2 higher-level sections hidden.
Educerie · IB Diploma · Physics
Theme D Fields · D.2 Electric and magnetic fields
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| State the direction of the force between like and unlike charges, and apply conservation of charge | SL, HL | Paper 1A, one mark; "state the charge on the sphere after…" |
| Explain charging by friction, by contact and by electrostatic induction, including earthing | SL, HL | "Explain how the sphere becomes positively charged" (3 to 4 marks) |
| Describe Millikan's experiment and explain how it shows that charge is quantised | SL, HL | Paper 1B: a table of drop charges; "deduce a value for e" (2 to 3 marks) |
| Use Coulomb's law F = kq₁q₂/r², with k = 1/4πε₀, and with other permittivities | SL, HL | Paper 1A ratio questions; Paper 2 calculations, 2 to 3 marks |
| Use E = F/q, and find the field of a point charge and the resultant field on a line between two charges | SL, HL | "Calculate the electric field strength at the midpoint" (3 marks) |
| Sketch and interpret electric field lines: point charges, two charges, a conducting sphere, parallel plates with edge effects | SL, HL | "Sketch the electric field pattern…" (2 to 3 marks) |
| Relate field line density to field strength | SL, HL | Paper 1A, and one mark inside a sketch question |
| Use E = V/d for the uniform field between parallel plates | SL, HL | 1 to 2 marks, often the first step of a D.3 question |
| Sketch magnetic field lines for a bar magnet, a straight wire, a flat coil and a solenoid, and find the field direction round a wire | SL, HL | "Draw the magnetic field pattern…", "state the direction of the field at P" |
| Define electric potential energy and electric potential; use Eₚ = kq₁q₂/r and Vₑ = kQ/r, with zero at infinity | HL only | "Calculate the work done to bring the charges together" (2 to 3 marks) |
| Use E = −ΔVₑ/Δr and W = qΔVₑ, in joules and in electronvolts | HL only | Paper 2 calculation from a V–r graph or a potential difference |
| Sketch and interpret equipotential surfaces and their relation to field lines | HL only | "Draw the 50 V equipotential"; "explain why no work is done…" |
Before you start
You need forces as vectors and Newton's second law from A.2, and work done by a force from A.3. Gravitational fields from D.1 are the closest relative of everything here: the same inverse-square law, the same idea of a field as force per unit of whatever feels it. Circuits from B.5 gave you charge in coulombs and potential difference in volts. The data booklet gives e = 1.60 × 10⁻¹⁹ C, k = 8.99 × 10⁹ N m² C⁻² and ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻².
1The idea in one paragraph
Electric charge comes in two kinds: like charges repel, unlike charges attract, with a force that falls as the square of the distance (Coulomb's law). Charge is never made or destroyed, only moved, and it comes only in whole-number multiples of one smallest charge, e, as Millikan showed with oil drops. Each charge sets up an electric field around it, measured as force per coulomb and drawn as field lines that crowd together where the field is strong. Currents and magnets set up a magnetic field, drawn the same way. At HL, each point in an electric field also has a potential, the work done per coulomb to bring a charge there from far away, and the field always points downhill in potential.
2Two kinds of charge, and charge is conserved
An atom has positive protons in its nucleus and negative electrons around it, and in a neutral atom the two charges cancel exactly. Everything charged in this subtopic is charged because electrons have been added or removed. A body with extra electrons is negatively charged; a body that has lost electrons is positively charged. Protons stay locked in nuclei and do not move in any of these processes.
The rule for the forces is short:
Like charges repel. Unlike charges attract.
Charge is measured in coulombs, C. One electron carries −e and one proton +e, where e = 1.60 × 10⁻¹⁹ C is the elementary charge.
The conservation of electric charge states that the total charge of an isolated system never changes. When you rub a balloon on a jumper, the balloon does not gain charge from nowhere: electrons move from the jumper to the balloon, and the jumper is left with exactly as much positive charge as the balloon has negative. If the balloon ends with −4.8 nC, the jumper has +4.8 nC, and the number of electrons that moved is
The balloon contains roughly 10²⁴ electrons, so charging moves only a tiny fraction of them.
3Charging: friction, contact, induction and earthing
The guide names three ways to move charge between bodies. All three work by moving electrons.
