D.2 Electric and Magnetic Fields Questions

1D-1A-04

Electric field and potential·D.2 Electric and magnetic fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Two point charges, each +Q, are fixed at two vertices of an equilateral triangle of side a. Point P is the third vertex of the triangle.

Which row gives the magnitude of the electric field strength E and the electric potential V_e at P?

E at PV_e at P

Show mark scheme

Marking point	Mark	Notes
Step 1Each charge is a distance a from P and gives a field of magnitude kQ/a², directed away from that charge along the side of the triangle.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The two sides meet at P at 60°, so the two field vectors are at 60° to each other: E = 2(kQ/a²) cos 30° = √3 kQ/a².	—
Step 3Potential is a scalar, so the two contributions simply add: V_e = kQ/a + kQ/a = 2kQ/a.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis adds the two field magnitudes as if both fields pointed the same way. The fields are at 60° to each other, so the resultant is less than their sum.
BCorrect: vector addition of two equal fields at 60° gives √3 kQ/a², and the scalar potentials add to 2kQ/a.
CThe field is right, but this resolves the potential as if it were a vector. Potential is a scalar: the two contributions add directly.
DThis takes the angle between the two field vectors as 120° (the exterior angle at P), so that 2 cos 60° = 1. The fields point away from the charges along the two sides, which meet at 60°.

Syllabus understandingD.2 — the electric field strength as given by E = F/q; the field between two point charges; D.2 (HL) — that the electric potential is a scalar quantity with zero defined at infinity; V_e = kQ/r Command term: Determine

2D-1A-12

Electric potential and work·D.2 Electric and magnetic fields

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

The graph shows how the electric potential V varies with distance x along a straight line in a region of electric field. An electron at x = 0 is to move along the line to x = 6.0 cm. Only the electric force acts on the electron.

What is the minimum kinetic energy that the electron must have at x = 0 to reach x = 6.0 cm?

Variation of electric potential with distance along a line.

Show mark scheme

Marking point	Mark	Notes
Step 1The electron has charge −e, so its electric potential energy is −eV: it is greatest where V is lowest.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2From x = 0 to x = 2.0 cm, V falls from +10 V to −20 V, so the electron's potential energy rises by e × 30 V = 30 eV.	—
Step 3Beyond 2.0 cm V rises and the electron speeds up again, so it reaches 6.0 cm provided it gets past the minimum of V: minimum kinetic energy 30 eV.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis compares only the end points: V at 6.0 cm (+25 V) is higher than at x = 0, so the electron loses potential energy overall. It must first climb the potential-energy barrier at 2.0 cm.
BThis uses the potential difference between the end points, 25 V − 10 V, ignoring the minimum of V in between and the sign of the electron's charge.
CThis uses the potential at the minimum, −20 V, measured from zero instead of from the starting potential of +10 V.
DCorrect: the electron must climb through a potential difference of 30 V to reach the minimum of V.

Syllabus understandingD.2 (HL) — the work done in moving a charge q in an electric field as given by W = qΔV_e; work done expressed in both joules and electronvolts Command term: Determine

3D-1A-15

Charged conducting sphere·D.2 Electric and magnetic fields

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A hollow conducting sphere of radius R carries a positive charge Q. Point X is inside the sphere, a distance R/2 from its centre. Point Y is outside the sphere, a distance 2R from its centre.

What is the work done by the electric field on a small positive charge q as it moves from X to Y?

Show mark scheme

Marking point	Mark	Notes
Step 1Inside a charged conducting sphere the field strength is zero, so the potential is constant and equal to its value at the surface: V_X = kQ/R.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Outside, the sphere acts as a point charge at its centre: V_Y = kQ/(2R).	—
Step 3Work done by the field = q(V_X − V_Y) = kQq/R − kQq/(2R) = kQq/(2R).	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis takes V_X = 0 because the field strength inside the sphere is zero. Zero field means a constant potential, not a zero potential.
BCorrect: V_X = kQ/R and V_Y = kQ/(2R).
CThis uses the correct V_X but ignores the potential at Y, as if Y were at infinity.
DThis treats X as if it were outside a point charge at the centre: V_X = kQ/(R/2) = 2kQ/R. Inside the conductor there is no field, so no work is done between X and the surface.

Syllabus understandingD.2 (HL) — the electric potential V_e = kQ/r with zero defined at infinity; equipotential surfaces inside and outside a hollow charged conducting sphere; W = qΔV_e Command term: Determine

4D-1A-18

Millikan's experiment·D.2 Electric and magnetic fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

In an experiment of Millikan's type, the charges on four oil drops are measured as 4.8 × 10⁻¹⁹ C, 8.0 × 10⁻¹⁹ C, 11.2 × 10⁻¹⁹ C and 14.4 × 10⁻¹⁹ C.

What is the largest value of the basic unit of charge that is consistent with these data?

Show mark scheme

Marking point	Mark	Notes
Step 1Every measured charge must be a whole-number multiple of the basic unit of charge, so the unit must divide exactly into all four values.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Test the candidate values: with 1.6 × 10⁻¹⁹ C the drops carry 3, 5, 7 and 9 units; 3.2 × 10⁻¹⁹ C gives 1.5, 2.5, 3.5, 4.5 and 4.8 × 10⁻¹⁹ C gives 8.0/4.8 = 1.67 — not whole numbers.	—
Step 3The largest unit that divides all four values exactly is 1.6 × 10⁻¹⁹ C (smaller units such as 0.80 × 10⁻¹⁹ C also divide them, but are not required by the data).	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis is consistent with the data (6, 10, 14 and 18 units), but it is not the largest such value: 1.6 × 10⁻¹⁹ C also fits every drop.
BCorrect: 4.8, 8.0, 11.2 and 14.4 are 3, 5, 7 and 9 times 1.6 (× 10⁻¹⁹ C).
CThis takes the constant difference between successive values (8.0 − 4.8 = 11.2 − 8.0 = 3.2) as the unit; but 4.8/3.2 = 1.5, so the charges are not whole multiples of 3.2 × 10⁻¹⁹ C.
DThis takes the smallest measured charge as the unit; but 8.0/4.8 = 1.67 is not a whole number.

Syllabus understandingD.2 — Millikan's experiment as evidence for quantization of electric charge Command term: Deduce

5D-1A-19

Magnetic field of a straight wire·D.2 Electric and magnetic fields

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

Two long straight parallel wires P and Q are perpendicular to the page and a distance d apart. P carries a current 2I out of the page and Q carries a current I into the page. The magnetic field strength at a distance r from a long straight wire carrying current I is proportional to I/r.

Points J, K, L and M lie on the line through the wires, as shown: K is midway between the wires, J is between the wires a distance d/3 from Q, L is a distance d beyond Q and M is a distance d beyond P.

At which point is the resultant magnetic field zero?

Two long parallel wires perpendicular to the page (⊙ current out of the page, ⊗ current into the page) and four points on the line through them.

Show mark scheme

Marking point	Mark	Notes
Step 1Right-hand grip rule: between the wires the fields of the two opposite currents point the same way, so they add. The resultant cannot be zero at J or K.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Outside the pair the two fields are opposite. They can be equal in magnitude only on the side of the smaller current, Q.	—
Step 3At L: P gives a field ∝ 2I/(2d) = I/d and Q gives a field ∝ I/d. These are equal and opposite, so the resultant is zero. At M, P gives ∝ 2I/d and Q only ∝ I/(2d), so they do not balance.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis is where the fields would cancel if both currents were in the same direction (2I/(2d/3) = I/(d/3)). With opposite currents the fields add between the wires.
BThis carries over "zero at the midpoint" from two equal currents in the same direction. Here the currents are opposite and unequal, and at the midpoint the two fields add.
CM is on the side of the larger current, where the field of P (∝ 2I/d) is always larger than the field of Q (∝ I/2d).
DCorrect: at L the fields of P and Q are equal in magnitude and opposite in direction.

Syllabus understandingD.2 — magnetic field lines of a current-carrying straight wire; the determination of the direction of the magnetic field based on the current direction (the proportionality B ∝ I/r is given in the stem) Command term: Deduce

6D-1A-23

Coulomb's law·D.2 Electric and magnetic fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Two small charged spheres in air exert an electric force F on each other. The permittivity of air may be taken as ε₀. The spheres are then immersed in an oil of permittivity 3.0ε₀ and their separation is halved.

What is the new electric force between the spheres?

Show mark scheme

Marking point	Mark	Notes
Step 1F = q₁q₂/(4πεr²), so F ∝ 1/(εr²).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Halving r multiplies the force by 4; increasing the permittivity to 3.0ε₀ divides it by 3.	—
Step 3New force = 4F/3.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis inverts the effect of the separation (dividing by 4 instead of multiplying) and also divides by 3.
BThis allows for the permittivity but ignores the halving of the separation.
CThis treats the force as inversely proportional to r rather than r²: 2/3.
DCorrect: × 4 for the separation and ÷ 3 for the permittivity.

Syllabus understandingD.2 — Coulomb's law as given by F = kq₁q₂/r² for charged bodies treated as point charges, where k = 1/(4πε); a range of permittivity values Command term: Determine

7D-1A-26

Equipotential surfaces·D.2 Electric and magnetic fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

The diagram shows the cross-sections of five plane equipotential surfaces in a region of electric field, with their potentials and a scale. Point P is midway between the 60 V and 40 V surfaces.

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

Cross-sections of equipotential surfaces (dashed) in a region of electric field.

Show mark scheme

Marking point	Mark	Notes
Step 1E = −ΔV_e/Δr: the field strength is the potential gradient at right angles to the equipotentials, so use the surfaces on either side of P.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2ΔV = 60 V − 40 V = 20 V across 4.5 cm − 2.5 cm = 2.0 cm.	—
Step 3E = 20/0.020 = 1.0 × 10³ V m⁻¹, directed from the 60 V surface towards the 40 V surface.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis divides the 20 V by the distance of P from the 100 V surface (3.5 cm) instead of by the spacing of the surfaces on either side of P.
BCorrect: 20 V/0.020 m = 1.0 × 10³ V m⁻¹.
CThis divides the potential at P (50 V) by its distance from the 100 V surface, as if E = V/d could be applied from a reference surface: 50/0.035.
DThis uses the 1.0 cm spacing of the first two surfaces. The surfaces are not equally spaced, so the field is not uniform and must be found near P.

Syllabus understandingD.2 (HL) — the electric field strength E as the electric potential gradient as given by E = −ΔV_e/Δr; equipotential surfaces for electric fields Command term: Determine

8D-1A-40

Charging by induction and earthing·D.2 Electric and magnetic fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

An uncharged aluminium sphere stands on an insulating base. A negatively charged PVC rod is held close to the sphere but never touches it. With the rod still in place, the sphere is briefly connected to earth by a wire. The wire is then removed, and finally the rod is taken away.

I. While the sphere is connected to earth, electrons flow from the sphere to earth.

II. After the rod is taken away, the sphere is positively charged.

III. If the rod had been taken away before the earth wire was removed, the sphere would also have been left positively charged.

Which statements are correct?

Show mark scheme

Marking point	Mark	Notes
Step 1The negative rod repels free electrons in the sphere; when the sphere is earthed, these electrons flow from the sphere to earth: I is correct.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The wire is removed while the rod is still there, so the electrons cannot return. The sphere has a deficit of electrons and is left positive: II is correct.	—
Step 3If the rod were removed first, with the earth wire still connected, electrons would flow back from earth and the sphere would be left uncharged: III is wrong.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: I and II only.
BThis accepts III. While the earth wire is still connected, removing the rod lets electrons flow back from earth, so the sphere ends uncharged. It also rejects II, which is correct.
CThis rejects I, as if electrons were attracted on to the sphere from earth. The negative rod repels electrons, so they flow away to earth.
DThis accepts III. The order of the last two steps matters: electrons can return through the earth wire for as long as it stays connected.

9D-1A-41

Conservation of charge·D.2 Electric and magnetic fields

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Three identical small conducting spheres X, Y and Z are mounted on insulating handles. At first X has a charge of +8.0 nC, Y has a charge of −2.0 nC and Z is uncharged.

X is touched against Y and they are then separated. Y is next touched against Z and they are separated.

What is the final charge on Z?

Show mark scheme

Marking point	Mark	Notes
Step 1Charge is conserved, and identical conducting spheres in contact share the total charge equally.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2X and Y: total = +8.0 + (−2.0) = +6.0 nC, so each has +3.0 nC after they are separated.	—
Step 3Y and Z: total = +3.0 + 0 = +3.0 nC, so Z ends with +1.5 nC.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: +6.0 nC is shared between X and Y, then Y's +3.0 nC is shared with Z.
BThis shares the total +6.0 nC equally among all three spheres, as if all three had touched at once. The order of the contacts is ignored.
CThis adds the magnitudes (8.0 + 2.0 = 10 nC) and ignores the sign of the charge on Y: 10/2 = 5.0 nC, then 5.0/2 = 2.5 nC.
DThis assumes that all of Y's charge (+3.0 nC) passes to Z. Identical spheres in contact share the charge equally.

10D-1A-42

Coulomb's law·D.2 Electric and magnetic fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Two small beads P and R are fixed 0.200 m apart on a horizontal insulating thread in air. P has a charge of +3.0 nC and R has a charge of −5.0 nC. A third bead S with a charge of −2.0 nC is placed on the line between them, 0.080 m from P.

What is the magnitude of the resultant electric force on S?

Show mark scheme

Marking point	Mark	Notes
Step 1P is positive and attracts S towards P. R and S are both negative, so R repels S — away from R, which is also towards P.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Force from P = 8.99 × 10⁹ × 3.0 × 10⁻⁹ × 2.0 × 10⁻⁹/0.080² = 8.43 × 10⁻⁶ N; force from R (0.120 m away) = 8.99 × 10⁹ × 5.0 × 10⁻⁹ × 2.0 × 10⁻⁹/0.120² = 6.24 × 10⁻⁶ N.	—
Step 3Both forces act towards P, so they add: 8.43 + 6.24 = 14.7 × 10⁻⁶ N ≈ 1.5 × 10⁻⁵ N.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis subtracts the forces, as if R attracted S towards R: 8.43 − 6.24 = 2.2 μN. R and S have the same sign, so R repels S.
BThis includes only the force from the nearer bead P.
CThis uses 0.200 m, the distance from P to R, as the distance from S to R: 8.43 + 2.25 = 10.7 μN.
DCorrect: the attraction towards P and the repulsion from R are in the same direction.

Syllabus understandingD.2 — the direction of forces between the two types of electric charge; Coulomb's law as given by F = kq₁q₂/r² for charged bodies treated as point charges Command term: Determine

11D-1A-43

Electric field strength·D.2 Electric and magnetic fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

In an experiment of Millikan's type, a charged oil drop is held stationary between two horizontal parallel plates. The potential difference between the plates is then tripled and, at the same moment, the separation of the plates is halved. Air resistance is negligible at the instant of the change.

What is the magnitude of the acceleration of the drop immediately after the change?

Show mark scheme

Marking point	Mark	Notes
Step 1Initially the electric force balances the weight: qE = qV/d = mg.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2E = V/d: tripling V and halving d multiply E, and so the electric force, by 6. The electric force becomes 6mg upwards.	—
Step 3Resultant force = 6mg − mg = 5mg upwards, so the acceleration is 5g.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis triples the field but ignores the halving of the separation: 3mg − mg = 2mg.
BCorrect: (6mg − mg)/m = 5g.
CThis forgets that the weight still acts: 6mg/m.
DThis adds the weight to the electric force, as if both acted in the same direction.

Syllabus understandingD.2 — the electric field strength as given by E = F/q; the uniform electric field strength between parallel plates as given by E = V/d; Millikan's experiment Command term: Determine

12D-1A-44

Electric field lines·D.2 Electric and magnetic fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

The diagram shows the electric field lines around two point charges X and Y. All the field lines that start or end on the charges are drawn.

Which row gives the sign of the charge on Y and the ratio (magnitude of charge on X)/(magnitude of charge on Y)?

Electric field lines around two point charges X and Y (the arrows show the direction of the field).

Sign of charge on YRatio

Show mark scheme

Marking point	Mark	Notes
Step 1Field lines start on positive charges and end on negative charges. The arrows point away from X and into Y, so X is positive and Y is negative.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The number of lines starting or ending on a charge is proportional to the size of the charge. 16 lines leave X and 8 end on Y.	—
Step 3Ratio = 16/8 = 2. The 8 lines from X that do not end on Y go off to large distances because the pair has a net positive charge.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis inverts the ratio: Y has half as many lines as X, so the charge on X is twice the charge on Y.
BThis assumes that equal and opposite charges produced the pattern. In that case every line from X would end on Y, but here half of them escape.
CCorrect: lines end on Y, so Y is negative, and 16 lines leave X while 8 end on Y.
DThis ignores the direction of the arrows. Lines point into Y and end there, which only happens at a negative charge.

Syllabus understandingD.2 — electric field lines; the relationship between field line density and field strength; the field between two point charges Command term: Deduce

13D-1A-45

Uniform electric fields·D.2 Electric and magnetic fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

One cell of an electrostatic air cleaner consists of two parallel metal plates with a potential difference between them. The plates are long compared with their separation. Point P is midway between the plates near their centre. Point Q is also near the centre, a quarter of the way from the negative plate to the positive plate. Point R is level with P but just beyond the edge of the plates, as shown.

An electron is placed in turn at P, Q and R. Which describes the magnitudes F_P, F_Q and F_R of the electric force on it?

Electric field between two oppositely charged parallel plates, including the edge region.

Show mark scheme

Marking point	Mark	Notes
Step 1Well inside the plates the field is uniform, E = V/d, and has the same value at every point, so F_P = F_Q.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2At the edges the field lines bow outwards; just beyond the edge they are further apart, so the field is weaker there but not zero.	—
Step 3F = eE, so F_P = F_Q > F_R > 0.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: uniform field between the plates and a weaker, non-zero field just beyond the edge.
BThis ignores the edge effect: E = V/d applies only well inside the plates, and beyond the edge the field lines spread out.
CThis assumes that the field is confined exactly to the space between the plates. The field lines bulge beyond the edges, so the field there is not zero.
DThis treats each plate like a point charge, with a stronger field closer to it. Between large parallel plates the field is uniform, so Q is no different from P.

