field inside?

Solution

B ∝ nIM1

UnchangedA1

6.9 Graded practice: charge and induction

WORKED EXAMPLE 32 Sharing charge (standard)

Two identical metal spheres carry +6.0 nC and −2.0 nC. They touch and are separated. Find the charge on each.

Solution

Total = +4.0 nC, shared equallyM1

Each = +2.0 nCA1

WORKED EXAMPLE 33 Force after sharing (exam level)

Before touching, the spheres in the previous example were 0.10 m apart. Compare the force before and after touching at the same separation.

Solution

Before: k(6.0)(2.0) × 10⁻¹⁸/0.010 = 1.08 × 10⁻⁵ N attractiveM1

After: k(2.0)(2.0) × 10⁻¹⁸/0.010 = 3.6 × 10⁻⁶ N repulsiveA1

WORKED EXAMPLE 34 Current as charge flow (standard)

A Van de Graaff dome gains 2.0 μC per second. What is the current carried by the belt?

Solution

I = Δq/ΔtM1

= 2.0 μAA1

6.10 Graded practice: explanations

WORKED EXAMPLE 35 Neutral attraction (exam level)

Explain why small pieces of paper are attracted to a charged comb.

Solution

The comb induces opposite charge on the near side of the paper (polarisation)R1

The attraction to the nearer opposite charge exceeds the repulsion from the more distant like chargeR1

WORKED EXAMPLE 36 Uniform field evidence (standard)

Explain how the field lines between parallel plates show the field is uniform.

Solution

Lines are parallel and equally spacedR1

Same direction and density everywhere between the plates (except near the edges)R1

WORKED EXAMPLE 37 Solenoid core (standard)

Explain why a soft iron core greatly increases the field of a solenoid.

Solution

The iron becomes magnetised: its domains align with the solenoid's fieldR1

Their fields add to the coil's fieldR1

6.11 Multi-step problem: electron between plates

WORKED EXAMPLE 38 Electron deflection preview (exam level)

An electron enters the uniform field between plates 1.0 cm apart with 200 V across them. Find (a) E; (b) the force on the electron; (c) its acceleration.

Solution

(a) E = 200/0.010 = 2.0 × 10⁴ V m⁻¹A1

(b) F = eE = 3.2 × 10⁻¹⁵ NA1

6.12 Self-test 2

Term	Meaning
elementary charge	e = 1.60 × 10⁻¹⁹ C
Coulomb's law	F = kq₁q₂/r²
electric field strength	force per unit positive charge
uniform field	E = V/d between parallel plates
magnetic flux density	B, measured in tesla
electric potential (HL)	work per unit charge from infinity, kQ/r
equipotential (HL)	surface of constant potential

Students often think…	Actually…
field lines can cross	they never cross
neutral objects are not attracted to charged ones	induced charges cause attraction
the field between plates is stronger near the plates	it is uniform
magnetic fields act on stationary charges	only on moving charges
potential and field strength are the same	field is the negative gradient of potential (HL)

Topic	Connection
D.1 Gravitational fields	inverse-square fields
D.3 Motion in EM fields	forces on charges in fields
B.5 Current and circuits	potential difference
D.4 Induction (HL)	changing magnetic fields

Part	Answer	Marks
(a)	E = 2000/0.040 = 5.0 × 10⁴ V m⁻¹	A1
(b)	qE = mg	M1
	q = 2.4 × 10⁻¹⁵ × 9.81/5.0 × 10⁴ = 4.7 × 10⁻¹⁹ C	A1
(c)	about 3	A1
(d)	parallel, equally spaced lines	A1
	from positive to negative plate; curved at edges	A1
(e)	F = 8.99 × 10⁹ × 6.0 × 10⁻¹⁸/0.010	M1
	5.4 × 10⁻⁶ N	A1

Idea	Remember
Charge	quantised, conserved; e = 1.6 × 10⁻¹⁹ C
Coulomb's law	F = kq₁q₂/r²
Field	E = F/q = kQ/r²
Uniform field	E = V/d
Magnetic fields	wires, solenoids, magnets
Comparisons	gravity vs electric vs magnetic