Friction. Rubbing two different insulators together transfers electrons from the material that holds them less tightly to the other. A polythene rod rubbed with wool takes electrons and becomes negative; an acetate rod rubbed with the same wool loses electrons and becomes positive. The two bodies always end with equal and opposite charges.
Contact. Touch a charged conductor against a neutral conductor and electrons flow between them until charge is shared. A sphere carrying −6 nC touched against an identical neutral sphere leaves each with −3 nC; for identical spheres the share is equal. Total charge is conserved, so the charges after contact always add up to the charge before.
Electrostatic induction. A charged body can charge a conductor without touching it. Figure 1 follows the four steps with a negatively charged rod and a metal sphere on an insulating stand.
- The rod is brought near. Its negative charge repels the free electrons in the metal to the far side of the sphere, leaving the near side positive. The sphere is still neutral overall; its charges have only separated. This separation alone is why a charged rod attracts a neutral object: the near, opposite charges are closer to the rod than the far, like charges, so the attraction wins.
- The sphere is earthed (grounded): connected by a wire to the Earth. The Earth is so large that it can give or take any number of electrons without its own charge changing noticeably. The electrons repelled by the rod flow down the wire to earth.
- The earth connection is broken while the rod is still in place. The electrons now have no way back.
- The rod is taken away. The sphere is left with a net positive charge, which spreads over its surface.
The charge induced is always opposite to the charge on the rod. Take the steps out of order, removing the rod before breaking the earth connection, and the electrons flow back up from the earth and the sphere ends neutral. With a positive rod the story runs in reverse: electrons are attracted up from the earth, and the sphere ends negative.
Earthing is also how charge is removed on purpose: a fuel tanker is earthed before it unloads, so charge built up by friction flows away instead of sparking.
4Millikan's experiment: charge comes in lumps
How do we know that charge comes in whole multiples of e rather than in any amount at all? The evidence the guide names is Millikan's experiment, carried out by Robert Millikan in the years around 1910. Figure 2 shows the idea.
A fine spray of oil drops enters the space between two horizontal metal plates. Friction in the spray gives some drops a small charge. A pd V across the plates, a distance d apart, sets up a uniform electric field E = V/d between them (section 8). Watching one drop through a microscope, the experimenter adjusts V until the drop hangs still. For that drop, the electric force balances the weight:
The mass of the drop was found separately, from how fast it fell when the field was switched off (a small drop falls at a steady terminal speed that depends on its size). Here is a worked example with invented numbers. A drop of mass 4.2 × 10⁻¹⁵ kg is held still between plates 6.0 mm apart when the pd is 510 V.
One drop proves nothing. The evidence is in the pattern across many drops. Figure 3 plots the charges found on sixteen drops.
Every charge sits close to 1e, 2e, 3e and so on, and none falls between. If charge could take any value, the dots would spread evenly along the line. They do not, so charge is quantised: it exists only in whole-number multiples of the elementary charge, and the step between clusters is e itself. To find e from data, divide every charge by the smallest one and look for whole-number ratios (if the ratios come out as 1.5 or 2.5, the smallest drop carried 2e, not e); then divide each charge by its whole number and average. Theme E adds two more quantised quantities: the energy of light and the energy of an electron in an atom.
5Coulomb's law
Coulomb's law gives the force between two point charges q₁ and q₂ a distance r apart:
F = kq₁q₂ ÷ r², where k = 1 ÷ 4πε₀
The constant ε₀ is the permittivity of free space, and k = 8.99 × 10⁹ N m² C⁻² in a vacuum. The force acts along the line joining the charges; it is repulsive for like charges and attractive for unlike ones. Use the sizes of the charges to find the size of the force, then decide its direction from the signs. Plugging in a negative sign and trying to interpret "a negative force" is where students go wrong.
Coulomb's law is exact for point charges and for spheres of charge, measured centre to centre. It is an inverse-square law, like Newton's law of gravitation in D.1: double the distance and the force falls to a quarter; triple it and it falls to a ninth.
Worked example 1. Two small charged spheres carry +3.0 nC and −5.0 nC, with their centres 4.0 cm apart in air. Find the force on each.
Each sphere feels 8.4 × 10⁻⁵ N towards the other: a Newton's third law pair, equal and opposite whatever the sizes of the charges.