Syllabus understandingD.2 — the uniform electric field strength between parallel plates as given by E = V/d; two oppositely charged parallel plates, including edge effects; the relationship between field line density and field strength Command term: Deduce

14D-1A-46

Magnetic field of a straight wire·D.2 Electric and magnetic fields

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A long straight vertical wire passes through a hole in a horizontal card. There is a steady current upwards in the wire. A small plotting compass is placed on the card due east of the wire. The Earth's magnetic field can be ignored.

In which direction does the north pole of the compass needle point?

Show mark scheme

Marking point	Mark	Notes
Step 1The field lines of a straight wire are concentric circles around the wire in a plane perpendicular to it.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Right-hand grip rule: with the thumb pointing up along the current, the fingers curl anticlockwise when the card is viewed from above.	—
Step 3At a point due east of the wire the anticlockwise circle points north, and the compass lines up with the field.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis treats the field as radial, pointing away from the wire like the electric field of a positive charge. Magnetic field lines of a wire go round the wire.
BThis treats the field as radial, pointing into the wire. The field lines of a wire are circles around it, not lines towards it.
CThis applies the grip rule with the wrong hand or with the current reversed, so the field direction comes out the wrong way round the circle.
DCorrect: the field circulates anticlockwise when viewed from above, which is northwards at a point east of the wire.

Syllabus understandingD.2 — magnetic field lines of a current-carrying straight wire; determination of the direction of the magnetic field based on the current direction Command term: Deduce

15D-1A-47

Magnetic field of a solenoid·D.2 Electric and magnetic fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A long air-core solenoid has ends P and Q. When the solenoid is viewed along its axis from end P, the current in the turns is clockwise.

Which row gives the magnetic polarity of end P and the direction of the magnetic field lines inside the solenoid?

Polarity of end PField inside the solenoid

Show mark scheme

Marking point	Mark	Notes
Step 1Right-hand grip rule for a solenoid: curl the fingers in the direction of the current and the thumb points along the field inside the solenoid.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2With a clockwise current seen from P, the thumb points away from the observer, so the field inside runs from P towards Q.	—
Step 3Field lines leave the solenoid at its north pole, so Q is a north pole and P, where the lines enter, is a south pole.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: the field inside runs from P to Q, and the lines enter the solenoid at P, which is therefore a south pole.
BP is a south pole, but inside a solenoid (as inside a bar magnet) the field lines run from the south pole to the north pole. They form closed loops and do not run from N to S inside.
CThis contradicts itself: lines leave a north pole, so if the field inside ran from P to Q, P would have to be a south pole.
DThis applies the grip rule backwards (clockwise gives north). The answer is self-consistent, but both parts are reversed.

Syllabus understandingD.2 — magnetic field lines; (guidance) magnetic field patterns will be restricted to a bar magnet, a current-carrying straight wire, a current-carrying circular coil and an air-core solenoid Command term: Deduce

16D-1A-48

Field and potential of several charges·D.2 Electric and magnetic fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Four point charges are fixed at the corners of a square of side a. The two upper corners each carry +q and the two lower corners each carry −q. M is the midpoint of the upper side of the square.

What is the electric potential at M?

Four point charges at the corners of a square of side a. M is the midpoint of the upper side.

Show mark scheme

Marking point	Mark	Notes
Step 1Potential is a scalar: add kq/r for every charge, with its sign. Each +q is a/2 from M; each −q is √(a² + (a/2)²) = (√5/2)a from M.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The two positive charges give 2 × kq/(a/2) = 4kq/a; the two negative charges give −2 × kq/((√5/2)a) = −(4/√5)kq/a.	—
Step 3V_e = 4(1 − 1/√5)kq/a ≈ 2.2kq/a.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis carries over the result for the centre O, where the four contributions cancel because all four charges are equidistant. M is closer to the positive charges, so the potential there is positive.
BThis uses the side a, instead of the slant distance (√5/2)a, as the distance from M to each negative charge: 4kq/a − 2kq/a.
CCorrect: 4kq/a from the near positive charges minus (4/√5)kq/a from the far negative charges.
DThis ignores the signs of the lower charges and adds all four contributions as if every charge were positive.

Syllabus understandingD.2 (HL) — that the electric potential is a scalar quantity with zero defined at infinity; V_e = kQ/r; equipotential surfaces for a collection of up to four point charges Command term: Determine

17D-1A-49

Electric potential and work·D.2 Electric and magnetic fields

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

An alpha particle (charge +2e) moves in an electrostatic analyser from a point where the electric potential is +1200 V to a point where the electric potential is −300 V. Only the electric force acts on it.

What is the increase in the kinetic energy of the alpha particle?

Show mark scheme

Marking point	Mark	Notes
Step 1Work done by the field = q × (fall in potential): W = qΔV_e.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Fall in potential = +1200 − (−300) = 1500 V.	—
Step 3Gain in kinetic energy = 2e × 1500 V = 3000 eV = 3.0 keV (4.8 × 10⁻¹⁶ J).	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis uses a charge of e instead of 2e, which gives 1500 eV.
BThis ignores the negative sign of the final potential and takes the difference as 1200 − 300 = 900 V, which gives 2 × 900 = 1800 eV.
CThis uses only the starting potential (1200 V), as if the final point were at zero potential: 2 × 1200 = 2400 eV.
DCorrect: 2e × 1500 V = 3.0 keV.

Syllabus understandingD.2 (HL) — the work done in moving a charge q in an electric field as given by W = qΔV_e; work done expressed in both joules and electronvolts Command term: Determine

18D-1A-63

Electric field and potential·D.2 Electric and magnetic fields

Paper 1AEasy1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

At point P, a distance r from an isolated point charge Q, the electric field strength is E and the electric potential is V.

The charge is replaced by a point charge 2Q and P is moved to a distance r/2 from it. Which row gives the new electric field strength and the new electric potential at P?

new field strengthnew potential

Show mark scheme

Marking point	Mark	Notes
Step 1E = kQ/r²: doubling the charge doubles E, and halving the distance multiplies it by 4, so the new field is 8E.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2V = kQ/r: doubling the charge doubles V, and halving the distance doubles it again, so the new potential is 4V.	✓ 1	Answer C

Answer: C · 2 stages of work, one mark

Every option, and why

AThis changes the distance only and ignores the doubling of the charge.
BThis treats the field strength as inversely proportional to r, like the potential.
CCorrect: E ∝ Q/r² and V ∝ Q/r.
DThis treats the potential as inversely proportional to r², like the field strength.

Syllabus understandingD.2 — the electric field strength as given by E = F/q (field of a point charge E = kQ/r²); D.2 (HL) — the electric potential V_e = kQ/r Command term: Deduce

19D-1A-64

Electric potential energy of a system of charges·D.2 Electric and magnetic fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Three point charges, +q, +q and −q, are fixed at the corners of an equilateral triangle of side a.

What is the electric potential energy of the system?

Show mark scheme

Marking point	Mark	Notes
Step 1The electric potential energy of the system is the work done to assemble it from infinite separation: the sum of kq₁q₂/r over the three pairs.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Pairs: (+q)(+q) = +q²; (+q)(−q) = −q²; (+q)(−q) = −q².	—
Step 3E_p = k(q² − q² − q²)/a = −kq²/a.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis treats all three pairs as attracting, ignoring the positive contribution of the two like charges.
BCorrect: one repulsive pair (+kq²/a) and two attractive pairs (−kq²/a each).
CThis reverses the sign of one of the two unlike pairs: q² + q² − q² = +q².
DThis adds the magnitudes of the three pair energies and ignores the signs of the charges.

Syllabus understandingD.2 (HL) — the electric potential energy E_p in terms of work done to assemble the system from infinite separation; the electric potential energy for a system of two charged bodies E_p = kq₁q₂/r Command term: Determine

20D-1A-65

Magnetic field patterns·D.2 Electric and magnetic fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

The diagram shows the magnetic field lines in a plane perpendicular to two long, straight, parallel wires X and Y. The wires carry currents of equal magnitude.

Which row gives the directions of the currents in X and in Y?

Magnetic field lines (with arrows showing their direction) in the plane of the page. The wires are perpendicular to the page.

current in Xcurrent in Y

Show mark scheme

Marking point	Mark	Notes
Step 1Midway between the wires there is a point of zero field, and far away the lines loop around both wires: the fields of the two wires oppose each other between them, so the currents are in the same direction.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The lines are directed clockwise around each wire (and around both).	—
Step 3Right-hand grip rule: a current into the page gives clockwise field lines, so both currents are into the page.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: equal currents in the same direction, into the page, give clockwise field lines.
BThis applies the grip rule the wrong way round: a current out of the page gives anticlockwise field lines.
COpposite currents would make the two fields add between the wires (crowded lines, no zero-field point) and no line would loop around both wires.
DOpposite currents would make the two fields add between the wires, and the line directions around X and Y would be opposite.

Syllabus understandingD.2 — magnetic field lines; the magnetic field pattern of a current-carrying straight wire; the determination of the direction of the magnetic field based on the current direction; sketching and interpretation of magnetic field lines Command term: Deduce

21D-1A-84

Equipotentials and field direction·D.2 Electric and magnetic fields

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

The diagram shows equipotential lines in the plane containing two point charges, +Q and −Q. Point P lies on the 0 V equipotential. Four arrows, A, B, C and D, are drawn at P.

Which arrow shows the direction of the electric field at P?

Equipotential lines (dashed) in the plane of two point charges +Q and −Q, and four arrows drawn at P.

Show mark scheme

Marking point	Mark	Notes
Step 1No work is done in moving a charge along an equipotential, so the field has no component along it: the field at P is perpendicular to the 0 V line. This rules out A and D.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2E = −ΔV_e/Δr: the field points in the direction in which the potential decreases, from the +10 V side of P towards the −10 V side.	—
Step 3Arrow C is perpendicular to the equipotential and points towards lower potential.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis draws the field along the equipotential. A charge moved along an equipotential has no work done on it, so the field can have no component in that direction.
BThis arrow is perpendicular to the equipotential but points towards higher potential. The field points from higher to lower potential (the minus sign in E = −ΔV_e/Δr).
CCorrect: the field is perpendicular to the equipotential through P and points towards lower potential.
DThis takes the field at P to be that of +Q alone, directed radially away from it. The field of −Q, directed towards −Q, also contributes; the resultant must be perpendicular to the equipotential, and arrow D is at 45° to it.

Syllabus understandingD.2 — electric field lines; D.2 (HL) — the relationship between equipotential surfaces and electric field lines; the electric field strength E as the electric potential gradient as given by E = −ΔV_e/Δr; Guidance: equipotential surfaces for a collection of up to four point charges Command term: Deduce

22D-1A-85

Charging by friction·D.2 Electric and magnetic fields

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A polythene rod and a woollen cloth are both uncharged. The rod is rubbed with the cloth, and afterwards the rod has a charge of −24 nC.

I. The cloth now has a charge of +24 nC.

II. About 1.5 × 10¹¹ electrons were transferred from the cloth to the rod.

III. Positive charge was transferred from the rod to the cloth.

Which statements are correct?

Show mark scheme

Marking point	Mark	Notes
Step 1Charge is conserved: the total charge of rod and cloth stays zero, so the cloth has +24 nC (I is correct).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Friction moves electrons, not protons. Number of electrons = 24 × 10⁻⁹/1.60 × 10⁻¹⁹ = 1.5 × 10¹¹, moved from the cloth to the rod (II is correct).	—
Step 3Protons are bound in nuclei and do not move; the cloth is positive because it has lost electrons (III is wrong).	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: charge is conserved, and the charge is carried by 1.5 × 10¹¹ electrons moving from the cloth to the rod.
BThis accepts III and rejects II. Only electrons move during charging by friction; the rod gains electrons from the cloth, and 24 nC/1.60 × 10⁻¹⁹ C is 1.5 × 10¹¹.
CThis rejects I, as if charge were created on the rod alone. Charge is conserved, so the cloth must carry an equal and opposite charge.
DThis accepts III. A positively charged object has lost electrons; protons are not transferred when bodies are rubbed together.

23D-1A-86

Magnetic field of a bar magnet·D.2 Electric and magnetic fields

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A bar magnet lies on a horizontal table. A small plotting compass is placed at P, beside the midpoint of the magnet, as shown. The Earth's magnetic field can be ignored.

In which direction does the north pole of the compass needle point?

A bar magnet and point P, beside the midpoint of the magnet (top view).

Show mark scheme

Marking point	Mark	Notes
Step 1Outside a bar magnet the field lines leave the north pole, curve round and enter the south pole.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Beside the midpoint of the magnet the lines run parallel to the magnet, from the N end towards the S end.	—
Step 3The north pole of the compass lines up with the field: it points along the magnet towards its south pole.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: outside the magnet the field lines run from N to S, so beside the midpoint they point along the magnet towards S.
BThis uses the direction of the field inside the magnet (from S to N). Field lines form closed loops: inside they run S to N, outside they return from N to S.
CThis treats the magnet like a charged body with field lines leaving its surface at right angles. Magnetic field lines leave the north pole and curve round to the south pole.
DThis assumes the field exists only near the poles. The field beside the midpoint is weaker than at the poles but not zero: the lines from N to S pass P.

24D-1A-87

Magnetic field of a circular coil·D.2 Electric and magnetic fields

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A flat circular coil lies in the plane of the page and carries an anticlockwise current, as shown. X is at the centre of the coil. Y is in the plane of the page, just outside the coil.

Which row gives the direction of the magnetic field at X and at Y?

A flat circular coil in the plane of the page carrying an anticlockwise current; X is at its centre and Y just outside it.

Field at XField at Y

Show mark scheme

Marking point	Mark	Notes
Step 1Right-hand grip rule for each part of the wire: with the thumb along the anticlockwise current, the fingers curl out of the page on the inside of the loop and into the page on the outside.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2All parts of the coil give a field out of the page at the centre X.	—
Step 3Field lines form closed loops: those that come out of the page inside the coil return into the page outside it, so at Y the field is into the page.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis assumes the field has the same direction everywhere around the coil. The lines that come out through the coil must return outside it, so at Y the field is reversed.
BThis applies the grip rule the wrong way round and also takes the field to be the same inside and outside the coil.
CThis applies the grip rule the wrong way round: an anticlockwise current, as seen, gives a field towards the observer (out of the page) at the centre.
DCorrect: out of the page inside the loop, and into the page just outside it.

25D-1A-88

Field and potential of two point charges·D.2 Electric and magnetic fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Two point charges, +q and +4q, are fixed a distance d apart. N is the point on the line joining them at which the resultant electric field strength is zero.

What is the electric potential at N?

Show mark scheme

Marking point	Mark	Notes
Step 1At N the two fields are equal and opposite: kq/x² = 4kq/(d − x)², so d − x = 2x and N is d/3 from +q (and 2d/3 from +4q).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Potential is a scalar: add the contributions. kq/(d/3) + 4kq/(2d/3) = 3kq/d + 6kq/d.	—
Step 3V_e = 9kq/d.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis assumes that zero field strength means zero potential. Both charges are positive, so each adds a positive potential at N; zero field means zero potential gradient, not zero potential.
BThis subtracts the two contributions (6 − 3) as if potentials, like fields, could cancel by pointing in opposite directions. Potential is a scalar and both terms are positive.
CCorrect: N is d/3 from +q, and the scalar potentials 3kq/d and 6kq/d add.
DThis places N at the midpoint, where the fields of two equal charges would cancel: 2kq/d + 8kq/d. (Taking x/(d − x) = 1/4 instead of 1/2, which puts N at d/5, gives the same wrong value.)

Syllabus understandingD.2 — the electric field strength as given by E = F/q; D.2 (HL) — that the electric potential is a scalar quantity with zero defined at infinity; that the electric potential V_e at a point is the work done per unit charge to bring a test charge from infinity to that point as given by V_e = kQ/r Command term: Determine

26D-1A-89

Electric potential energy from a graph·D.2 Electric and magnetic fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Two small spheres carry charges of the same sign. The graph shows how the electric potential energy E_p of the pair varies with the separation r of their centres.

What is the magnitude of the electric force between the spheres when r = 10 cm?

Electric potential energy E_p of the pair against separation r (drawn to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1E_p = kq₁q₂/r and F = kq₁q₂/r², so F = E_p/r.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2From the graph, E_p = 0.25 mJ at r = 10 cm = 0.10 m.	—
Step 3F = 0.25 × 10⁻³/0.10 = 2.5 × 10⁻³ N. (This equals the magnitude of the gradient of the tangent to the curve at 10 cm.)	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis divides E_p by r in centimetres (0.25 × 10⁻³/10). The separation must be in metres.
BThis uses the gradient of the chord from 10 cm to 20 cm, (0.25 − 0.125) mJ/0.10 m = 1.25 × 10⁻³ N ≈ 1.3 × 10⁻³ N. The curve is steeper at 10 cm than this average.
CCorrect: F = E_p/r = 0.25 mJ/0.10 m.
DThis divides E_p by r², as if E_p were kq₁q₂: 0.25 × 10⁻³/0.010.

Syllabus understandingD.2 (HL) — the electric potential energy for a system of two charged bodies as given by E_p = kq₁q₂/r; D.2 — Coulomb’s law as given by F = kq₁q₂/r² for charged bodies treated as point charges Command term: Determine

27D-1A-90

Equipotentials of a charged conductor·D.2 Electric and magnetic fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

An isolated pear-shaped metal conductor carries a positive charge. The charges on it are at rest.

I. Every point inside the conductor and on its surface is at the same electric potential.

II. The electric field strength just outside the surface is the same at every point of the surface.

III. Just outside the surface, the electric field lines are perpendicular to the surface.

Which statements are correct?

Show mark scheme

Marking point	Mark	Notes
Step 1The charges are at rest, so there is no field inside the metal (free electrons would otherwise move). With E = 0 the potential gradient is zero: the whole conductor, including its surface, is one equipotential (I is correct).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Field lines are perpendicular to equipotential surfaces, so they leave the surface at right angles (III is correct). Equivalently, a field component along the surface would move the surface charges.	—
Step 3Equal potential on the surface does not mean equal field strength: the field depends on how quickly the potential changes with distance from the surface, and the equipotentials outside are crowded together near the narrow, more sharply curved end (II is wrong).	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis accepts II. The surface is an equipotential, but the field strength just outside is not the same everywhere: it is greatest where the surface is most sharply curved. It also rejects III, which follows from the surface being an equipotential.
BCorrect: the conductor is a single equipotential and field lines meet it at right angles, but the field strength varies over the surface.
CThis rejects I. With no field inside the conductor, no work is done moving a charge from one point of it to another, so all points are at the same potential.
DThis accepts II. A constant potential over the surface does not make the potential gradient outside it the same everywhere.

Syllabus understandingD.2 (HL) — equipotential surfaces for electric fields; the relationship between equipotential surfaces and electric field lines; the electric field strength E as the electric potential gradient as given by E = −ΔV_e/Δr; Guidance: the field inside and outside a single spherical conducting body Command term: Deduce

28D-1A-91

Solid and hollow conducting spheres·D.2 Electric and magnetic fields

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A solid metal sphere and a hollow, thin-walled metal sphere have the same radius R. Each sphere is isolated and carries the same positive charge Q.

Which statement about the centres of the two spheres is correct?

Show mark scheme

Marking point	Mark	Notes
Step 1In a conductor the charge moves to the outer surface, so the charge is on the surface of both spheres, whether solid or hollow.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Inside each sphere the field strength is zero, so the potential does not change from the surface to the centre.	—
Step 3Both centres are at V_e = kQ/R, so the potentials are the same.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis assumes that zero field strength means zero potential. Zero field means a constant potential, equal to the value kQ/R at the surface.
BThis treats the charge on the solid sphere as spread through its volume. In a conductor the charge resides on the outer surface, so the two spheres behave in the same way.
CThis takes the empty interior of the hollow sphere to be at zero potential because it encloses no charge. Enclosing no charge makes the field inside zero, so the potential inside is constant and equal to the surface value kQ/R, not zero.
DCorrect: all the charge is on the outer surface of each sphere, so both have zero field inside and the same potential kQ/R.

Syllabus understandingD.2 (HL) — that the electric potential V_e at a point is the work done per unit charge to bring a test charge from infinity to that point as given by V_e = kQ/r; equipotential surfaces for electric fields; Guidance: the field inside and outside a single spherical conducting body; Guidance: equipotential surfaces inside and outside a solid and a hollow charged conducting sphere Command term: Deduce

29D-1A-92

Work done to change a system of charges·D.2 Electric and magnetic fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Two small spheres carrying equal charges are held a distance r apart. The electric potential energy of the pair is U.

How much work must be done to move the spheres slowly until they are r/3 apart?

Show mark scheme

Marking point	Mark	Notes
Step 1E_p = kq²/r, so E_p ∝ 1/r: at a separation r/3 the potential energy is 3U.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The work done moving the spheres slowly equals the increase in electric potential energy.	—
Step 3W = 3U − U = 2U.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis multiplies the initial force U/r by the distance moved, 2r/3, treating the force as constant. The force increases as the spheres approach, so more work is needed.
BCorrect: the potential energy rises from U to 3U.
CThis gives the final potential energy, forgetting that the pair already had potential energy U at the start.
DThis takes E_p ∝ 1/r², like the force, so that the energy rises from U to 9U.

30D-1A-93

Potential gradient and force on an electron·D.2 Electric and magnetic fields

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

The graph shows how the electric potential V varies with position x along a straight line. The electric field is directed along the line. An electron is placed in turn at the points P, Q, R and S.

At which point is the electric force on the electron greatest in the positive x-direction?

Electric potential V against position x along the line (drawn to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1E = −ΔV/Δx, and the force on an electron is F = −eE = +e ΔV/Δx.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2So the force on the electron is in the +x-direction only where V increases with x. At P and R the gradient is zero (no force); at Q V decreases steeply (force in the −x-direction).	—
Step 3Only at S does V increase with x, so S is the only point with a force in the +x-direction.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis chooses the highest potential. The force depends on the potential gradient, which is zero at the maximum P.
BAt Q the field strength is greatest and the field points in the +x-direction, but the electron is negative, so the force on it is in the −x-direction.
CThis chooses the lowest potential. The gradient is zero at the minimum R, so the force there is zero.
DCorrect: V rises with x at S, so the field points in the −x-direction and the force on the negative electron points in the +x-direction.

Syllabus understandingD.2 (HL) — the electric field strength E as the electric potential gradient as given by E = −ΔV_e/Δr; D.2 — the electric field strength as given by E = F/q; the direction of forces between the two types of electric charge Command term: Deduce

31D-1A-94

Earthing and induced charge·D.2 Electric and magnetic fields

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A metal sphere of radius R, with centre O, is connected to earth. A small sphere with a fixed charge +q is held a distance d from O, as shown. The earth connection remains in place.

What is the total charge on the metal sphere?

An earthed metal sphere of radius R and a point charge +q a distance d from its centre O (not to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1The sphere is earthed, so it is at zero potential, and inside a conductor the potential is the same everywhere: the potential at O is zero.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The induced charge Q_ind is on the surface, and every part of it is a distance R from O. Potential is a scalar, so its contribution at O is kQ_ind/R however unevenly it is spread.	—
Step 3At O: kq/d + kQ_ind/R = 0, so Q_ind = −qR/d.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis sets the field at O to zero using kQ_ind/R². The induced charge is crowded on the side nearer +q, so its field at O is not that of a uniform shell; the scalar potential gives the answer without knowing the distribution.
BCorrect: zero potential at the centre requires kQ_ind/R = −kq/d.
CThis assumes that earthing draws in a charge equal and opposite to the inducing charge. That would be true only if the conductor surrounded +q; here only enough charge flows to bring the sphere to zero potential.
DThis assumes that an earthed conductor is uncharged. Earthing fixes the potential at zero, not the charge: electrons flow up from earth because of the nearby +q.

Syllabus understandingD.2 — that the electric charge can be transferred between bodies using friction, electrostatic induction and by contact, including the role of grounding (earthing); D.2 (HL) — that the electric potential is a scalar quantity with zero defined at infinity; that the electric potential V_e at a point is the work done per unit charge to bring a test charge from infinity to that point as given by V_e = kQ/r; Guidance: the field inside and outside a single spherical conducting body Command term: Determine

32D-1A-95

Magnetic field patterns·D.2 Electric and magnetic fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

Two identical bar magnets lie on a straight line with their north poles facing each other, as shown. X is midway between the magnets. Y is a short distance from X on the perpendicular bisector of the line joining them.

Which row describes the magnetic field strength at X and the direction of the magnetic field at Y?

Two identical bar magnets with their north poles facing each other; X is midway between them and Y is on the perpendicular bisector.

Field strength at XDirection of field at Y

Show mark scheme

Marking point	Mark	Notes
Step 1Field lines leave north poles. At X the two N poles are equidistant and their fields are equal and opposite, so the field at X is zero (a neutral point).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2At Y the field of each N pole points away from that pole, one up and to the right, the other up and to the left; the components along the axis cancel by symmetry.	—
Step 3The resultant at Y points along the bisector, away from X (the lines from the two N poles turn aside and spread out along the bisector).	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: X is a neutral point, and at Y the fields of the two N poles combine to point away from X.
BThis makes field lines converge into the neutral point. Lines leave the N poles, so near X they are pushed out sideways, away from X.
CThis adds the field magnitudes of the two magnets without allowing for their opposite directions at X.
DThis adds the magnitudes at X and also reverses the direction at Y.

33D-1A-96

Electric field strength·D.2 Electric and magnetic fields

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A small test charge of −2.0 nC at point P in an electric field experiences an electric force of 3.0 × 10⁻⁵ N directed towards the north.

The test charge is replaced by a charge of +6.0 nC at P. The field is unchanged. What is the electric force on the new charge?

Show mark scheme

Marking point	Mark	Notes
Step 1E = F/q = 3.0 × 10⁻⁵/2.0 × 10⁻⁹ = 1.5 × 10⁴ N C⁻¹. The force on the negative charge is opposite to the field, so the field points south.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Force on +6.0 nC: F = qE = 6.0 × 10⁻⁹ × 1.5 × 10⁴ = 9.0 × 10⁻⁵ N.	—
Step 3A positive charge is pushed along the field: to the south.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis inverts the ratio of the charges (2.0/6.0 instead of 6.0/2.0).
BThis reverses the direction but keeps the same magnitude, as if the force were set by the field alone. F = qE, so tripling the size of the charge triples the force.
CThis scales the force correctly but ignores the change of sign of the charge: a positive charge is pushed in the opposite direction to a negative one.
DCorrect: the field is 1.5 × 10⁴ N C⁻¹ to the south, and F = 6.0 × 10⁻⁹ × 1.5 × 10⁴.

Syllabus understandingD.2 — the electric field strength as given by E = F/q; the direction of forces between the two types of electric charge Command term: Determine

34D-1B-02

Coulomb's law·D.2 Electric and magnetic fields

Paper 1BHard7 marks

Data-based question6 steps to full marksDetermine

Two identical small conducting spheres, each of radius 0.50 cm, are given equal charges. One sphere is fixed on an insulating stand; the other is mounted on an insulating rod attached to a digital force sensor, which measures the electrostatic force F between the spheres. The separation r of their centres is set with a scale, and its uncertainty is negligible. Each value of F has an absolute uncertainty of ±0.10 mN.

To test whether F ∝ rⁿ with n = −2, the student plots lg(F / mN) against lg(r / cm). The graph shows five of the six points with their error bars.

r / cm	F / mN	lg(r / cm)	lg(F / mN)
2.0	13.24	0.301	1.122
3.0	6.11	0.477	0.786
4.0	3.42	0.602	0.534
5.0	2.19	0.699
6.0	1.58	0.778	0.199
8.0	0.89	0.903	−0.051

Graph drawn to scale. Error bars show ±0.10 mN.

(a)

The line of best fit.

(i)

Complete the table.

(1)

(ii)

Plot the missing point and draw the line of best fit. Determine the gradient of the line.

(2)

(b)

Testing the law.

(i)

Draw the lines of maximum and minimum gradient consistent with the error bars. Hence state n with its uncertainty.

(2)

(ii)

Discuss whether these data support Coulomb's law.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
lg 2.19 = 0.340	✓ 1	Accept 0.34.
Part (a)(ii)
Point plotted at (0.699, 0.340) and a single straight line of best fit drawn through all six points	✓ 1	Allow ECF from (a)(i) for the plotted point.
Gradient = −1.95, from a large triangle	✓ 1	Accept −1.92 to −1.98. A positive value scores 0.
Part (b)(i)
Steepest line ≈ −1.97 and least steep line ≈ −1.91, each passing through every error bar (the bars at small r are very short, so both lines must pass almost exactly through the first two points)	✓ 1	Accept −1.96 to −1.99 and −1.89 to −1.92. Lines that miss any error bar (e.g. lines joining opposite ends of the bars at r = 2.0 cm and 8.0 cm) score 0 for this mark.
n = −1.95 ± 0.03	✓ 1	Uncertainty = half the difference of the two gradients. Allow ECF from (a)(ii) and from the candidate's lines. Accept ±0.02 to ±0.05; the uncertainty to 1 s.f. and the value to the same decimal place.
Part (b)(ii)
−2 lies outside the range found in (b)(i) (about −1.97 to −1.91), so the data do not follow an exact inverse-square law for these spheres	✓ 1	The conclusion must agree with the candidate's own range (ECF from (b)(i)): a candidate whose range includes −2 may conclude that the data are consistent with an inverse-square law.
This does not disprove Coulomb's law, which applies to point charges: at small separations the like charges repel each other to the far sides of the spheres, so the effective separation is larger than r, the force at small r is less than the point-charge value and the graph is less steep than −2	✓ 1	Accept: the spheres cannot be treated as point charges when r is only a few radii. Do not accept "human error" or "air resistance".

Answers: (a)(i) 0.340 · (a)(ii) −1.95 · (b)(i) −1.95 ± 0.03 (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 — Coulomb's law as given by F = kq₁q₂/r² for charged bodies treated as point charges; Tools 3 — logarithmic graphs to test a power law; uncertainty bars; lines of maximum and minimum gradient and the uncertainty in a gradient; Inquiry 3 — comparing a result with the accepted scientific context Command term: Determine

35D-1B-07

Millikan's experiment·D.2 Electric and magnetic fields

Paper 1BMedium7 marks

Data-based question6 steps to full marksDetermine

In a Millikan-type experiment, oil drops of density 875 kg m⁻³ are viewed between two horizontal metal plates 5.00 mm apart. With no potential difference between the plates, drop A falls at constant (terminal) speed. A student times it with a hand-held stopwatch over 1.00 mm of the eyepiece scale and records 8.35 s.

A potential difference V is then applied and adjusted until drop A is stationary: V = 457 V. The viscosity of air is 1.82 × 10⁻⁵ Pa s, and the buoyancy force of the air on a drop may be neglected. The charges on four other drops, B to E, were found in the same way.

Drop	B	C	D	E
q / 10⁻¹⁹ C	3.19	7.93	6.42	9.50

(a)

Drop A.

(i)

Show that the radius of drop A is about 1.1 × 10⁻⁶ m.

(2)

(ii)

Determine the charge on drop A.

(2)

(b)

Evaluation.

(i)

Millikan's hypothesis is that every charge is a whole-number multiple of e = 1.60 × 10⁻¹⁹ C. Evaluate this hypothesis using drop A and at least two other drops.

(2)

(ii)

Suggest how the student could reduce the percentage uncertainty in the speed of drop A, and hence in its radius, while still using the same stopwatch.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
At terminal speed the drag balances the weight: 6πηrv = (4/3)πr³ρg, with v = 1.00 × 10⁻³/8.35 = 1.20 × 10⁻⁴ m s⁻¹	✓ 1
r = √(9ηv/(2ρg)) = √(9 × 1.82 × 10⁻⁵ × 1.20 × 10⁻⁴/(2 × 875 × 9.81)) = 1.07 × 10⁻⁶ m	✓ 1	The answer must be given to at least 3 s.f., or the full substitution shown.
Part (a)(ii)
m = (4/3)π(1.07 × 10⁻⁶)³ × 875 = 4.5 × 10⁻¹⁵ kg	✓ 1	Allow ECF from (a)(i). Using 1.1 × 10⁻⁶ m gives 4.9 × 10⁻¹⁵ kg.
qV/d = mg, so q = mgd/V = 4.48 × 10⁻¹⁵ × 9.81 × 5.00 × 10⁻³/457 = 4.8 × 10⁻¹⁹ C	✓ 1	Accept 4.7–4.9 × 10⁻¹⁹ C; using 1.1 × 10⁻⁶ m gives 5.2 × 10⁻¹⁹ C (allow ECF).
Part (b)(i)
q/e for at least three drops: A 3.00, B 1.99, C 4.96, D 4.01, E 5.94	✓ 1	Allow ECF from (a)(ii). Award only if drop A and at least two other drops are used.
Every ratio is within about 1 % of a whole number (3, 2, 5, 4, 6), so the data support the hypothesis	✓ 1	The conclusion must be consistent with the candidate's ratios; with 5.2 × 10⁻¹⁹ C (3.3 e) for A, "A does not fit; the radius was rounded" is acceptable.
Part (b)(ii)
Time the drop over a longer distance (several millimetres of the scale): the reaction-time uncertainty stays the same, so it is a smaller percentage of the longer time; r ∝ √v, so the percentage uncertainty in r is half that in v	✓ 1	Accept: time several successive falls of the same drop and use the mean. Do not accept "use light gates" (a different instrument).

Answers: (a)(ii) 4.8 × 10⁻¹⁹ C (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 — Millikan's experiment as evidence for quantization of electric charge; the uniform electric field strength between parallel plates E = V/d; A.2 — viscous drag F_d = 6πηrv and weight; Tools 3 — evaluating a hypothesis; Inquiry 3 — improvements using the same apparatus Command term: Determine

36D-1B-15

Potential around a charged sphere·D.2 Electric and magnetic fields

Paper 1BMedium7 marks

Data-based question5 steps to full marksDetermine

A hollow aluminium sphere of radius 0.060 m stands on an insulating base and carries a charge Q. A student measures the electric potential V at distances r from the centre of the sphere using a flame probe connected to an electrostatic voltmeter. Before taking the readings, the student touches the probe against an earthed metal plate: the voltmeter then reads −45 V.

The student's hypothesis is that V is inversely proportional to r, as it would be for a point charge at the centre of the sphere. The uncorrected voltmeter readings are shown. k = 8.99 × 10⁹ N m² C⁻².

r / m	0.080	0.100	0.150	0.200	0.250	0.300
Voltmeter reading / V	3330	2650	1760	1300	1040	850

(a)

Testing the hypothesis.

(i)

Determine the potential at r = 0.100 m, corrected for the zero error.

(1)

(ii)

Test the hypothesis, using at least three corrected readings.

(2)

(b)

The charge and its limit.

(i)

Determine Q.