Other materials. The guide expects a range of permittivities. In a material, replace ε₀ by the permittivity of that material, ε, so k = 1/4πε. The material's molecules partly line up in the field and weaken it, so ε is larger than ε₀ and the force is smaller. Air is close enough to a vacuum to use ε₀. Water's permittivity is about 80 times ε₀, so the same two spheres under water attract with about 1/80 of the force, roughly 1.1 × 10⁻⁶ N; that is why ionic solids fall apart so readily in water.
How strong is it? The guide asks about the relative strengths of the fundamental forces. Compare the electric and the gravitational attraction between the proton and the electron in a hydrogen atom, 5.3 × 10⁻¹¹ m apart:
Both forces go as 1/r², so the ratio is the same at every separation. Gravity still runs the solar system because planets are almost exactly neutral: their electric forces cancel, while gravity only ever adds.
6Electric field strength
A field is a region in which a body feels a force. An electric field is a region in which a charge feels a force. To measure how strong the field is at a point, place a small positive test charge q there and measure the force on it:
E = F ÷ q
The electric field strength E is the force per unit charge on a small positive test charge. Its unit is N C⁻¹, which is the same as V m⁻¹. It is a vector: its direction is the direction of the force on a positive charge. A negative charge in the same field is pushed the opposite way. "Small" matters, because a large test charge would push the other charges around and change the field it was meant to measure.
For a point charge Q, put Coulomb's law into E = F/q and the test charge cancels:
So the field of a point charge also falls with the square of the distance, and it points away from a positive Q and towards a negative one. Figure 4 draws both.
Two charges on a line. Fields are vectors, so at any point the field of two charges is the vector sum of the two separate fields. The guide keeps the calculations to points on the line joining the charges, where the vectors point along the same line and you add or subtract.
Worked example 2. A charge of +4.0 nC sits at A and a charge of +1.0 nC sits at B, 0.30 m from A. Find the field at the midpoint M, and find the point between them where the field is zero.
For the zero, the two fields must be equal in size and opposite in direction, which happens between two like charges. Call the distance from A x:
The neutral point is nearer the smaller charge, which makes sense: to match the larger charge's field, you must be closer to the smaller one.
7Field lines, and what they tell you
An electric field line shows the direction of the force on a small positive charge at every point along it. Field lines follow four rules, and every sketch is marked on them:
- Lines start on positive charges and end on negative charges (or run off to infinity).
- The arrow shows the direction of the force on a positive charge.
- Lines never cross. If they did, a charge at the crossing would be pushed two ways at once.
- Where the lines are close together the field is strong; where they spread out it is weak.
The fourth rule is the guide's relationship between field line density and field strength. Around a point charge the lines spread out as they go, and the number crossing each square metre falls as 1/r², exactly as E does. Lines leave or enter a surface at right angles when the surface is a conductor.
Figure 5 shows the two pairs you must know. For +Q and −Q, every line from the positive charge curves round and ends on the negative one. For two equal positive charges, the lines push away from each other and there is a neutral point X midway between them, where the fields cancel and no line passes.
Parallel plates. Two flat plates with equal and opposite charges produce the field in Figure 6. In the middle the lines are straight, parallel and evenly spaced, so the field strength is the same everywhere: a uniform field. Near the ends the lines bow outwards and the field weakens. These are the edge effects, and the guide expects them in your sketch.
A conducting sphere. Put charge on a metal sphere and it spreads out until it is at rest. That can only happen when there is no field inside the metal, because any field would push the free electrons along. So all the charge ends up on the outer surface, and the field inside is zero, whether the sphere is solid or hollow. Outside, the field lines leave the surface at right angles and point radially outwards, exactly as if all the charge were a point charge at the centre. Figure 7 shows both.
So for r ≥ R, E = kQ/r²; for r < R, E = 0. The largest field is just outside the surface, where r = R.
The algebra and the pictures say the same things: E = kQ/r² weakens with distance and the lines spread out; the vector sum in worked example 2 finds a zero between like charges and Figure 5(b) shows the gap. A strong answer uses both.
8The uniform field between parallel plates
Between two parallel plates a distance d apart, with a pd V across them, the field is uniform and its strength is
E = V ÷ d
This is why N C⁻¹ and V m⁻¹ are the same unit. The field points from the positive plate to the negative plate, and it has the same size near the positive plate as near the negative plate, which students often doubt.
Worked example 3. Two plates 2.5 cm apart have a pd of 500 V across them. Find the field strength, and compare the electric force on an electron between the plates with its weight.