(2)

(ii)

The air next to the sphere becomes conducting, and the sphere discharges, when the electric field strength reaches 3.0 × 10⁶ V m⁻¹. Determine the field strength at the surface of the sphere and hence the factor by which Q could be increased before the sphere discharges.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
V = 2650 − (−45) = 2695 V ≈ 2.70 × 10³ V	✓ 1	Subtracting 45 V (2605 V) scores 0.
Part (a)(ii)
Corrected Vr for at least three readings, e.g. 3375 × 0.080 = 270, 2695 × 0.100 = 270, 895 × 0.300 = 269 (V m)	✓ 1	Allow ECF from (a)(i). Uncorrected readings give Vr falling from 266 to 255 V m: [1 max] for the part with a consistent conclusion.
Vr is constant to within about 1 %, so V ∝ 1/r and the hypothesis is supported	✓ 1
Part (b)(i)
V = kQ/r, so Q = Vr/k with Vr = 270 V m (mean)	✓ 1
Q = 270/8.99 × 10⁹ = 3.0 × 10⁻⁸ C	✓ 1	Allow ECF from (a)(ii). Accept 2.95–3.05 × 10⁻⁸ C.
Part (b)(ii)
Outside the sphere the field is that of a point charge at its centre: E = kQ/R² = 8.99 × 10⁹ × 3.0 × 10⁻⁸/0.060² = 7.5 × 10⁴ V m⁻¹	✓ 1	Allow ECF from (b)(i).
Factor = 3.0 × 10⁶/7.5 × 10⁴ = 40	✓ 1	Allow ECF. Accept 39–41.

Answers: (a)(i) 2.70 × 10³ V · (a)(ii) Vr ≈ 270 V m (constant) · (b)(i) 3.0 × 10⁻⁸ C · (b)(ii) 7.5 × 10⁴ V m⁻¹; 40 (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 (HL) — the electric potential V_e = kQ/r with zero defined at infinity; the field inside and outside a spherical conducting body; Tools 1/3 — zero-error correction; evaluating a hypothesis with several data points Command term: Determine

37D-1B-16

Mapping a uniform field·D.2 Electric and magnetic fields

Paper 1BMedium7 marks

Data-based question6 steps to full marksDetermine

Two long, straight, parallel copper strips are painted on a sheet of conducting paper 8.0 cm apart and connected to a 6.00 V supply; the negative strip is at 0 V. A probe connected to a voltmeter is used to measure the potential V on the paper at distances x from the negative strip, along a line halfway between the ends of the strips.

The graph shows the data with the line of best fit, extended to both strips.

x / cm	1.0	2.0	3.0	4.0	5.0	6.0	7.0
V / V	0.92	1.57	2.31	3.03	3.68	4.42	5.09

Graph drawn to scale.

(a)

The field.

(i)

Determine the electric field strength between the strips.

(2)

(ii)

Calculate the percentage difference between your answer to (a)(i) and the value predicted by E = V/d.

(1)

(b)

Explaining the difference.

(i)

Use the graph to explain the difference in (a)(ii).

(2)

(ii)

The two contacts and the paper are in series with the supply. Estimate the ratio (resistance of one contact)/(resistance of the paper between the strips).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Large triangle on the line, e.g. (0, 0.21) to (8.0, 5.80): gradient = 5.59/8.0 = 0.70 V cm⁻¹	✓ 1
E = 70 V m⁻¹	✓ 1	Accept 68–72 V m⁻¹. The conversion to V m⁻¹ is required for this mark.
Part (a)(ii)
E = 6.00/0.080 = 75 V m⁻¹; difference = (75 − 70)/75 × 100 = 7 %	✓ 1	Allow ECF from (a)(i). Accept 6–8 %.
Part (b)(i)
The line meets the V axis at about 0.21 V, not 0, and reaches only about 5.8 V at x = 8.0 cm	✓ 1	Accept 0.18–0.24 V and 5.75–5.85 V.
About 0.2 V is lost across the contact between each strip and the paper (contact resistance), so only about 5.6 V is across the 8.0 cm of paper: 5.6/0.080 = 70 V m⁻¹, less than 75 V m⁻¹	✓ 1	Allow ECF from (a)(i). "Voltmeter zero error" explains the intercept but not the smaller gradient: do not award the second mark for it.
Part (b)(ii)
The current is the same in components in series, so their resistances are in the ratio of the potential differences across them	✓ 1
Ratio ≈ 0.2/5.6 ≈ 0.04	✓ 1	Allow ECF from (b)(i). Accept 0.03–0.05.

Answers: (a)(i) 70 V m⁻¹ · (a)(ii) 7 % · (b)(ii) ≈ 0.04 (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 — the uniform electric field strength between parallel plates as given by E = V/d; (HL) E = −ΔV_e/Δr; B.5 — resistors in series; Tools 3 — gradient with a unit conversion, interpreting an intercept as a systematic offset; Inquiry 3 — comparing with theory and explaining the discrepancy Command term: Determine

38D-1B-17

Magnetic field of a circular coil·D.2 Electric and magnetic fields

Paper 1BMedium6 marks

Data-based question5 steps to full marksDetermine

A flat circular coil of N turns is sealed inside a plastic housing, so its radius R cannot be measured directly. The coil carries a constant current of 1.50 A. A magnetic field sensor measures the field strength B at distances x from the centre of the coil along its axis.

The field on the axis is given by B = μ₀NIR²/[2(R² + x²)^3/2], and at the centre by B₀ = μ₀NI/(2R). These can be combined to give (B₀/B)^2/3 = 1 + x²/R². The graph shows six of the seven points.

x / m	0.000	0.040	0.060	0.080	0.100	0.120	0.140
B / mT	0.471	0.341	0.237	0.170	0.112	0.082	0.057
x² / 10⁻³ m²	0	1.6	3.6	6.4	10.0	14.4	19.6
(B₀/B)^2/3	1.000	1.240	1.581	1.973	2.605		4.087

Graph drawn to scale.

(a)

The graph.

(i)

Complete the table.

(1)

(ii)

Plot the missing point, draw the line of best fit and determine its gradient, including its unit.

(2)

(b)

The coil.

(i)

Hence determine R.

(1)

(ii)

Determine N.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
(0.471/0.082)^2/3 = 3.21	✓ 1	Accept 3.2.
Part (a)(ii)
Point plotted correctly; gradient from a large triangle, e.g. (0, 1.00) to (20.0 × 10⁻³, 4.13): 3.13/0.0200	✓ 1	Allow ECF from (a)(i).
Gradient = 1.6 × 10² m⁻²	✓ 1	Accept 150–163 m⁻². Unit required.
Part (b)(i)
Gradient = 1/R², so R = 1/√157 = 0.080 m	✓ 1	Allow ECF from (a)(ii). Accept 0.078–0.082 m.
Part (b)(ii)
N = 2RB₀/(μ₀I) = 2 × 0.080 × 0.471 × 10⁻³/(4π × 10⁻⁷ × 1.50)	✓ 1	Allow ECF from (b)(i).
N = 40 (39.9, so a whole number of turns)	✓ 1	Accept 39–41.

Answers: (a)(i) 3.21 · (a)(ii) 1.6 × 10² m⁻² · (b)(i) 0.080 m · (b)(ii) 40 (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 — magnetic field lines and the magnetic field of a current-carrying circular coil (the magnitude is found from the relationship given in the stem); Tools 3 — linearising a relationship, gradient with its unit Command term: Determine

39D-1B-22

Uniform electric fields·D.2 Electric and magnetic fields

Paper 1BMedium7 marks

Data-based question6 steps to full marksDetermine

A small insulated sphere carrying a fixed charge q hangs between two large, vertical, parallel metal plates. The potential difference between the plates is kept at 4.00 kV. A force sensor measures the horizontal electric force F on the sphere. The plates are moved apart in steps, keeping the sphere midway between them, and for each position the student records the reading d_s of a scale fixed to one of the plates.

The student expects that F = qV/d, where d is the separation of the plates, and plots a graph of 1/F against d_s. The graph shows five of the six points.

d_s / cm	F / mN	(1/F) / mN⁻¹
2.0	0.405	2.47
3.0	0.283	3.53
4.0	0.224	4.46
5.0	0.181
6.0	0.155	6.45
7.0	0.133	7.52

Graph drawn to scale. The dashed line marks ds = 0.

(a)

The graph.

(i)

Complete the table.

(1)

(ii)

Plot the missing point, draw the line of best fit and determine its gradient.

(2)

(b)

The charge and the scale.

(i)

Determine q.

(2)

(ii)

The line does not pass through the origin. Determine the intercept of the line on the d_s axis and deduce what this shows about the readings d_s.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
1/0.181 = 5.52 mN⁻¹	✓ 1	Accept 5.5 mN⁻¹.
Part (a)(ii)
Point plotted at (5.0, 5.52) and a single straight line of best fit drawn	✓ 1	Allow ECF from (a)(i).
Gradient = 1.00 mN⁻¹ cm⁻¹	✓ 1	Accept 0.96–1.04 mN⁻¹ cm⁻¹.
Part (b)(i)
1/F = d/(qV), so the gradient is 1/(qV) = 1.00 mN⁻¹ cm⁻¹ = 1.00 × 10⁵ N⁻¹ m⁻¹	✓ 1	Allow ECF from (a)(ii). The conversion 1 mN⁻¹ cm⁻¹ = 10⁵ N⁻¹ m⁻¹ is required for this mark.
q = 1/(1.00 × 10⁵ × 4.00 × 10³) = 2.5 × 10⁻⁹ C	✓ 1	Accept 2.4–2.6 × 10⁻⁹ C.
Part (b)(ii)
Line extrapolated: intercept on the d_s axis = −0.5 cm	✓ 1	Allow ECF from the candidate's line. Accept −0.4 to −0.6 cm.
1/F should be zero when the true separation is zero, so d = d_s + 0.5 cm: every reading is about 0.5 cm less than the true separation (a systematic, zero-offset error that does not affect the gradient or q)	✓ 1	Accept: the scale zero is about 0.5 cm away from the surface of the plate. Do not accept "random error" or "parallax".

Answers: (a)(i) 5.52 mN⁻¹ · (a)(ii) 1.00 mN⁻¹ cm⁻¹ · (b)(i) 2.5 × 10⁻⁹ C · (b)(ii) −0.5 cm (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 — the uniform electric field strength between parallel plates as given by E = V/d; the electric field strength as given by E = F/q; Tools 3 — linearising a relationship, gradient with a unit conversion, interpreting an intercept as a systematic error Command term: Determine

40D-1B-27

Magnetic field inside a solenoid·D.2 Electric and magnetic fields

Paper 1BEasy7 marks

Data-based question5 steps to full marksDetermine

A student investigates the magnetic field inside a long air-core solenoid. The solenoid has 460 turns wound uniformly over a length of 0.400 m and is fixed with its axis horizontal. A magnetic field sensor, aligned with the axis, is placed at the centre of the solenoid. The student records the sensor reading B for currents I from 10 mA to 60 mA. The uncertainty in each reading of B is ±1.0 μT; the uncertainty in I is negligible.

Theory predicts that the magnetic field strength inside a long solenoid is B = μ₀nI, where n is the number of turns per unit length. The graph shows the data.

I / mA	10	20	30	40	50	60
B / μT	34.3	47.5	62.6	76.1	91.4	104.5

Sensor reading B against current I. Error bars show ±1.0 μT (drawn to scale).

(a)

The gradient.

(i)

Draw the line of best fit and determine its gradient, including its unit.

(2)

(ii)

Determine n from your gradient. Compare your answer with the value calculated from the number of turns and the length of the solenoid.

(2)

(b)

The intercept.

(i)

The line does not pass through the origin. State the intercept on the B axis and suggest a reason for it.

(2)

(ii)

Suggest how the student could remove this systematic error using the same apparatus, and state its effect on the value of n found in (a)(ii).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Single straight line drawn through all six error bars; gradient from a large triangle, e.g. (0, 19.8) to (60, 104.8)	✓ 1
Gradient = 1.42 μT mA⁻¹	✓ 1	Accept 1.37–1.47 μT mA⁻¹ (or 1.37–1.47 × 10⁻³ T A⁻¹). Unit required for this mark.
Part (a)(ii)
n = gradient/μ₀ = 1.42 × 10⁻³/(4π × 10⁻⁷) = 1.13 × 10³ m⁻¹	✓ 1	Allow ECF from (a)(i). 1 μT mA⁻¹ = 10⁻³ T A⁻¹: a power-of-ten error in this conversion scores 0 for this mark. Accept 1.09–1.17 × 10³ m⁻¹.
460/0.400 = 1150 m⁻¹; the values differ by about 2 %, so they agree closely (within the uncertainty of the gradient)	✓ 1	The comparison must be consistent with the candidate's value; accept a correct percentage difference with a sensible comment (e.g. the field at the centre of a solenoid of finite length is slightly less than μ₀nI).
Part (b)(i)
Intercept = 20 μT	✓ 1	Accept 18–22 μT.
The sensor also measures the component of the Earth's magnetic field along the axis, which is present even when I = 0 and adds the same amount to every reading (a systematic error)	✓ 1	Accept a zero error of the sensor. "Random error" or "the solenoid is not long enough" scores 0.
Part (b)(ii)
Record the reading with zero current and subtract it from every reading (or reverse the current and use half the difference of the two readings); the gradient, and so n, is unchanged	✓ 1	Both the method and "no effect on n" are needed for the mark.

Answers: (a)(i) 1.42 μT mA⁻¹ · (a)(ii) 1.13 × 10³ m⁻¹, about 2 % less than 1150 m⁻¹ · (b)(i) 20 μT; the Earth's magnetic field (or a zero offset of the sensor) · (b)(ii) subtract the zero-current reading; n is unaffected (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 — magnetic field lines; Guidance: magnetic field patterns of a bar magnet, a current-carrying straight wire, a current-carrying circular coil and an air-core solenoid; sketching and interpretation of magnetic field lines; Tools — gradient and intercept of a straight-line graph with units, systematic error and its removal Command term: Determine

41D-1B-28

Potential gradient on conducting paper·D.2 Electric and magnetic fields

Paper 1BMedium7 marks

Data-based question5 steps to full marksDetermine

A sheet of conducting paper has a small circular copper electrode of radius 0.50 cm painted at its centre and a thin copper ring of radius 8.0 cm painted around it, with the same centre. The central electrode is connected to the +10.0 V terminal of a supply and the ring to the 0 V terminal. A probe connected to a digital voltmeter is used to measure the potential V at distances r from the centre along one radius. The uncertainty in each reading of V is ±0.01 V; the uncertainty in r is negligible.

The student calculates the average electric field strength E between neighbouring readings from E = −ΔV/Δr, and assigns each value to the midpoint r_m of the interval. Because the current spreads out in a flat sheet, the student's hypothesis is that E ∝ 1/r.

r / cm	1.0	2.0	3.0	4.0	5.0	6.0	7.0
V / V	7.51	5.00	3.53	2.51	1.70	1.03	0.49

(a)

The field strength.

(i)

Show that the average electric field strength between r = 3.0 cm and r = 4.0 cm is about 100 V m⁻¹, and state its direction.

(2)

(ii)

The hypothesis predicts that Er_m is constant. Test the hypothesis using at least three intervals.

(2)

(b)

Uncertainties and equipotentials.

(i)

Determine the percentage uncertainty in E for the interval from r = 6.0 cm to r = 7.0 cm. Hence comment on the spread of the values of Er_m.

(2)

(ii)

State and explain the shape of the equipotential lines on the paper.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
E = −(2.51 − 3.53)/(0.010) = 102 V m⁻¹	✓ 1	The answer must be given to at least 3 s.f. (102), or the full substitution shown in SI units.
Radially outwards, away from the central electrode (from higher to lower potential)	✓ 1
Part (a)(ii)
Er_m calculated for at least three intervals, e.g. 251 × 0.015 = 3.76 V, 102 × 0.035 = 3.57 V, 54 × 0.065 = 3.51 V	✓ 1	Allow ECF from (a)(i). All six values: 3.76, 3.67, 3.57, 3.65, 3.69, 3.51 V.
The values are constant to within about ±4 % (3.51–3.76 V), so the data support E ∝ 1/r	✓ 1	The conclusion must agree with the candidate's values. Comparing E values alone, or using only two intervals, scores 0 for this mark.
Part (b)(i)
ΔV = 0.54 V with an uncertainty of ±0.02 V (two readings), so E has an uncertainty of 0.02/0.54 × 100 = 3.7 %	✓ 1	Accept 4 %. Using ±0.01 V (1.9 %): [0] for this mark.
This is about the same as the spread of the Er_m values (about ±4 %), so the spread can be explained largely by the uncertainty in the readings; the hypothesis is still supported	✓ 1	Allow ECF from (a)(ii) and from the candidate's percentage. Also accept, for this mark, the observation that the first value (3.76 V) differs from the mean by about 3 %, more than its own uncertainty (about 0.8 %), because E changes rapidly across that interval and the average E is not the value at the midpoint (a systematic effect), with the conclusion that the hypothesis is still supported.
Part (b)(ii)
Circles centred on the central electrode: by symmetry V depends only on r, so the field lines are radial and the equipotentials are perpendicular to them	✓ 1	Both "circles (concentric with the electrode)" and a reason (symmetry, or perpendicular to radial field lines) are needed.

Answers: (a)(i) 102 V m⁻¹, radially outwards · (a)(ii) Er_m ≈ 3.64 V (constant): supported · (b)(i) 3.7 %; the spread is consistent with the uncertainties · (b)(ii) concentric circles, perpendicular to the radial field lines (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 (HL) — the electric field strength E as the electric potential gradient as given by E = −ΔV_e/Δr; equipotential surfaces for electric fields; the relationship between equipotential surfaces and electric field lines; Tools — testing a hypothesis from data, uncertainty in a difference of two readings Command term: Determine

42D-1B-29

Sharing charge by contact·D.2 Electric and magnetic fields

Paper 1BMedium7 marks

Data-based question5 steps to full marksDetermine

Two small identical conducting spheres, A and B, are mounted on insulating rods. B stands on the pan of a digital balance and A is clamped vertically above it, with their centres 6.0 cm apart. A and B are given equal charges q₀ of the same sign. The repulsion increases the balance reading by Δm. B is then touched with a third identical, uncharged sphere C, held on an insulating rod, and C is taken away and discharged. Δm is recorded again. This is repeated, and n is the number of times B has been touched by C. The charge on A does not change. The uncertainty in each value of Δm is ±0.2 mg.