The electric force is more than 10¹⁴ times the weight, which is why gravity is ignored for electrons and ions in fields. It was not ignored for Millikan's oil drops, which are about 10¹⁵ times heavier than an electron. D.3 takes these forces and works out the motion.
9Magnetic field lines
A magnetic field is a region in which a magnet, a moving charge or a current-carrying wire feels a force. Its lines follow the electric rules: the direction is the way a compass needle's north pole points, lines never cross, and closer lines mean a stronger field. One rule differs: magnetic field lines never start or end. They are closed loops, running N to S outside a magnet and S back to N inside it.
The guide limits the patterns to four, shown in Figure 8.
A bar magnet (a). Lines leave the N pole and curve round into the S pole, most crowded at the poles, where the field is strongest.
A straight wire (b). The lines are concentric circles round the wire, in planes at right angles to it, spreading out further from the wire because the field weakens. The direction comes from the right-hand grip rule: grip the wire with your right hand, thumb pointing along the conventional current, and your fingers curl the way the field goes. In a flat drawing, a dot ⊙ means the current comes out of the page towards you (you see the point of the arrow), and a cross ⊗ means it goes into the page (you see its tail feathers). A current out of the page gives anticlockwise field lines; into the page, clockwise.
A flat circular coil (c). The slice through the coil's centre cuts the wire twice, once coming out of the page and once going in. Near each, the field circles the wire as round a straight wire; in the middle every part of the loop contributes the same way, so the field is strong and runs straight through along the axis.
A solenoid (d). A long coil of many turns, with no iron inside, is an air-core solenoid. Inside, the lines are straight, parallel and evenly spaced: the field is uniform, just like the electric field between parallel plates. Outside, the pattern is that of a bar magnet. To find which end is the north pole, grip the solenoid with your right hand with your fingers curling the way the current flows round the turns; your thumb points to the N end. Equivalently, look at one end: if the current flows anticlockwise as you look at it, that end is N.
D.3 puts moving charges and currents into these fields.
10HLElectric potential energy and electric potential
SL students can skip to section 12.
Electric potential energy. Pushing two positive charges together takes work, and that work is stored in the arrangement. The electric potential energy Eₚ of a system of charges is the work done to assemble it from infinite separation. For two point charges a distance r apart:
Eₚ = kq₁q₂ ÷ r
Here the signs of the charges go in. Like charges give a positive Eₚ: work had to be done to push them together. Unlike charges give a negative Eₚ: they pull together on their own, and energy must be supplied to separate them again. Eₚ = 0 means infinitely far apart.
Worked example 4. How much work is needed to bring a +2.0 μC charge from far away to a point 0.10 m from a fixed +3.0 μC charge?
Released, the charges fly apart and the 0.54 J becomes kinetic energy.
Electric potential. Just as field strength is force per unit charge, electric potential is potential energy per unit charge. The electric potential Vₑ at a point is the work done per unit charge in bringing a small positive test charge from infinity to that point. For a point charge Q:
Vₑ = kQ ÷ r
The unit is the volt, one joule per coulomb, and potential is zero at infinity. Potential is a scalar: it has a sign but no direction, positive near a positive charge and negative near a negative one. The potential due to several charges is the plain sum of their separate potentials, signs included, with no components or angles.
Worked example 5. Charges of +2.0 nC, +2.0 nC, −1.0 nC and +3.0 nC sit at the four corners of a square of side 0.20 m. Find the potential at the centre.
Work done moving a charge. Moving a charge q between two points whose potentials differ by ΔVₑ takes work
W = qΔVₑ
This is the V = W/q of circuits in B.5. A positive charge released in a field moves from high to low potential and gains kinetic energy qΔV; a negative charge moves from low to high.
The electronvolt. For single particles, joules are awkwardly large. The electronvolt, eV, is the energy transferred when a charge of e moves through a pd of 1 V:
A particle of charge e gains as many eV as the volts it falls through. An electron accelerated from rest through 2.0 kV gains 2.0 keV = 2000 × 1.60 × 10⁻¹⁹ = 3.2 × 10⁻¹⁶ J of kinetic energy; a proton moving from +300 V to +100 V gains 200 eV.