The student's hypothesis is that each contact halves the charge on B. The graph shows lg(Δm / mg) against n for five of the six readings.

n	Δm / mg	lg(Δm / mg)
0	159.7	2.203
1	74.5	1.872
2	35.3	1.548
3	16.4
4	7.9	0.898
5	3.5	0.544

lg(Δm / mg) against n. The point for n = 3 is not plotted. Error bars show ±0.2 mg (drawn to scale).

(a)

The graph.

(i)

Complete the table.

(1)

(ii)

Plot the missing point, draw the line of best fit and determine its gradient.

(2)

(b)

Interpreting the results.

(i)

Deduce, using your gradient, the fraction of its charge that B keeps at each contact. Comment on the student's hypothesis.

(2)

(ii)

Determine q₀.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
lg 16.4 = 1.215	✓ 1	Accept 1.21 or 1.215.
Part (a)(ii)
Point plotted at (3, 1.215) and a single straight line of best fit drawn through all six points	✓ 1	Allow ECF from (a)(i).
Gradient = −0.330	✓ 1	Accept −0.32 to −0.34. A positive gradient scores 0.
Part (b)(i)
Δm ∝ the charge on B, so the fraction kept = 10^gradient = 10^−0.330 = 0.47	✓ 1	Allow ECF from (a)(ii). Accept 0.46–0.48.
Halving would need a gradient of lg 0.5 = −0.301 (fraction 0.50); the error bars are far too small to allow this, so the data do not support exact halving: B loses slightly more than half, e.g. charge also leaks into the air or along the insulating rods	✓ 1	The conclusion must be consistent with the candidate's fraction. Accept any sensible loss mechanism, or C being slightly larger than B. "Charge is not conserved" scores 0.
Part (b)(ii)
Intercept = 2.205, so Δm₀ = 10^2.205 = 160 mg and F₀ = 160 × 10⁻⁶ kg × 9.81 = 1.57 × 10⁻³ N	✓ 1	Allow ECF from (a)(ii). Using the first reading, 159.7 mg, is acceptable. A unit error (mg used as g) scores 0 for this mark.
q₀ = √(Fr²/k) = √(1.57 × 10⁻³ × 0.060²/8.99 × 10⁹) = 2.5 × 10⁻⁸ C	✓ 1	Accept 2.4–2.6 × 10⁻⁸ C.

Answers: (a)(i) 1.215 · (a)(ii) −0.330 · (b)(i) 0.47; not exact halving (extra charge lost each time) · (b)(ii) 2.5 × 10⁻⁸ C (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 — the conservation of electric charge; that the electric charge can be transferred between bodies using friction, electrostatic induction and by contact, including the role of grounding (earthing); Coulomb’s law as given by F = kq₁q₂/r² for charged bodies treated as point charges; Tools — logarithmic linearisation, gradient and intercept, testing a hypothesis; A.2 — weight and the balance reading Command term: Determine

43D-2-02

Electric potential and field from a graph·D.2 Electric and magnetic fields

Paper 2Hard13 marks

Short answer & extended response8 steps to full marksDetermine

Two small charged spheres are fixed on a straight, horizontal insulating track: P, of charge +3.0 nC, at x = 0, and Q, of unknown positive charge, at x = 0.30 m. The graph shows how the electric potential V varies with x between x = 0.03 m and x = 0.24 m. Ignore gravitational effects.

Graph drawn to scale. P is at x = 0 and Q at x = 0.30 m.

(a)

The minimum of the potential.

(i)

State and explain what the graph shows about the electric field strength at x = 0.10 m.

(2)

(ii)

Hence deduce, without calculating any potential, that the charge on Q is 12 nC.

(2)

(b)

The field at x = 0.20 m.

(i)

By drawing a tangent, determine the magnitude of the electric field strength at x = 0.20 m and state its direction.

(2)

(ii)

Confirm your answer to (b)(i) by calculation.

(2)

(c)

A charged bead of charge +2.0 nC, free to slide along the track without friction, is released from rest at x = 0.05 m.

(i)

Use the graph to determine the maximum kinetic energy of the bead.

(2)

(ii)

Identify the other position at which the bead is momentarily at rest.

(1)

(iii)

The bead oscillates between these two positions. Explain why its motion is not simple harmonic.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
The gradient of the graph is zero at the minimum, so E = −ΔV/Δx = 0	✓ 1
The fields of P and Q are equal in magnitude and opposite in direction there	✓ 1	Accept "the field of P (in the +x direction) cancels the field of Q (in the −x direction)".
Part (a)(ii)
kq_P/0.10² = kq_Q/0.20²	✓ 1	Allow ECF from the position of the minimum read in (a)(i). The distance from Q (0.20 m) must be used.
q_Q = q_P(0.20/0.10)² = 4 × 3.0 nC = 12 nC	✓ 1
Part (b)(i)
Tangent drawn at x = 0.20 m; \|E\| = \|gradient\| ≈ 1.0 × 10⁴ V m⁻¹	✓ 1	Accept 0.85–1.2 × 10⁴ V m⁻¹.
In the −x direction, towards P (the potential falls towards P)	✓ 1	Direction must be justified from the gradient or from the larger, nearer charge Q.
Part (b)(ii)
E = k(q_P/0.20² − q_Q/0.10²) = 8.99 × 10⁹ × (3.0 × 10⁻⁹/0.040 − 12 × 10⁻⁹/0.010)	✓ 1	Allow ECF from (a)(ii).
E = −1.0 × 10⁴ V m⁻¹ (in the −x direction), in agreement with (b)(i)	✓ 1	The comparison with (b)(i) is required.
Part (c)(i)
ΔV = V(0.05) − V(0.10) = 971 − 809 = 162 V	✓ 1	Accept readings within ±10 V of each value.
E_k = qΔV = 2.0 × 10⁻⁹ × 162 = 3.2 × 10⁻⁷ J	✓ 1	Accept 2.8–3.6 × 10⁻⁷ J.
Part (c)(ii)
Where V again equals V(0.05) ≈ 971 V: x ≈ 0.17 m	✓ 1	Allow ECF from the reading in (c)(i). Accept 0.16–0.18 m.
Part (c)(iii)
For simple harmonic motion the restoring force (acceleration) must be proportional to the displacement from equilibrium, a = −ω²x	✓ 1
The graph is not symmetrical about x = 0.10 m (it is steeper on the side of Q), so the force (gradient) does not increase in proportion to the displacement	✓ 1	Accept: the turning points are at different distances (0.05 m and about 0.07 m) from the equilibrium position.

Answers: (a)(ii) 12 nC · (b)(i) 1.0 × 10⁴ V m⁻¹, towards P · (b)(ii) −1.0 × 10⁴ V m⁻¹ · (c)(i) 3.2 × 10⁻⁷ J · (c)(ii) x ≈ 0.17 m (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 (HL) — the electric field strength as the electric potential gradient E = −ΔV_e/Δr; V_e = kQ/r and superposition; the work done in moving a charge W = qΔV_e; D.2 — the field between two point charges; C.1 — conditions that lead to simple harmonic motion Command term: Determine

44D-2-07

Electric fields in the atmosphere·D.2 Electric and magnetic fields

Paper 2Medium12 marks

Short answer & extended response8 steps to full marksDetermine

During a thunderstorm the base of a cloud carries a large negative charge, and the ground below it becomes positively charged. The base of the cloud is 1.2 km above the ground, and the region between them may be modelled as the uniform field between two parallel plates. The electric field strength in this region is 5.0 × 10⁴ V m⁻¹.

Water droplets are spheres of density 1000 kg m⁻³. Viscosity of air = 1.8 × 10⁻⁵ Pa s; density of air = 1.2 kg m⁻³.

(a)

The field below the cloud.

(i)

Outline how the ground below the cloud becomes positively charged.

(2)

(ii)

Calculate the potential difference between the base of the cloud and the ground.

(1)

(b)

A droplet of radius 10 μm in this region carries a positive charge of 4.8 × 10⁻¹⁶ C.

(i)

Calculate the number of electrons that the droplet has lost.

(1)

(ii)

Determine the magnitude and the direction of the electric force on the droplet.

(2)

(iii)

Determine the terminal speed of the droplet.

(3)

(c)

A lightning flash transfers a charge of 20 C between the cloud and the ground.

(i)

Estimate the energy transferred by the flash.

(2)

(ii)

State the assumption made in your estimate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
The negative charge on the cloud repels free electrons in the ground (electrostatic induction)	✓ 1
The electrons move away through the conducting ground to the rest of the Earth (earthing), leaving the surface below the cloud with a net positive charge	✓ 1	Do not accept "positive charges move towards the cloud".
Part (a)(ii)
V = Ed = 5.0 × 10⁴ × 1200 = 6.0 × 10⁷ V	✓ 1
Part (b)(i)
N = 4.8 × 10⁻¹⁶/1.60 × 10⁻¹⁹ = 3.0 × 10³	✓ 1
Part (b)(ii)
F = qE = 4.8 × 10⁻¹⁶ × 5.0 × 10⁴ = 2.4 × 10⁻¹¹ N	✓ 1
Upwards: the field points from the positive ground to the negative cloud, and the force on a positive charge is along the field	✓ 1	The direction must be justified.
Part (b)(iii)
Weight = ρ(4/3)πr³g = 1000 × (4/3)π × (1.0 × 10⁻⁵)³ × 9.81 = 4.1 × 10⁻¹¹ N; the buoyancy (about 5 × 10⁻¹⁴ N) is negligible	✓ 1
At the terminal speed: weight = electric force + 6πηrv	✓ 1	Allow ECF from (b)(ii).
v = (4.1 × 10⁻¹¹ − 2.4 × 10⁻¹¹)/(6π × 1.8 × 10⁻⁵ × 1.0 × 10⁻⁵) = 5.0 × 10⁻³ m s⁻¹	✓ 1	Accept 4.9–5.2 × 10⁻³ m s⁻¹. Award [2 max] for 1.2 × 10⁻² m s⁻¹ (electric force ignored) or 1.9 × 10⁻² m s⁻¹ (forces added).
Part (c)(i)
W = QV with V from (a)(ii)	✓ 1	Allow ECF from (a)(ii).
W = 20 × 6.0 × 10⁷ = 1.2 × 10⁹ J	✓ 1	ALT: if the potential difference is taken to fall to zero during the flash, ½QV = 6 × 10⁸ J; accept either value when it is consistent with the assumption in (c)(ii).
Part (c)(ii)
The potential difference stays constant during the flash (the charge transferred is a small fraction of the charge on the cloud) / the whole charge moves through the full potential difference	✓ 1	Must be consistent with the value in (c)(i).

Answers: (a)(ii) 6.0 × 10⁷ V · (b)(i) 3.0 × 10³ · (b)(ii) 2.4 × 10⁻¹¹ N upwards · (b)(iii) 5.0 × 10⁻³ m s⁻¹ · (c)(i) 1.2 × 10⁹ J (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 — charge transfer by electrostatic induction, including the role of grounding (earthing); the uniform electric field E = V/d between parallel plates; E = F/q; Millikan's experiment as evidence for the quantization of charge; A.2 — viscous drag F_d = 6πηrv and buoyancy F_b = ρVg Command term: Determine

45D-2-12

Charged sphere in a uniform field·D.2 Electric and magnetic fields

Paper 2Hard13 marks

Short answer & extended response8 steps to full marksDetermine

A small sphere of mass 2.00 × 10⁻⁴ kg, carrying a positive charge q, hangs from an insulating thread of length 4.0 cm. The top of the thread is fixed midway between two vertical parallel plates that are 5.00 cm apart. When a potential difference of 2.50 kV is applied to the plates, the sphere comes to rest with the thread at 30.0° to the vertical. The radius of the sphere is negligible.

(a)

Equilibrium.

(i)

Calculate the electric field strength between the plates.

(1)

(ii)

Show that q is about 2.3 × 10⁻⁸ C.

(2)

(iii)

Calculate the tension in the thread.

(1)

(b)

The thread is cut.

(i)

Determine the magnitude and direction of the acceleration of the sphere immediately after the thread is cut.

(2)

(ii)

Explain why the sphere then moves in a straight line.

(2)

(iii)

Determine the time taken for the sphere to reach the plate.

(3)

(c)

(i)

The sphere touches the negative plate. Explain why it is then pushed back across the gap.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
E = V/d = 2.50 × 10³/0.0500 = 5.00 × 10⁴ V m⁻¹	✓ 1
Part (a)(ii)
T sin 30° = qE and T cos 30° = mg, so qE = mg tan 30°	✓ 1
q = 2.00 × 10⁻⁴ × 9.81 × tan 30.0°/5.00 × 10⁴ = 2.27 × 10⁻⁸ C	✓ 1	Allow ECF from (a)(i). Full substitution or an answer to at least 3 s.f. is required.
Part (a)(iii)
T = mg/cos 30° = 1.962 × 10⁻³/0.866 = 2.27 × 10⁻³ N	✓ 1
Part (b)(i)
The resultant of the weight and the electric force is equal and opposite to the former tension, 2.27 × 10⁻³ N, so a = 2.27 × 10⁻³/2.00 × 10⁻⁴ = 11.3 m s⁻²	✓ 1	Allow ECF from (a)(iii). ALT: a = g/cos 30°.
At 30° to the vertical, downwards and towards the negative plate (along the line of the thread)	✓ 1
Part (b)(ii)
Both forces (weight and electric force) are constant, so the resultant force and the acceleration are constant in magnitude and direction	✓ 1
The sphere starts from rest, so it moves along the line of the resultant force	✓ 1	Do not accept "there is no horizontal force".
Part (b)(iii)
The sphere starts L sin 30° = 2.0 cm from the midline, so it is 2.5 − 2.0 = 0.50 cm from the plate	✓ 1	This geometrical step is required.
Horizontal acceleration = qE/m = 5.66 m s⁻²	✓ 1	Allow ECF from (a)(ii). ALT: a distance of 1.0 cm along the line at 11.3 m s⁻².
t = √(2 × 0.0050/5.66) = 0.042 s	✓ 1	Award [2 max] for 0.094 s (2.5 cm used).
Part (c)(i)
Charge is transferred by contact: electrons flow from the plate onto the sphere, which becomes negatively charged	✓ 1
The electric force on a negative charge is opposite to the field, towards the positive plate	✓ 1

Answers: (a)(i) 5.00 × 10⁴ V m⁻¹ · (a)(ii) 2.27 × 10⁻⁸ C · (a)(iii) 2.27 × 10⁻³ N · (b)(i) 11.3 m s⁻² · (b)(iii) 0.042 s (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 — the uniform electric field between parallel plates E = V/d; E = F/q; charge transfer by contact; D.3 — the motion of a charged particle in a uniform electric field; A.2 — free-body diagrams and translational equilibrium; A.1 — equations of motion for uniform acceleration Command term: Determine

46D-2-15

Field and potential of two point charges·D.2 Electric and magnetic fields

Paper 2Medium11 marks

Short answer & extended response7 steps to full marksDetermine

Two point charges, A of +4.00 μC and B of −2.00 μC, are fixed 0.300 m apart. Point P is 0.250 m from each charge, and M is the midpoint of AB, as shown.

Not to scale. M is the midpoint of AB. (k = 8.99 × 10⁹ N m² C⁻²)

(a)

The field at P.

(i)

Calculate the electric field strength at P due to charge A alone, and state its direction.

(1)

(ii)

Show that the resultant electric field strength at P is about 5.7 × 10⁵ N C⁻¹.

(2)

(iii)

Determine the angle between the resultant field at P and the line AB.

(1)

(b)

Potential and work.

(i)

Calculate the electric potential at P.

(1)

(ii)

Determine the work done by the electric field when a charge of +1.0 nC moves from P to M.

(2)

(c)

Zero potential.

(i)

Determine the position on the line AB, between the charges, at which the electric potential is zero.

(2)

(ii)

Explain why the electric field strength is not zero at this point, and state, with a reason, on which side of B the field strength is zero.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
E_A = kq/r² = 8.99 × 10⁹ × 4.00 × 10⁻⁶/0.250² = 5.75 × 10⁵ N C⁻¹, directed away from A along AP	✓ 1	The direction is required for the mark.
Part (a)(ii)
E_B = 2.88 × 10⁵ N C⁻¹ towards B; with cos θ = 0.600 and sin θ = 0.800, the components parallel to AB are (5.75 + 2.88) × 0.600 = 5.18 × 10⁵ and perpendicular to AB (5.75 − 2.88) × 0.800 = 2.30 × 10⁵	✓ 1	Allow ECF from (a)(i).
E = √(5.18² + 2.30²) × 10⁵ = 5.67 × 10⁵ N C⁻¹	✓ 1	The answer must be given to at least 3 s.f.
Part (a)(iii)
tan⁻¹(2.30/5.18) = 24°, pointing away from AB on the side of P and in the direction from A to B	✓ 1	Allow ECF from (a)(ii).
Part (b)(i)
V_e = k(q_A + q_B)/r = 8.99 × 10⁹ × 2.00 × 10⁻⁶/0.250 = 7.19 × 10⁴ V	✓ 1
Part (b)(ii)
V_e_M = 8.99 × 10⁹ × (4.00 − 2.00) × 10⁻⁶/0.150 = 1.20 × 10⁵ V	✓ 1
W = q(V_e_P − V_e_M) = 1.0 × 10⁻⁹ × (7.19 × 10⁴ − 1.20 × 10⁵) = −4.8 × 10⁻⁵ J: the field does negative work	✓ 1	Allow ECF from (b)(i). Accept +4.8 × 10⁻⁵ J only if it is clearly stated as the work done by an external agent.
Part (c)(i)
k(4.00)/x = k(2.00)/(0.300 − x)	✓ 1
x = 0.200 m from A (0.100 m from B)	✓ 1
Part (c)(ii)
Between the charges both fields point the same way (away from A and towards B), so they add: zero potential does not imply zero field	✓ 1
Beyond B, on the side away from A: there the fields are opposite and the smaller charge B is nearer, so the two can be equal	✓ 1

Answers: (a)(i) 5.75 × 10⁵ N C⁻¹ · (a)(ii) 5.67 × 10⁵ N C⁻¹ · (a)(iii) 24° · (b)(i) 7.19 × 10⁴ V · (b)(ii) −4.8 × 10⁻⁵ J · (c)(i) 0.200 m from A (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 — Coulomb's law; the electric field strength E = F/q and electric field lines; the field between two point charges; D.2 (HL) — the electric potential as a scalar quantity; V_e = kQ/r; W = qΔV_e Command term: Determine

47D-2-22

Electrostatic powder coating·D.2 Electric and magnetic fields

Paper 2Medium12 marks

Short answer & extended response7 steps to full marksExplain

In electrostatic powder coating, paint powder is blown through a spray gun lined with PTFE (a plastic). The particles are charged by friction as they rub against the lining and leave the gun positively charged. They then travel between a flat metal grid at a potential of +30 kV and a flat earthed metal panel that is to be coated. The grid and the panel are parallel and 0.25 m apart.