Field strength as a potential gradient. The field and the potential are two descriptions of one thing, and they are linked by
E = −ΔVₑ ÷ Δr
The field strength is minus the potential gradient. On a graph of V against r, E is minus the gradient of the tangent at a point. The minus sign says the field points the way the potential falls: downhill. If the potential drops from 900 V to 600 V over 5.0 cm along a line, the field along that line is 300 ÷ 0.050 = 6.0 × 10³ V m⁻¹, pointing towards the lower potential. Between parallel plates the potential falls steadily from one plate to the other, so its gradient is constant, and its size is V/d: the E = V/d of section 8 is this rule in its simplest case.
The same ideas hold for gravity in D.1, with one difference worth stating: gravitational potential is always negative, because gravity only attracts; electric potential can be either sign.
11HLEquipotential surfaces
An equipotential surface is a surface on which every point has the same potential. On a flat diagram it appears as a line. Two facts follow from the definition, and both are regularly examined.
No work is done moving a charge along an equipotential, because ΔV = 0, so W = qΔV = 0.
Equipotentials are perpendicular to field lines everywhere. If the field had any component along an equipotential, moving a charge along it would take work, which the first fact rules out. So field lines cross equipotentials at right angles, and the field points from higher to lower equipotentials.
Figure 9 shows the pattern for a point charge and for a pair of equal and opposite charges.
For the point charge of 1.0 nC, V = kQ/r gives 90 V at 0.10 m, 60 V at 0.15 m, 45 V at 0.20 m and 30 V at 0.30 m. The equipotentials are concentric spheres, and for equal steps of potential they get further apart as you move out, because the field, which is the potential gradient, is getting weaker. Close equipotentials mean a strong field, just as close field lines do.
The guide lists the cases to recognise:
- A point charge: concentric spheres centred on the charge, spacing growing outwards (Figure 9a).
- A collection of up to four point charges: add the potentials of the charges as scalars. Very close to any one charge its own potential dominates, so the equipotentials are small near-circles round it. For +Q and −Q the plane halfway between them is at V = 0 (Figure 9b). Far from a group, the equipotentials become circles again, as if the total charge sat at the centre.
- A solid charged conducting sphere, and a hollow one: there is no field inside the metal or inside the cavity, so there is no potential gradient, and the whole conductor, inside included, is one equipotential at V = kQ/R. Outside, the equipotentials are concentric spheres, the same as for a point charge. Figure 10 draws V and E against distance from the centre.
- Two oppositely charged parallel plates: flat planes parallel to the plates, equally spaced for equal steps of potential, because the field is uniform. Near the edges they bend round, following the bowing field lines.
In Figure 10, inside the sphere the potential is flat, so its gradient and E are zero; outside, V falls as 1/r and E, minus its gradient, falls as 1/r². This is how you map a field using potential: draw field lines at right angles to the equipotentials, pointing downhill, and read the strength from how tightly the equipotentials are packed.
12Where marks are lost
Putting the signs of the charges into Coulomb's law and then misreading the answer. For the force, use sizes, then state attract or repel from the signs. (At HL, Eₚ and V are different: there the signs go in, because they are scalars.)
Forgetting to square r, or using the diameter. F and E go as 1/r², with r measured from centre to centre. Doubling r quarters F.
Defining E without "positive" or "per unit charge". E is the force per unit charge on a small positive test charge. "The force on a charge" scores nothing.
Adding field strengths as numbers when they point opposite ways. E is a vector. Draw an arrow for each charge's field at the point, then add or subtract.
Crossing field lines, or drawing them without arrows. Lines never cross, always carry arrows, and meet a conductor's surface at right angles.
Saying the field inside a charged conducting sphere is largest at the centre. It is zero everywhere inside, solid or hollow; the charge sits on the outer surface.
Taking the rod away before breaking the earth connection in induction. The electrons flow back and the sphere ends neutral. The order is: rod near, earth, remove earth, remove rod.
HL: treating potential as a vector. Potentials add as plain numbers with their signs; there are no components. Field strengths add as vectors.
13Draw it right
- Electric field lines start on + and end on −, carry arrows, never cross, and meet a conductor's surface at 90°.
- Point charge: evenly spaced radial lines, outward for +, inward for −. Two charges: + to − lines curving between unlike charges; like charges push their lines apart round an empty neutral point, nearer the smaller charge.
- Parallel plates: straight, parallel, evenly spaced lines from + to −, bowing outwards at both ends. Conducting sphere: nothing inside, radial lines outside.
- Magnetic fields: closed loops with arrows, N to S outside a magnet; concentric circles round a wire with spacing increasing outwards, direction by the right-hand grip rule; a coil with circles round each side and a strong field through the middle; a solenoid with uniform lines inside and a bar-magnet pattern outside, N and S labelled.
- HL equipotentials: dashed, at right angles to every field line; circles getting further apart round a point charge; equally spaced straight lines between plates.
- Label charges with signs, poles with N and S, and currents with ⊙ or ⊗.
14Try it
Marks in brackets. Use k = 8.99 × 10⁹ N m² C⁻², e = 1.60 × 10⁻¹⁹ C, g = 9.81 m s⁻². Answers and marker's notes are at the end.
Q1. (Paper 1A style) Two point charges repel each other with a force F. One charge is doubled and the distance between them is tripled. What is the new force? 1 mark
A. 2F/3 B. 2F/9 C. 4F/9 D. 6F
Q2. A metal sphere on an insulating stand is neutral. Explain how it can be given a negative charge using a positively charged rod, without the rod touching it. 4 marks
Q3. (Paper 1B style) In an oil-drop experiment, the magnitudes of the charges on six drops are found. The data are invented.
| Drop | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| q / 10⁻¹⁹ C | 6.41 | 9.59 | 4.83 | 12.78 | 8.02 | 3.19 |
(a) Explain how these data support the idea that charge is quantised. 2 marks
(b) Determine a value for the elementary charge from the data. 2 marks
(c) Another drop, of mass 3.3 × 10⁻¹⁵ kg, is held stationary between horizontal plates 5.0 mm apart when the pd between them is 506 V. Calculate the charge on the drop, and state how many elementary charges it carries. 3 marks
Q4. A point charge of +6.0 nC is fixed at A and a point charge of −2.0 nC is fixed at B, 0.40 m from A. M is the midpoint of AB.
(a) Calculate the electric field strength at M, and state its direction. 3 marks
(b) (HL) Calculate the electric potential at M. 2 marks
(c) (HL) An electron is placed at M. Determine the minimum energy, in eV and in J, needed to move it to infinity. 2 marks
Q5. (a) Sketch the magnetic field pattern inside and around a current-carrying air-core solenoid. Label the north pole and show the current direction. 3 marks
(b) A long straight vertical wire carries a current upwards. Point P is 5 cm due east of the wire. State the direction of the magnetic field due to the wire at P. 1 mark
Q6. (HL) Two parallel plates 4.0 cm apart are held at 0 V and +200 V.
(a) Calculate the electric field strength between the plates. 1 mark
(b) Describe the equipotentials at 50 V, 100 V and 150 V between the plates. 2 marks
(c) A proton moves from the 150 V equipotential to the 50 V equipotential. Calculate the work done on it by the field, in eV and in J. 2 marks
(d) Explain why no work is done when the proton moves along the 100 V equipotential. 1 mark
15In one breath
Like charges repel, unlike attract, and charge is conserved: bodies are charged by moving electrons, by friction, by contact, or by induction, where a nearby rod separates charge and earthing lets electrons leave or arrive before the earth and then the rod are removed. Millikan balanced oil drops, qV/d = mg, and every charge was a whole multiple of e = 1.60 × 10⁻¹⁹ C: charge is quantised. Coulomb's law, F = kq₁q₂/r² with k = 1/4πε₀, is inverse-square, and a larger permittivity weakens it. E = F/q is force per unit positive charge, kQ/r² for a point charge, added as vectors. Field lines run + to −, never cross and crowd where E is strong; between plates they are uniform, E = V/d, with edge effects; inside a charged conductor E = 0. Magnetic field lines are closed loops: N to S outside a magnet, circles round a wire by the right-hand grip rule, straight through a coil, uniform inside a solenoid. HL: Eₚ = kq₁q₂/r is the work to assemble charges from infinity; Vₑ = kQ/r is work per unit charge from infinity, a scalar; W = qΔV in J or eV; E = −ΔV/Δr; equipotentials meet field lines at right angles, and moving along one takes no work.
Answers
Q1. B. F ∝ q₁q₂/r². Doubling one charge doubles F; tripling r divides it by 9. New force = 2F/9. B only. A forgets to square r; C doubles both charges; D multiplies by 3 instead of dividing by 9.