A typical powder particle has a mass of 2.0 × 10⁻¹⁰ kg and a charge of +2.4 × 10⁻¹³ C.

(a)

Charging.

(i)

State the direction in which electrons are transferred when a particle is charged by friction against the lining.

(1)

(ii)

Explain, with reference to the conservation of charge, why the lining of the gun must be earthed.

(2)

(b)

The field between the grid and the panel.

(i)

Calculate the number of electrons removed from a typical particle.

(1)

(ii)

Determine the electric force on a typical particle between the grid and the panel.

(2)

(iii)

Compare this force with the weight of the particle and comment on the path of the particle.

(2)

(c)

Coating the panel.

(i)

Determine the speed with which a particle that starts from rest at the grid would reach the panel if air resistance were negligible.

(2)

(ii)

The coating is found to be thicker along the edges of the panel. Explain this with reference to the electric field lines near the edges.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
From the powder particles to the PTFE lining	✓ 1
Part (a)(ii)
Charge is conserved, so the lining gains a negative charge equal in magnitude to the total positive charge carried away by the powder	✓ 1
Without earthing this negative charge would build up (until sparking occurs, or it attracts the particles back and stops further charging); the earth connection lets the excess electrons flow away	✓ 1
Part (b)(i)
N = 2.4 × 10⁻¹³/1.60 × 10⁻¹⁹ = 1.5 × 10⁶	✓ 1
Part (b)(ii)
E = V/d = 30 × 10³/0.25 = 1.2 × 10⁵ V m⁻¹	✓ 1
F = qE = 2.4 × 10⁻¹³ × 1.2 × 10⁵ = 2.9 × 10⁻⁸ N	✓ 1
Part (b)(iii)
Weight = 2.0 × 10⁻¹⁰ × 9.81 = 2.0 × 10⁻⁹ N, so the electric force is about 15 times larger	✓ 1	Allow ECF from (b)(ii). Accept 14–15.
Gravity has little effect: the particles follow paths close to the electric field lines, from the grid to the panel	✓ 1
Part (c)(i)
Work done by the field = qV = 2.4 × 10⁻¹³ × 3.0 × 10⁴ = 7.2 × 10⁻⁹ J = gain in kinetic energy	✓ 1	Allow ECF from the charge and potential difference used in (b)(ii).
v = √(2 × 7.2 × 10⁻⁹/2.0 × 10⁻¹⁰) = 8.5 m s⁻¹	✓ 1	ALT: a = F/m and v² = 2as.
Part (c)(ii)
Near the edges of the panel the field lines crowd together, so the field line density and hence the field strength are greatest there	✓ 1
More field lines end on the edges and the force on the particles is larger, so more particles are guided there and the coating is thicker	✓ 1

Answers: (b)(i) 1.5 × 10⁶ · (b)(ii) 2.9 × 10⁻⁸ N · (b)(iii) about 15 × the weight · (c)(i) 8.5 m s⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 — the transfer of charge by friction, including the role of grounding (earthing); the conservation of electric charge; E = F/q; E = V/d between parallel plates; electric field lines and the relationship between field line density and field strength; A.2 — weight F_g = mg; A.3 — the work done by a force as a transfer of energy Command term: Explain

48D-2-23

Electric potential energy of a system of charges·D.2 Electric and magnetic fields

Paper 2Hard13 marks

Short answer & extended response9 steps to full marksDetermine

In the alpha-cluster model of the carbon-12 nucleus, the nucleus is pictured as three alpha particles at the corners of an equilateral triangle of side 3.0 × 10⁻¹⁵ m. Each alpha particle is treated as a point charge +2e. Consider only the electric interactions.

(a)

Electric potential energy.

(i)

Outline what is meant by the electric potential energy of a system of charges.

(1)

(ii)

Show that the electric potential energy of one pair of alpha particles in this model is about 1.9 MeV.

(2)

(iii)

Determine the total electric potential energy of the three-alpha system, in MeV.

(1)

(b)

The centre of the triangle.

(i)

Calculate the electric potential at the centre of the triangle due to the three alpha particles.

(2)

(ii)

State the electric field strength at the centre of the triangle, and explain your answer.

(1)

(iii)

The potential at the centre is not zero although the field strength there is zero. Explain this, and explain, with reference to E = −ΔV_e/Δr, why the potential at the centre is a maximum along the line through the centre perpendicular to the plane of the triangle.

(2)

(c)

A proton probe.

(i)

A proton approaches the nucleus from a large distance along the line through the centre of the triangle, perpendicular to its plane. Determine, in MeV and in J, the minimum initial kinetic energy the proton needs to reach the centre. Assume the alpha particles stay fixed.

(2)

(d)

Stability.

(i)

The answer to (a)(iii) is positive. Explain what this implies about the forces needed to hold the nucleus together.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
The work done (by an external agent) to assemble the system by bringing the charges from infinite separation to their positions	✓ 1
Part (a)(ii)
E_p = kq₁q₂/r = 8.99 × 10⁹ × (3.20 × 10⁻¹⁹)²/3.0 × 10⁻¹⁵ = 3.07 × 10⁻¹³ J	✓ 1
= 3.07 × 10⁻¹³/1.60 × 10⁻¹³ MeV = 1.92 MeV ≈ 1.9 MeV	✓ 1	Answer to at least 3 s.f. or a clear conversion required for "show that".
Part (a)(iii)
There are three pairs, so E_p = 3 × 1.92 = 5.8 MeV	✓ 1	Accept 5.75 MeV. Award 0 for 1.9 MeV or 3.8 MeV. Allow ECF from (a)(ii).
Part (b)(i)
Distance from each corner to the centre = a/√3 = 3.0 × 10⁻¹⁵/√3 = 1.73 × 10⁻¹⁵ m	✓ 1
V = 3 × k(2e)/r = 3 × 8.99 × 10⁹ × 3.20 × 10⁻¹⁹/1.73 × 10⁻¹⁵ = 5.0 × 10⁶ V	✓ 1	Accept 4.98 × 10⁶ V. Using the side length gives 2.9 × 10⁶ V: award the second mark only.
Part (b)(ii)
Zero: the three fields are equal in magnitude and point away from the three corners at 120° to one another, so their vector sum is zero	✓ 1
Part (b)(iii)
Potential is a scalar: the three positive contributions add and cannot cancel, while field strength is a vector and the three contributions cancel by symmetry	✓ 1
Along the axis the field at the centre is zero, so the potential gradient −ΔV_e/Δr is zero there; on either side of the plane the field points away from the triangle, so V falls with distance from the centre in both directions: the centre is a maximum of V along the axis	✓ 1
Part (c)(i)
The proton must gain electric potential energy eV: along this line the potential rises continuously up to its value at the centre, so this is the minimum energy needed	✓ 1
E_k = 1 × 5.0 MV = 5.0 MeV = 5.0 × 10⁶ × 1.60 × 10⁻¹⁹ = 8.0 × 10⁻¹³ J	✓ 1	Accept 4.98 MeV, 7.97 × 10⁻¹³ J; ECF from (b)(i).
Part (d)(i)
A positive electric potential energy means the electric forces are repulsive: work would have to be done to push the alpha particles together, and on their own they would fly apart	✓ 1	Allow ECF from the sign of the answer to (a)(iii).
An additional attractive force must act that is stronger than the electric repulsion at these separations: the short-range strong nuclear force between the nucleons	✓ 1

Answers: (a)(ii) 1.92 MeV · (a)(iii) 5.8 MeV · (b)(i) 5.0 × 10⁶ V · (b)(ii) 0 · (c)(i) 5.0 MeV; 8.0 × 10⁻¹³ J (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 (HL) — the electric potential energy E_p in terms of work done to assemble the system from infinite separation; E_p = kq₁q₂/r; the electric potential is a scalar quantity with zero defined at infinity; V_e = kQ/r; E = −ΔV_e/Δr; W = qΔV_e in joules and electronvolts; E.3 — the strong nuclear force Command term: Determine

49D-2-24

Charged conducting sphere·D.2 Electric and magnetic fields

Paper 2Hard14 marks

Short answer & extended response8 steps to full marksExplain

A hollow, thin-walled metal sphere A of radius 0.120 m is mounted on an insulating stand and carries a charge of +36.0 nC. (k = 8.99 × 10⁹ N m² C⁻²)

(a)

The isolated sphere.

(i)

Calculate the electric potential of sphere A.

(1)

(ii)

Sketch graphs to show how the electric potential and the electric field strength vary with distance r from the centre of A, from r = 0 to r = 0.36 m. Label the values at r = 0.120 m.

(2)

(b)

Connecting a second sphere.

(i)

Sphere A is now connected by a long thin wire to a distant, uncharged metal sphere B of radius 0.030 m. Explain why charge flows along the wire and why the flow stops.

(2)

(ii)

Determine the charge on each sphere after the flow stops.

(2)

(iii)

Determine the common potential of the spheres.

(1)

(iv)

Show that the electric field strength at the surface of B is four times that at the surface of A.

(2)

(v)

Air near a conductor becomes ionised when the field strength there is very large. Use (b)(iv) to explain why charge leaks into the air most readily from sharply pointed parts of a charged conductor.

(2)

(c)

Work.

(i)

State and explain the work done in moving a small positive test charge from the surface of B to the surface of A.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
V_e = kQ/R = 8.99 × 10⁹ × 36.0 × 10⁻⁹/0.120 = 2.70 × 10³ V	✓ 1
Part (a)(ii)
Potential: constant at 2.70 kV from r = 0 to 0.120 m, then falling as 1/r (a curve, not a straight line) to 0.90 kV at 0.36 m, approaching zero at infinity	✓ 1
Field strength: zero inside, jumping to a maximum of V/R = 2.25 × 10⁴ V m⁻¹ at the surface, then falling as 1/r² (to 2.5 × 10³ V m⁻¹ at 0.36 m)	✓ 1	Values at 0.36 m not required; shape and labelled surface values are.
Part (b)(i)
Charge flows because there is a potential difference between the spheres (A at 2.70 kV, B at 0 V), so there is a field along the wire that moves the free electrons	✓ 1
The flow stops when the two spheres (joined as one conductor) are at the same potential, so the potential difference and the field along the wire are zero	✓ 1
Part (b)(ii)
Same potential: kQ_A/0.120 = kQ_B/0.030, so Q_A = 4Q_B; charge is conserved: Q_A + Q_B = 36.0 nC	✓ 1	Spheres far apart, so each is treated as isolated.
Q_A = 28.8 nC, Q_B = 7.2 nC	✓ 1
Part (b)(iii)
V = 8.99 × 10⁹ × 28.8 × 10⁻⁹/0.120 = 2.16 × 10³ V	✓ 1	Or using B: 8.99 × 10⁹ × 7.2 × 10⁻⁹/0.030; ECF from (b)(ii).
Part (b)(iv)
At the surface of a sphere E = kQ/R² = V/R. Both spheres have the same V, so E ∝ 1/R	✓ 1
E_B/E_A = 0.120/0.030 = 4 (E_A = 1.80 × 10⁴ V m⁻¹, E_B = 7.19 × 10⁴ V m⁻¹)	✓ 1	Allow ECF from (b)(iii).
Part (b)(v)
A sharp point behaves like a part of the conductor with a very small radius of curvature. The whole conductor is at one potential, so the field there (∝ 1/radius) is very large, and the field lines are most closely packed there	✓ 1	Allow ECF from (b)(iv): the field at the surface is inversely proportional to the radius.
The strong field ionises the air near the point, and the ions carry charge away, so charge leaks most readily from sharp points	✓ 1
Part (c)(i)
Zero	✓ 1
The two surfaces are at the same potential (the connected spheres form one equipotential), so W = qΔV_e = 0	✓ 1

Answers: (a)(i) 2.70 × 10³ V · (b)(ii) 28.8 nC and 7.2 nC · (b)(iii) 2.16 × 10³ V · (b)(iv) ratio 4 · (c)(i) 0 (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 (HL) — V_e = kQ/r; fields and equipotential surfaces inside and outside a hollow and a solid charged conducting sphere; the relationship between field strength and potential gradient; W = qΔV_e; D.2 — the conservation of electric charge Command term: Explain

50D-2-25

Magnetic field patterns·D.2 Electric and magnetic fields

Paper 2Medium12 marks

Short answer & extended response7 steps to full marksDescribe

A long, straight, horizontal overhead cable of a direct-current power line runs north–south, 10.0 m above level ground. It carries a steady current of 600 A towards the north.

The magnetic field strength at a distance r from a long straight wire carrying current I is B = μ₀I/(2πr), with μ₀ = 4π × 10⁻⁷ T m A⁻¹. The horizontal component of the Earth's magnetic field at the site is 1.8 × 10⁻⁵ T, directed towards geographic north.

(a)

The field of the cable.

(i)

Sketch the magnetic field lines of the cable in a vertical plane perpendicular to it, as seen by an observer standing to the south of the cable and looking north.

(2)

(ii)

Determine the magnitude and the direction of the magnetic field due to the cable at the ground directly below it.

(2)

(b)

A compass under the cable.

(i)

A compass is placed on the ground directly below the cable. Determine the angle between the compass needle and geographic north.

(2)

(ii)

The current in the cable is reversed. State and explain the new reading of the compass.

(1)

(c)

Other field patterns.

(i)

A long air-core solenoid carries a steady current. Sketch the magnetic field lines inside and outside the solenoid.

(2)

(ii)

State one similarity and one difference between the field of the solenoid and the field of a bar magnet.

(2)

(iii)

A single flat circular coil is viewed from one side, and the current in it is anticlockwise. State the magnetic polarity of the face of the coil nearest the observer.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Concentric circles centred on the cable, with the spacing between the circles increasing with distance from the cable	✓ 1
Clockwise, as seen by the observer (the current flows away from the observer)	✓ 1
Part (a)(ii)
B = 4π × 10⁻⁷ × 600/(2π × 10.0) = 1.2 × 10⁻⁵ T	✓ 1
Horizontal, towards the west	✓ 1	By the right-hand grip rule: clockwise as seen from the south means westwards below the cable.
Part (b)(i)
The needle lines up with the resultant of 1.8 × 10⁻⁵ T north and 1.2 × 10⁻⁵ T west: tan θ = 1.2/1.8	✓ 1
θ = 34° west of north	✓ 1	Accept 33.7°. Direction (west) required. Allow ECF from (a)(ii).
Part (b)(ii)
34° east of north: reversing the current reverses the cable's field (now eastwards) without changing its magnitude, while the Earth's field is unchanged	✓ 1	Allow ECF from (b)(i).
Part (c)(i)
Inside: straight, parallel, equally spaced lines along the axis (uniform field), with the line density greatest inside	✓ 1
Outside: lines leave one end and loop round to the other end, spreading out (weaker field), forming closed loops; direction consistent with the grip rule	✓ 1
Part (c)(ii)
Similarity: the external field pattern is the same as that of a bar magnet, with a north pole at one end and a south pole at the other	✓ 1
Difference: e.g. the field of the solenoid can be switched off or reversed and its strength changed by changing the current; inside the solenoid there is a uniform field in an accessible (air) space	✓ 1	Any one valid difference.
Part (c)(iii)
North: by the right-hand grip rule, with the fingers following the anticlockwise current the thumb points towards the observer, so the field lines emerge from the near face	✓ 1	Accept a correct argument from the "anticlockwise = N" end rule.

Answers: (a)(ii) 1.2 × 10⁻⁵ T, towards the west · (b)(i) 34° west of north · (b)(ii) 34° east of north (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 — magnetic field lines of a bar magnet, a current-carrying straight wire, a current-carrying circular coil and an air-core solenoid; determination of the direction of the magnetic field based on the current direction (magnitude from the relationship given in the stem) Command term: Describe

51D-2-26

Potential gradient and permittivity·D.2 Electric and magnetic fields

Paper 2Hard13 marks

Short answer & extended response9 steps to full marksDetermine

The membrane of a nerve cell is modelled as a layer of lipid 7.0 nm thick with conducting fluid on each side. The graph shows how the electric potential V varies with distance x measured from the outer surface of the membrane into the cell. The fluid outside the cell is at 0 V.

In a medium of permittivity ε, Coulomb's law is F = q₁q₂/(4πεr²). The permittivity of water is 7.1 × 10⁻¹⁰ C² N⁻¹ m⁻² and that of lipid is 1.8 × 10⁻¹¹ C² N⁻¹ m⁻². Body temperature is 310 K.

Potential across the membrane of a nerve cell. The dashed lines mark the surfaces of the membrane.

(a)

The field in the membrane.

(i)

Determine the magnitude and the direction of the electric field strength in the membrane.

(2)

(ii)

State the electric field strength in the fluid inside the cell, with a reason from the graph.

(1)

(iii)

Calculate the electric force on a sodium ion (Na⁺, charge +e) inside the membrane.

(1)

(b)

An ion crossing the membrane.