Q2. The positive rod, held near, attracts free electrons to the near side of the sphere, leaving the far side positive. Earth the sphere with the rod still near: electrons flow up from the earth onto the sphere, attracted by the rod. Break the earth connection with the rod still in place, so the electrons cannot leave. Remove the rod: the sphere keeps a net negative charge, spread over its surface. 1 for the rod attracting electrons to the near side / separating charge, 1 for earthing with electrons flowing from earth to the sphere, 1 for removing the earth before the rod, 1 for the final charge being negative and spreading out. "Positive charge flows to earth" loses the second mark: only electrons move.
Q3. (a) Every charge is close to a whole-number multiple of one value, about 1.6 × 10⁻¹⁹ C (ratios 4, 6, 3, 8, 5, 2), and none falls between multiples, so charge comes in discrete amounts. Note that the ratios to the smallest charge are 2, 3, 1.5, 4, 2.5, 1: the smallest drop carried 2e. (b) Dividing each charge by its whole number: 6.41/4 = 1.603, 9.59/6 = 1.598, 4.83/3 = 1.610, 12.78/8 = 1.598, 8.02/5 = 1.604, 3.19/2 = 1.595. Mean e ≈ 1.60 × 10⁻¹⁹ C. (c) q = mgd/V = 3.3 × 10⁻¹⁵ × 9.81 × 5.0 × 10⁻³ ÷ 506 = 3.2 × 10⁻¹⁹ C, which is 2 elementary charges. (a) 1 for charges being multiples of a common value, 1 for no values in between. (b) M1 for dividing each value by an integer, A1 for 1.60 × 10⁻¹⁹ C. Taking 3.19 × 10⁻¹⁹ C as the smallest charge and calling it e loses the A1. (c) M1 for qV/d = mg, A1 for 3.2 × 10⁻¹⁹ C, A1 for 2.
Q4. (a) At M, 0.20 m from each charge. Field of A: E = 8.99 × 10⁹ × 6.0 × 10⁻⁹ ÷ 0.20² = 1349 N C⁻¹, pointing away from A, towards B. Field of B: 8.99 × 10⁹ × 2.0 × 10⁻⁹ ÷ 0.20² = 450 N C⁻¹, pointing towards B (a negative charge attracts a positive test charge). Both point towards B, so they add: E = 1.8 × 10³ N C⁻¹ towards B. (b) V = k(QA + QB)/r = 8.99 × 10⁹ × (6.0 − 2.0) × 10⁻⁹ ÷ 0.20 = 1.8 × 10² V. (c) The electron has Eₚ = qV = −e × 180 V = −180 eV at M and zero at infinity, so it needs at least 180 eV = 2.9 × 10⁻¹⁷ J. (a) M1 for either field value, A1 for 1.8 × 10³ N C⁻¹, A1 for the direction; subtracting the fields loses both A marks. (b) M1 for adding the potentials with signs, A1 for 180 V. (c) 1 for 180 eV, 1 for 2.9 × 10⁻¹⁷ J.
Q5. (a) Uniform, parallel, evenly spaced lines inside along the axis; outside, a bar-magnet pattern of closed loops from one end to the other. Lines emerge from the N end, where the current, seen from that end, flows anticlockwise. (b) Seen from above the current comes towards you, so the field circles anticlockwise; due east of the wire that direction is north. (a) 1 for uniform parallel lines inside, 1 for a bar-magnet pattern outside with arrows forming closed loops, 1 for the N pole consistent with the current direction shown. (b) 1 for north.
Q6. (a) E = V/d = 200 ÷ 0.040 = 5.0 × 10³ V m⁻¹. (b) Flat planes parallel to the plates, equally spaced at 1.0 cm, 2.0 cm and 3.0 cm from the 0 V plate, perpendicular to the field lines, curving near the edges. (c) W = qΔV = e × (150 − 50) V = 100 eV = 1.6 × 10⁻¹⁷ J, done by the field as the proton moves to lower potential. (d) Every point on it is at the same potential, so ΔV = 0 and W = qΔV = 0; the field is perpendicular to the motion. (a) 1. (b) 1 for parallel to the plates, 1 for equally spaced at the correct distances. (c) 1 for 100 eV, 1 for 1.6 × 10⁻¹⁷ J. (d) 1 for ΔV = 0 linked to W = qΔV.
Educerie · written from the published IB Diploma Programme Physics guide, first assessment 2025, section D.2 Electric and magnetic fields. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.