(i)

A sodium ion passes through a channel in the membrane from the outside to the inside of the cell. Determine the work done on the ion by the electric field, in eV and in J.

(2)

(ii)

Compare your answer to (b)(i) with the average random kinetic energy of an ion at body temperature.

(2)

(c)

Why ions dissolve in water but not in lipid.

(i)

Calculate the electric force between a sodium ion and a chloride ion (Cl⁻, charge −e) 0.50 nm apart in water.

(2)

(ii)

By calculating the work needed to separate the two ions from 0.50 nm to infinity in water and in lipid, explain why ions separate readily in water but not in lipid.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
E = −ΔV/Δx: magnitude = 0.070/7.0 × 10⁻⁹ = 1.0 × 10⁷ V m⁻¹	✓ 1
Directed from outside to inside (in the +x direction), from higher to lower potential	✓ 1
Part (a)(ii)
Zero: the potential is constant inside the cell, so the potential gradient is zero	✓ 1
Part (a)(iii)
F = qE = 1.60 × 10⁻¹⁹ × 1.0 × 10⁷ = 1.6 × 10⁻¹² N (into the cell)	✓ 1	ECF from (a)(i).
Part (b)(i)
W = qΔV_e: the ion falls through 70 mV, so the field does +0.070 eV of work (the electric potential energy decreases by 0.070 eV)	✓ 1
= 0.070 × 1.60 × 10⁻¹⁹ = 1.1 × 10⁻²⁰ J	✓ 1	Accept 1.12 × 10⁻²⁰ J. Sign or statement that the field does positive work is required for the first mark.
Part (b)(ii)
E_k = (3/2)k_BT = 1.5 × 1.38 × 10⁻²³ × 310 = 6.4 × 10⁻²¹ J (0.040 eV)	✓ 1
The work done by the field is about 1.7 times (accept 1.7–1.8) the average thermal kinetic energy, so the same order of magnitude: the field has a large effect on ion movement, but thermal motion is also significant	✓ 1	Allow ECF from (b)(i).
Part (c)(i)
F = (1.60 × 10⁻¹⁹)²/(4π × 7.1 × 10⁻¹⁰ × (0.50 × 10⁻⁹)²)	✓ 1
F = 1.1 × 10⁻¹¹ N (attractive)	✓ 1	Accept 1.15 × 10⁻¹¹ N.
Part (c)(ii)
Work to separate = electric potential energy magnitude q₁q₂/(4πεr): in water 5.7 × 10⁻²¹ J (0.036 eV)	✓ 1	Or F × r in water.
In lipid, 2.3 × 10⁻¹⁹ J (1.4 eV), about 40 times larger because the permittivity is about 40 times smaller	✓ 1
In water the separation energy is comparable to the average thermal energy (6.4 × 10⁻²¹ J), so collisions easily pull the ions apart and they stay dissolved. In lipid it is about 35 times the thermal energy, so ion pairs are not separated and ions do not dissolve in the lipid layer	✓ 1	Allow ECF from the thermal energy found in (b)(ii).

Answers: (a)(i) 1.0 × 10⁷ V m⁻¹, inwards · (a)(ii) 0 · (a)(iii) 1.6 × 10⁻¹² N · (b)(i) 0.070 eV = 1.1 × 10⁻²⁰ J · (b)(ii) 6.4 × 10⁻²¹ J · (c)(i) 1.1 × 10⁻¹¹ N · (c)(ii) 5.7 × 10⁻²¹ J; 2.3 × 10⁻¹⁹ J (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 (HL) — the electric field strength as the electric potential gradient E = −ΔV_e/Δr; W = qΔV_e in joules and electronvolts; E_p = kq₁q₂/r; D.2 — Coulomb's law with a range of permittivity values; B.3 — average kinetic energy of particles E_k = (3/2)k_BT Command term: Determine

52D-2-34

Electric potential energy in the hydrogen atom·D.2 Electric and magnetic fields

Paper 2Medium14 marks

Short answer & extended response9 steps to full marksDetermine

In a simple model of the hydrogen atom, the electron (charge −e) moves in a circular orbit of radius 5.29 × 10⁻¹¹ m around a stationary proton (charge +e). Only the electric force between them is considered.

(a)

Electric potential energy.

(i)

Outline why the electric potential energy of the electron–proton system is negative.

(1)

(ii)

Show that the electric potential energy of the system is about −27 eV.

(2)

(b)

Energy of the electron in its orbit.

(i)

Show that the kinetic energy of the electron is ke²/(2r).

(2)

(ii)

Hence determine, in eV, the total energy of the atom.

(2)

(iii)

State the ionisation energy of the atom in this model.

(1)

(iv)

Calculate the speed of the electron.

(2)

(c)

Ionisation and gravity.

(i)

Determine the longest wavelength of electromagnetic radiation that can ionise the atom, and identify the region of the electromagnetic spectrum to which it belongs.

(3)

(ii)

Outline why the gravitational force between the electron and the proton can be ignored in this model.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
E_p is the work done to assemble the system from infinite separation; the force between the opposite charges is attractive, so this work is negative	✓ 1	Accept: Ep = kq₁q₂/r with q₁q₂ < 0, zero at infinity.
Part (a)(ii)
E_p = −ke²/r = −8.99 × 10⁹ × (1.60 × 10⁻¹⁹)²/5.29 × 10⁻¹¹ = −4.35 × 10⁻¹⁸ J	✓ 1
−4.35 × 10⁻¹⁸/1.60 × 10⁻¹⁹ = −27.2 eV	✓ 1	Full substitution or an answer to at least 3 s.f. is required.
Part (b)(i)
The electric force provides the centripetal force: ke²/r² = mv²/r	✓ 1
E_k = ½mv² = ke²/(2r)	✓ 1
Part (b)(ii)
E_k = −½E_p = +13.6 eV	✓ 1	Allow ECF from (a)(ii).
E = E_k + E_p = 13.6 − 27.2 = −13.6 eV	✓ 1
Part (b)(iii)
13.6 eV: the energy needed to raise the total energy to zero	✓ 1	Allow ECF from (b)(ii).
Part (b)(iv)
v = √(2E_k/m_e) = √(2 × 2.18 × 10⁻¹⁸/9.11 × 10⁻³¹)	✓ 1	Allow ECF from (b)(ii).
v = 2.2 × 10⁶ m s⁻¹	✓ 1
Part (c)(i)
Photon energy = 13.6 × 1.60 × 10⁻¹⁹ = 2.18 × 10⁻¹⁸ J	✓ 1	Allow ECF from (b)(iii).
λ = hc/E = 6.63 × 10⁻³⁴ × 3.00 × 10⁸/2.18 × 10⁻¹⁸ = 9.1 × 10⁻⁸ m	✓ 1
Ultraviolet	✓ 1	Must be consistent with the candidate's wavelength.
Part (c)(ii)
The ratio of the electric force to the gravitational force, ke²/(Gm_em_p), is about 2 × 10³⁹, so the gravitational force is negligible	✓ 1	A comparison of the two forces (by calculation or by order of magnitude) is required; do not accept "gravity is weak" on its own.

Answers: (a)(ii) −27.2 eV · (b)(ii) −13.6 eV · (b)(iii) 13.6 eV · (b)(iv) 2.2 × 10⁶ m s⁻¹ · (c)(i) 9.1 × 10⁻⁸ m, ultraviolet (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 (HL) — the electric potential energy E_p in terms of work done to assemble the system from infinite separation; E_p = kq₁q₂/r; work in joules and electronvolts; D.2 — Coulomb's law; A.2 — centripetal force; E.1 — photon energy E = hf; D.1 — Newton's universal law of gravitation Command term: Determine

53D-2-42

Charging a balloon·D.2 Electric and magnetic fields

Paper 2Easy11 marks

Short answer & extended response8 steps to full marksExplain

A rubber balloon is rubbed with a woollen cloth. Both are uncharged at first. After rubbing, the balloon has a charge of −60 nC.

(a)

Charging by friction.

(i)

State the charge on the cloth after rubbing.

(1)

(ii)

Calculate the number of electrons transferred from the cloth to the balloon.

(1)

(b)

The balloon is placed against a vertical wall, which is uncharged. The balloon stays at rest on the wall.

(i)

Explain why the uncharged wall attracts the balloon.

(2)

(ii)

The mass of the balloon is 2.5 g, and the coefficient of static friction between the balloon and the wall is 0.40. Determine the minimum electric force of attraction between the balloon and the wall.

(2)

(c)

The balloon is now held well away from the wall. It is modelled as a sphere of radius 0.12 m with its charge spread uniformly over its surface.

(i)

Calculate the magnitude of the electric field strength at a point 0.30 m from the centre of the balloon.

(1)

(ii)

A small water droplet with 50 excess electrons is at this point. Determine the magnitude and the direction of the electric force on the droplet.

(2)

(iii)

The droplet is a sphere of radius 2.0 μm. The density of water is 1000 kg m⁻³. Compare the electric force on the droplet with its weight.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
+60 nC (charge is conserved)	✓ 1	Sign required.
Part (a)(ii)
N = 60 × 10⁻⁹/1.60 × 10⁻¹⁹ = 3.8 × 10¹¹	✓ 1	Accept 3.7–3.8 × 10¹¹.
Part (b)(i)
The negative balloon repels electrons in the surface of the wall (or displaces the charges within its molecules), so the part of the wall nearest the balloon becomes positive and negative charge is displaced further away (charging by induction)	✓ 1	Accept a labelled sketch of the separated charges.
The positive charge is closer to the balloon than the negative charge; since the force decreases with distance (F ∝ 1/r²), the attraction is greater than the repulsion, so the resultant force is attractive	✓ 1	"Opposite charges attract" alone scores 0 for this mark: the comparison of distances is required.
Part (b)(ii)
Friction must support the weight: μ_sF_N ≥ mg, and the normal force F_N equals the electric attraction (horizontal equilibrium)	✓ 1
F = 2.5 × 10⁻³ × 9.81/0.40 = 0.061 N	✓ 1	Accept 0.061 N. Answers that use F = μmg (0.025 N) score [0] for this mark.
Part (c)(i)
E = kQ/r² = 8.99 × 10⁹ × 60 × 10⁻⁹/0.30² = 6.0 × 10³ N C⁻¹	✓ 1	Accept 6 × 10³ N C⁻¹. Using 0.30 − 0.12 = 0.18 m scores 0.
Part (c)(ii)
F = qE = 50 × 1.60 × 10⁻¹⁹ × 6.0 × 10³ = 4.8 × 10⁻¹⁴ N	✓ 1	Allow ECF from (c)(i).
Away from the balloon: the field points towards the negative balloon and the droplet is negative, so the force is opposite to the field (like charges repel)	✓ 1	Accept "radially outwards from the balloon".
Part (c)(iii)
m = 1000 × (4/3)π(2.0 × 10⁻⁶)³ = 3.4 × 10⁻¹⁴ kg, so the weight is 3.3 × 10⁻¹³ N	✓ 1
The weight is about 7 times the electric force, so gravity dominates the motion of the droplet, although the electric force (about 15 % of the weight) is not negligible	✓ 1	Allow ECF from (c)(ii). The comparison must be consistent with the candidate's values. Accept a ratio of 6.8–7.0 (or the electric force as 14–15 % of the weight).

Answers: (a)(i) +60 nC · (a)(ii) 3.8 × 10¹¹ · (b)(ii) 0.061 N · (c)(i) 6.0 × 10³ N C⁻¹ · (c)(ii) 4.8 × 10⁻¹⁴ N, away from the balloon · (c)(iii) weight 3.3 × 10⁻¹³ N ≈ 7 × the electric force (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 — the conservation of electric charge; that the electric charge can be transferred between bodies using friction, electrostatic induction and by contact, including the role of grounding (earthing); Coulomb’s law as given by F = kq₁q₂/r² for charged bodies treated as point charges; the electric field strength as given by E = F/q; the direction of forces between the two types of electric charge; A.2 — static friction F_f ≤ μ_sF_N and equilibrium Command term: Explain

54D-2-43

Magnetic field of a circular coil·D.2 Electric and magnetic fields

Paper 2Medium14 marks

Short answer & extended response8 steps to full marksDetermine

In a laboratory, the Earth's magnetic field has a horizontal component of 18 μT directed towards geographic north and a vertical component of 45 μT directed downwards. A flat circular coil of 50 turns and radius 0.15 m is mounted in a vertical plane with its axis horizontal and along the north–south line, as shown in the plan view.

The magnetic field strength at the centre of a flat circular coil of N turns and radius R carrying a current I is B = μ₀NI/(2R).

Plan view (from above) of the vertical coil, with the compass C at its centre (not to scale).

(a)

The field of the coil.

(i)

Sketch the magnetic field lines of the coil alone in the horizontal plane that contains its axis.

(2)

(ii)

Show that a current of about 0.09 A in the coil makes the horizontal component of the resultant magnetic field at the centre of the coil zero.

(2)

(iii)

The coil is viewed by the observer O, looking north. State and explain whether the current in the coil must be clockwise or anticlockwise as seen by O.

(2)

(b)

A small compass, free to turn only in a horizontal plane, is placed at the centre of the coil.

(i)

The current is set to the value in (a)(ii). Describe the behaviour of the compass.

(1)

(ii)

The current is doubled. Determine the magnitude and direction of the horizontal component of the magnetic field at the centre, and state the direction in which the compass now points.

(2)

(c)

The coil is switched off. Electrons travel horizontally towards the east through the centre of the coil at a speed of 2.0 × 10⁷ m s⁻¹.

(i)

Calculate the magnitude of the magnetic force on one electron due to the horizontal component of the Earth's magnetic field, and state its direction.

(2)

(ii)

The coil is switched on again with the current in (a)(ii). Explain why the electrons are still deflected, and determine the magnitude and the direction of the magnetic force on one electron.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Line(s) through the centre of the coil straight along the axis; other lines form closed loops around each side of the coil, with no lines crossing	✓ 1	The loops must enclose the points where the wire crosses the plane.
Directions consistent: all lines pass through the coil in the same direction and return outside it (the loops round the two sides circulate in opposite senses)	✓ 1	Arrows required. Award this mark for a consistent pattern even if the current direction is not stated.
Part (a)(ii)
I = 2RB/(μ₀N) with B = 18 μT	✓ 1
I = 2 × 0.15 × 18 × 10⁻⁶/(4π × 10⁻⁷ × 50) = 0.0859 A	✓ 1	The answer must be given to at least 2 s.f. (0.086 A), or the full substitution shown.
Part (a)(iii)
The field of the coil at its centre must point south (towards O), opposite to the Earth's horizontal component	✓ 1
Right-hand grip rule: a field towards the observer needs an anticlockwise current as seen by O	✓ 1	The direction mark depends on a correct reason.
Part (b)(i)
There is no horizontal resultant field, so the needle has no preferred direction: it stays pointing in whatever direction it is set (the vertical component cannot turn it in a horizontal plane)	✓ 1	Allow ECF from (a)(ii)–(a)(iii): the candidate must refer to zero horizontal field.
Part (b)(ii)
Coil field = 2 × 18 = 36 μT to the south; resultant = 36 − 18 = 18 μT to the south	✓ 1	Allow ECF from (a)(ii) and (a)(iii).
The north pole of the needle points south (geographic)	✓ 1
Part (c)(i)
F = evB = 1.60 × 10⁻¹⁹ × 2.0 × 10⁷ × 18 × 10⁻⁶ = 5.8 × 10⁻¹⁷ N	✓ 1
Vertically downwards	✓ 1	Velocity east, field north: the force on a positive charge would be upwards; the electron is negative.
Part (c)(ii)
The coil cancels only the horizontal component; the vertical component (45 μT) is not cancelled and is perpendicular to the velocity	✓ 1	Allow ECF from (a)(ii).
F = evB = 1.60 × 10⁻¹⁹ × 2.0 × 10⁷ × 45 × 10⁻⁶ = 1.4 × 10⁻¹⁶ N	✓ 1
The force is horizontal, towards the south	✓ 1	Velocity east, field down: the force on a positive charge would be to the north; the electron is negative.

Answers: (a)(iii) anticlockwise · (b)(i) it points in any direction in which it is set · (b)(ii) 18 μT to the south; the compass points south · (c)(i) 5.8 × 10⁻¹⁷ N, downwards · (c)(ii) 1.4 × 10⁻¹⁶ N, horizontal, towards the south (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 — magnetic field lines; Guidance: magnetic field patterns of a bar magnet, a current-carrying straight wire, a current-carrying circular coil and an air-core solenoid; sketching and interpretation of magnetic field lines; D.3 — the magnitude and direction of the force on a charge moving in a magnetic field as given by F = qvB sin θ Command term: Determine

55D-2-44

Charged conducting sphere·D.2 Electric and magnetic fields

Paper 2Hard13 marks

Short answer & extended response9 steps to full marksDetermine

The dome of a Van de Graaff generator is a hollow metal sphere of radius 0.20 m on an insulating column. The graph shows how the electric field strength E due to the charge on the dome varies with distance r from its centre.

Electric field strength E against distance r from the centre of the dome (drawn to scale).

(a)

The charge on the dome.

(i)

Outline why the electric field strength is zero for r < 0.20 m.

(1)

(ii)

Show that the charge on the dome is about 3 μC.

(2)

(b)

Potential difference.

(i)

Use the graph to estimate the potential difference between the surface of the dome and a point 0.60 m from its centre.

(3)

(ii)

Use V_e = kQ/r to check your answer to (b)(i).

(1)

(c)

A proton is released from rest just outside the surface of the dome.

(i)

Determine the speed of the proton when it is 0.60 m from the centre of the dome.

(2)

(d)

The air around the dome becomes conducting, and the dome discharges, when the electric field strength at its surface reaches the value E_b.

(i)

Show that the greatest potential to which a dome of radius R can be charged is V_max = E_bR.

(2)

(ii)

E_b = 3.0 × 10⁶ V m⁻¹. Calculate the greatest potential to which this dome can be charged.

(1)

(iii)

A dome is to be charged to a potential of 1.0 MV. Deduce the least radius it can have.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
The charges on the conducting dome are at rest, so they have moved to the outer surface until there is no field inside the conductor; the hollow dome encloses no charge, so the field inside it is zero	✓ 1	Accept "all the charge is on the outer surface of a conductor and the field inside a charged hollow conducting sphere is zero".
Part (a)(ii)
E = 7.5 × 10⁵ V m⁻¹ at r = 0.20 m, and outside the dome E = kQ/r²	✓ 1	Accept 7.4–7.6 × 10⁵ V m⁻¹ read from the graph.
Q = 7.5 × 10⁵ × 0.20²/8.99 × 10⁹ = 3.34 × 10⁻⁶ C	✓ 1	The answer must be given to at least 2 s.f. (3.3 × 10⁻⁶ C).
Part (b)(i)
E = −ΔV/Δr, so the potential difference is the area under the E–r graph between r = 0.20 m and r = 0.60 m	✓ 1
Area estimated by counting squares or by trapezia (e.g. one small square = 0.025 m × 0.25 × 10⁵ V m⁻¹ = 625 V; about 160 squares)	✓ 1	A single rectangle or triangle under the whole curve scores 0 for this mark.
ΔV ≈ 1.0 × 10⁵ V	✓ 1	Accept 0.90–1.1 × 10⁵ V.
Part (b)(ii)
ΔV = kQ(1/0.20 − 1/0.60) = 8.99 × 10⁹ × 3.34 × 10⁻⁶ × 3.33 = 1.0 × 10⁵ V	✓ 1	Allow ECF from (a)(ii). Using 3 μC gives 9.0 × 10⁴ V.
Part (c)(i)
Gain in kinetic energy = work done by the field = eΔV: ½mv² = 1.60 × 10⁻¹⁹ × 1.0 × 10⁵	✓ 1	Allow ECF from (b).
v = √(2 × 1.60 × 10⁻¹⁴/1.673 × 10⁻²⁷) = 4.4 × 10⁶ m s⁻¹	✓ 1	Accept 4.2–4.6 × 10⁶ m s⁻¹.
Part (d)(i)
At the surface V = kQ/R and E = kQ/R², so V = ER	✓ 1
The field at the surface may not exceed E_b, so V_max = E_bR	✓ 1
Part (d)(ii)
V_max = 3.0 × 10⁶ × 0.20 = 6.0 × 10⁵ V	✓ 1	Allow ECF from (d)(i).
Part (d)(iii)
R = 1.0 × 10⁶/3.0 × 10⁶ = 0.33 m	✓ 1	Allow ECF from (d)(i).

Answers: (b)(i) ≈ 1.0 × 10⁵ V · (b)(ii) 1.0 × 10⁵ V · (c)(i) 4.4 × 10⁶ m s⁻¹ · (d)(ii) 6.0 × 10⁵ V · (d)(iii) 0.33 m (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 — the electric field strength as given by E = F/q; Guidance: the field inside and outside a single spherical conducting body; D.2 (HL) — the electric field strength E as the electric potential gradient as given by E = −ΔV_e/Δr; that the electric potential V_e at a point is the work done per unit charge to bring a test charge from infinity to that point as given by V_e = kQ/r; the work done in moving a charge q in an electric field as given by W = qΔV_e; A.3 — conservation of energy and kinetic energy Command term: Determine

56D-2-45

Oscillation of a charge between two fixed charges·D.2 Electric and magnetic fields

Paper 2Hard14 marks

Short answer & extended response8 steps to full marksDetermine

Two small spheres, each with a fixed charge Q = +40 nC, are fixed to the ends of a straight, horizontal insulating rod. Each sphere is a distance d = 0.100 m from the midpoint M of the rod. A small bead of mass 0.50 g and charge q = +8.0 nC can slide along the rod without friction.

Two fixed charged spheres and a charged bead free to slide on a horizontal rod (not to scale).

(a)

The bead at M.

(i)

Explain why the bead can remain at rest at M.

(1)

(ii)

Show that the electric potential at M is about 7 kV.

(1)

(b)

The bead is displaced a small distance x from M towards one of the spheres and released.

(i)

Show that the resultant electric force on the bead is directed towards M and has magnitude F = kQq[1/(d − x)² − 1/(d + x)²].

(2)

(ii)

Show that, for x ≪ d, F ≈ 4kQqx/d³.

(2)

(iii)

Hence explain why the bead performs simple harmonic motion, and determine its period.

(3)

(c)

The bead is now released from rest at x = 0.050 m.

(i)

Determine the speed of the bead as it passes M.

(3)

(ii)

Calculate the speed at M predicted by the simple harmonic model of (b)(iii), and explain why it differs from your answer to (c)(i).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
The two spheres exert equal and opposite forces on the bead at M (equal charges at equal distances), so the resultant force is zero (the field strength at M is zero)	✓ 1
Part (a)(ii)
V_e = 2kQ/d = 2 × 8.99 × 10⁹ × 40 × 10⁻⁹/0.100 = 7192 V	✓ 1	Full substitution or an answer to at least 2 s.f. (7.2 kV) is required. 0 V (from zero field) scores 0.
Part (b)(i)
The nearer sphere repels the bead towards M with a force kQq/(d − x)²; the further sphere repels it away from M with a force kQq/(d + x)²	✓ 1
d − x < d + x, so the force towards M is the larger and the resultant is their difference, directed towards M	✓ 1
Part (b)(ii)
1/(d − x)² − 1/(d + x)² = [(d + x)² − (d − x)²]/(d² − x²)² = 4dx/(d² − x²)²	✓ 1	Allow ECF from (b)(i).
For x ≪ d, (d² − x²)² ≈ d⁴, so F ≈ 4kQqx/d³	✓ 1
Part (b)(iii)
The acceleration is −(4kQq/md³)x: proportional to the displacement and directed towards M, so the motion is simple harmonic with ω² = 4kQq/(md³)	✓ 1	Allow ECF from (b)(ii). "Force towards M" alone scores 0: proportionality to x is required.
ω = √(4 × 8.99 × 10⁹ × 40 × 10⁻⁹ × 8.0 × 10⁻⁹/(0.50 × 10⁻³ × 0.100³)) = 4.80 rad s⁻¹	✓ 1	Using m = 0.50 kg gives ω = 0.15 rad s⁻¹: [0] for this mark.
T = 2π/ω = 1.31 s	✓ 1	Allow ECF. Accept 1.3 s.
Part (c)(i)
Potential at x = 0.050 m: kQ(1/0.050 + 1/0.150) = 9589 V	✓ 1	Accept 9.6 kV.
Loss of electric potential energy = qΔV = 8.0 × 10⁻⁹ × (9589 − 7192) = 1.9 × 10⁻⁵ J	✓ 1	Allow ECF from (a)(ii).
½mv² = 1.9 × 10⁻⁵, so v = 0.28 m s⁻¹	✓ 1	Accept 0.27–0.28 m s⁻¹.
Part (c)(ii)
v_max = ωx₀ = 4.80 × 0.050 = 0.24 m s⁻¹	✓ 1	Allow ECF from (b)(iii).
At this amplitude x is not ≪ d: the actual force (e.g. 1.0 × 10⁻³ N at x = 0.050 m) is larger than 4kQqx/d³ (5.8 × 10⁻⁴ N), so more work is done on the bead and it is faster at M than the model predicts	✓ 1	Accept "the restoring force increases faster than in proportion to x". The direction of the difference must agree with the candidate's values.

Answers: (b)(iii) 1.31 s · (c)(i) 0.28 m s⁻¹ · (c)(ii) 0.24 m s⁻¹ (smaller, because the real force exceeds the linear approximation) (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 — Coulomb’s law as given by F = kq₁q₂/r² for charged bodies treated as point charges; the direction of forces between the two types of electric charge; D.2 (HL) — that the electric potential is a scalar quantity with zero defined at infinity; that the electric potential V_e at a point is the work done per unit charge to bring a test charge from infinity to that point as given by V_e = kQ/r; the work done in moving a charge q in an electric field as given by W = qΔV_e; C.1 — the conditions that lead to simple harmonic motion; a = −ω²x and T = 2π/ω Command term: Determine

57D-2-46

Equipotentials of two point charges·D.2 Electric and magnetic fields

Paper 2Medium12 marks

Short answer & extended response8 steps to full marksDetermine

Two point charges, X and Y, are fixed 12.0 cm apart. The diagram shows some of the equipotential lines in a plane containing the charges, with their potentials. The grid squares are 1.0 cm × 1.0 cm.

Equipotential lines in the plane of two point charges X and Y, labelled with their potentials in volts, drawn to scale on a grid of 1.0 cm squares.

(a)

The charges.

(i)

Deduce the sign of each charge, and which charge has the larger magnitude.

(2)

(ii)

The 0 V equipotential crosses the line XY at point Z. Determine the ratio (magnitude of the charge on X)/(magnitude of the charge on Y).

(2)

(b)

The field at S.

(i)

Describe the direction of the electric field at S.

(1)

(ii)

Estimate the magnitude of the electric field strength at S.

(2)

(c)

An electron is moved from T to U.

(i)

Determine the work done by the electric field on the electron. Give your answer in eV and in J.

(2)

(ii)

The electron is instead moved from T to U along a path that passes through S. State the work done on it by the electric field, and explain your answer.

(1)

(iii)

An electron is released from rest at U. Only the electric force acts on it. Determine its speed when it reaches the +200 V equipotential.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
X is positive and Y is negative: the potential is large and positive near X and negative near Y	✓ 1
X has the larger magnitude: the 0 V line is a small loop close to Y (most of the plane is at positive potential)	✓ 1	Accept: at the 0 V line kq/r is equal for both, and that line is much closer to Y than to X.
Part (a)(ii)
Z is 9.6 cm from X and 2.4 cm from Y	✓ 1	Accept 9.4–9.8 cm and 2.2–2.6 cm.
At Z, kq_X/r_X = k\|q_Y\|/r_Y, so the ratio = 9.6/2.4 = 4.0	✓ 1	Allow ECF from (a)(i) for which charge is larger. Accept 3.5–4.5. The inverse ratio (0.25) scores [1 max].
Part (b)(i)
Perpendicular to the +400 V line at S, pointing away from X, towards the +300 V line (towards lower potential)	✓ 1	Both "perpendicular to the equipotential" and the sense are needed.
Part (b)(ii)
Distance between the +500 V and +300 V lines, measured through S perpendicular to them, ≈ 2.8 cm	✓ 1	Accept 2.4–3.2 cm.
E ≈ ΔV/Δr = 200/0.028 = 7.1 × 10³ V m⁻¹	✓ 1	Allow ECF from the candidate's distance. Accept 6.0–8.5 × 10³ V m⁻¹. A distance left in cm (71 V cm⁻¹ quoted as V m⁻¹) scores 0 for this mark.
Part (c)(i)
W = q(V_T − V_U) = (−e)(200 − (−200)) = −400 eV	✓ 1	A positive answer: [1 max]. Accept "the field does −400 eV of work" or "400 eV of work is done against the field".
−400 × 1.60 × 10⁻¹⁹ = −6.4 × 10⁻¹⁷ J	✓ 1	Allow ECF from the first mark.
Part (c)(ii)
−400 eV again: the work done is qΔV_e, which depends only on the potentials at T and U, not on the path between them	✓ 1	Allow ECF from (c)(i). The value and the reason are both needed.
Part (c)(iii)
The field does +400 eV = 6.4 × 10⁻¹⁷ J of work on the electron, which becomes kinetic energy: ½m_ev² = 6.4 × 10⁻¹⁷ J	✓ 1	Allow ECF from (c)(i) (magnitude).
v = √(2 × 6.4 × 10⁻¹⁷/9.11 × 10⁻³¹) = 1.2 × 10⁷ m s⁻¹	✓ 1	Accept 1.2 × 10⁷ m s⁻¹.

Answers: (a)(i) X positive, Y negative; |charge on X| larger · (a)(ii) 4.0 · (b)(i) perpendicular to the equipotential, towards lower potential · (b)(ii) ≈ 7.1 × 10³ V m⁻¹ · (c)(i) −400 eV = −6.4 × 10⁻¹⁷ J · (c)(ii) −400 eV (unchanged) · (c)(iii) 1.2 × 10⁷ m s⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 (HL) — equipotential surfaces for electric fields; the relationship between equipotential surfaces and electric field lines; that the electric potential is a scalar quantity with zero defined at infinity; that the electric potential V_e at a point is the work done per unit charge to bring a test charge from infinity to that point as given by V_e = kQ/r; the electric field strength E as the electric potential gradient as given by E = −ΔV_e/Δr; the work done in moving a charge q in an electric field as given by W = qΔV_e; Guidance: equipotential surfaces for a collection of up to four point charges; A.3 — kinetic energy Command term: Determine

58D-2-59

Electrostatic levitation of lunar dust·D.2 Electric and magnetic fields

Paper 2Hard18 marks

Short answer & extended response11 steps to full marksDetermine

On the sunlit side of the Moon, ultraviolet radiation from the Sun releases photoelectrons from the surface, which becomes positively charged. An electric field forms in the region just above the surface. The graph shows how the electric potential V varies with height h above the surface, relative to a point far above it.

Mass of the Moon = 7.35 × 10²² kg; radius of the Moon = 1.74 × 10⁶ m.

Graph drawn to scale.

(a)

Charging the surface.

(i)

Most of the photoelectrons are released by ultraviolet photons of wavelength 122 nm. The work function of the surface material is 5.0 eV. Determine, in eV, the maximum kinetic energy of the photoelectrons.

(2)

(ii)

Explain why the potential of the surface does not rise much above about 5 V.

(2)

(b)

Levitating dust.

(i)

Use the graph to determine the electric field strength at h = 0.50 m and state its direction.

(3)

(ii)

Show that the gravitational field strength at the surface of the Moon is about 1.6 N kg⁻¹.

(1)

(iii)

A dust grain of mass 2.0 × 10⁻¹⁵ kg floats at rest at h = 0.50 m. Determine the number of electrons that the grain has lost.

(3)

(iv)

Explain whether the equilibrium of the grain is stable for small vertical displacements.

(2)

(c)

(i)

A second grain of the same material has a larger radius. Its charge is proportional to its surface area. Deduce, without calculation, whether it floats higher or lower than the grain in (b)(iii).

(2)

(d)

(i)

Estimate the work done by the electric field on the grain in (b)(iii) as it rose from the surface to h = 0.50 m.

(2)

(ii)

State one assumption made in your estimate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Photon energy = hc/λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸/122 × 10⁻⁹ = 1.63 × 10⁻¹⁸ J = 10.2 eV	✓ 1
E_max = 10.2 − 5.0 = 5.2 eV	✓ 1	The answer must be in eV.
Part (a)(ii)
As the surface becomes positive, the emitted electrons are attracted back: to escape an electron must do work eΔV against the field	✓ 1
When the surface potential reaches about 5.2 V, even the fastest photoelectrons (5.2 eV) cannot escape, so the surface stops losing electrons; this agrees with the 5.0 V at h = 0 on the graph	✓ 1	Allow ECF from (a)(i).
Part (b)(i)
Tangent drawn to the curve at h = 0.50 m	✓ 1
\|E\| = \|gradient\| ≈ 3.0 V m⁻¹	✓ 1	Accept 2.6–3.5 V m⁻¹.
Vertically upwards, away from the surface (the potential decreases with height)	✓ 1
Part (b)(ii)
g = GM/R² = 6.67 × 10⁻¹¹ × 7.35 × 10²²/(1.74 × 10⁶)² = 1.62 N kg⁻¹	✓ 1	Full substitution or 3 s.f. required.
Part (b)(iii)
In equilibrium the electric force balances the weight: qE = mg	✓ 1	The grain is treated as remaining at the same height above a locally flat surface.
q = 2.0 × 10⁻¹⁵ × 1.62/3.0 = 1.1 × 10⁻¹⁵ C	✓ 1	Allow ECF from (b)(i) and (b)(ii).
N = q/e ≈ 6.7 × 10³	✓ 1	Accept 5.8–7.9 × 10³.
Part (b)(iv)
The field strength (the gradient of the graph) decreases with height	✓ 1	Allow ECF from the shape of the field found in (b)(i).
If the grain rises, qE < mg and the resultant force is downwards; if it falls, qE > mg and the resultant force is upwards: the force is restoring, so the equilibrium is stable	✓ 1
Part (c)(i)
m ∝ r³ and q ∝ r², so q/m ∝ 1/r: the larger grain needs a stronger field, E = mg/q, to float	✓ 1
The field is stronger closer to the surface, so the larger grain floats lower	✓ 1
Part (d)(i)
W = qΔV with ΔV = 5.0 − 3.0 = 2.0 V read from the graph	✓ 1	Accept 1.9–2.1 V.
W = 1.1 × 10⁻¹⁵ × 2.0 = 2.2 × 10⁻¹⁵ J	✓ 1	Allow ECF from (b)(iii). Accept 2.0–2.3 × 10⁻¹⁵ J.
Part (d)(ii)
The charge on the grain stays constant as it rises	✓ 1	Accept: the grain starts from the surface with the same charge.

Answers: (a)(i) 5.2 eV · (b)(i) 3.0 V m⁻¹ upwards · (b)(ii) 1.62 N kg⁻¹ · (b)(iii) 6.7 × 10³ · (c)(i) lower · (d)(i) 2.2 × 10⁻¹⁵ J (the remaining parts are explanations — see the table above)

Syllabus understandingD.2 (HL) — the electric field strength as the electric potential gradient E = −ΔV_e/Δr; W = qΔV_e in joules and electronvolts; D.2 — E = F/q; D.1 — g = GM/r²; E.2 (HL) — the photoelectric effect, E_max = hf − Φ; A.2 — translational equilibrium Command term: Determine

D.2 Electric and magnetic fields: IB Physics HL exam-style questions

32 practice questions on D.2 Electric and magnetic fields

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