D.2 Electric and magnetic fields – IB Physics exam questions with mark scheme

1Core4 marks

Two parallel metal plates are 12 mm apart in a vacuum and connected to a 240 V supply, as shown.

(a) Calculate the magnitude of the electric field strength between the plates. [2]

(b) An electron is released at rest midway between the plates. Calculate the force on it and state its direction. [2]

(a) E = V/d with d converted to metres (0.012 m)(1)
(a) E = 2.0 × 10⁴ V m⁻¹ (N C⁻¹)(1)
(b) F = eE = 1.60 × 10⁻¹⁹ × 2.0 × 10⁴ = 3.2 × 10⁻¹⁵ N(1)
(b) Direction: towards the positive (upper) plate, i.e. opposite to the field(1)
Answer: (a) 2.0 × 10⁴ V m⁻¹; (b) 3.2 × 10⁻¹⁵ N towards the positive plate
Examiner tip: Convert mm to m first – a factor of 1000 is the usual lost mark; an electron moves against the field.

2Core5 marks

A polythene rod is rubbed with a woollen cloth. The rod becomes negatively charged.

(a) Explain, in terms of charge transfer, why the rod becomes negative and state the charge gained by the cloth. [2]

(b) An uncharged metal sphere stands on an insulating base. Describe how the rod can be used to give the sphere a positive charge by induction, without the rod touching the sphere. [3]

(a) electrons are transferred from the wool/cloth to the rod (by friction) ; ignore "charge is created" only if electron transfer stated; do not award if positive charges are said to move(1)
(a) cloth gains an equal positive charge, as (total) charge is conserved ; accept "equal and opposite charge"(1)
(b) rod brought near the sphere repels (free) electrons to the far side of the sphere / sphere becomes polarised (near side positive)(1)
(b) far side of the sphere is earthed/grounded (e.g. touched with a finger or earthed wire) while the rod is held near, so electrons flow from the sphere to earth(1)
(b) earth connection removed before the rod is taken away, leaving the sphere with a net positive charge ; order must be clear(1)
Answer: (a) Electrons move from the wool to the rod; cloth gains an equal positive charge. (b) Bring rod near, earth the sphere, remove earth, then remove rod.
Examiner tip: Positive charge never moves in a metal: say electrons leave the sphere to earth, and remove the earth before the rod or the electrons flow back.

3Core5 marks

Two identical small conducting spheres, X and Y, are held with their centres 4.0 cm apart in air. X carries a charge of +3.0 nC and Y carries a charge of −5.0 nC.

(a) Calculate the magnitude of the electrostatic force between the spheres and state whether it is attractive or repulsive. [3]

(b) The spheres are touched together and then returned to a separation of 4.0 cm. Determine the new magnitude of the force between them. [2]

(a) $F=\dfrac{kq_1q_2}{r^2}$ with $r=0.040$ m, i.e. $\dfrac{8.99\times10^{9}\times3.0\times10^{-9}\times5.0\times10^{-9}}{0.040^2}$(1)
(a) $F=8.4\times10^{-5}$ N ; accept $8.43\times10^{-5}$ N; allow $8.5\times10^{-5}$ N from $k=9\times10^9$(1)
(a) attractive (opposite charges)(1)
(b) total charge $-2.0$ nC shared equally, so each sphere has $-1.0$ nC(1)
(b) $F=\dfrac{8.99\times10^{9}\times(1.0\times10^{-9})^2}{0.040^2}=5.6\times10^{-6}$ N (repulsive) ; allow ECF for an incorrect shared charge(1)
Answer: (a) $8.4\times10^{-5}$ N, attractive; (b) $5.6\times10^{-6}$ N (now repulsive)
Examiner tip: Charge is conserved when the spheres touch: add the charges with their signs (+3 + (−5) = −2 nC) before sharing, not their magnitudes.

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4Standard6 marks

In a version of Millikan's experiment, a charged oil drop of mass $4.24\times10^{-15}$ kg is held stationary between two horizontal parallel plates 6.0 mm apart. The potential difference between the plates is 520 V.

(a) Calculate the charge on the drop. [3]

(b) The drop is negatively charged. State which plate is positive. [1]

(c) Charges on five other drops were measured, in units of $10^{-19}$ C: 3.21, 4.79, 6.38, 8.03, 11.2. Explain how these data support the idea that charge is quantised, and estimate the elementary charge. [2]

(a) electric force balances weight: $qE=mg$ (or $\dfrac{qV}{d}=mg$)(1)
(a) $E=\dfrac{520}{6.0\times10^{-3}}=8.7\times10^{4}$ V m$^{-1}$ (seen or implied in substitution)(1)
(a) $q=\dfrac{mgd}{V}=\dfrac{4.24\times10^{-15}\times9.81\times6.0\times10^{-3}}{520}=4.8\times10^{-19}$ C(1)
(b) the upper plate (electric force on the negative drop must be upwards)(1)
(c) every value is (close to) a whole-number multiple of one smallest value / the charges are 2, 3, 4, 5 and 7 times about $1.6\times10^{-19}$ C ; do not accept "the values are all different"(1)
(c) elementary charge $\approx1.6\times10^{-19}$ C ; allow any value from 1.59 to $1.61\times10^{-19}$ C with working(1)
Answer: (a) $4.8\times10^{-19}$ C (= 3e); (b) upper plate; (c) all charges are integer multiples of $\approx1.6\times10^{-19}$ C
Examiner tip: The smallest measured charge (3.21) is not e itself; look for the largest common factor, since 3.21 is 2e.

5Core5 marks

The diagram shows, on the left, a long straight wire perpendicular to the page carrying a current out of the page. Points P and Q are the same distance from the wire. On the right, a solenoid is viewed end-on from its end X; the arrow shows the direction of the current in the turns as seen by the observer.

(a) A small plotting compass is placed at P and then at Q. State the direction in which the north pole of the compass points at each position. Earth's field may be ignored. [2]

(b) State how the magnetic field strength at a point twice as far from the wire as P compares with that at P. [1]

(d) Describe the magnetic field pattern inside the solenoid, away from its ends. [1]

(a) at P: to the left (field lines are anticlockwise circles about the wire)(1)
(a) at Q: towards the top of the page / upwards ; award 1 mark only if both directions reversed (clockwise field)(1)
(b) weaker / smaller ; accept "half as strong" ; do not accept "the same"(1)
(c) north pole (current anticlockwise as viewed from that end)(1)
(d) (field lines) parallel, equally spaced and along the axis, so the field is uniform ; ignore direction(1)
Answer: (a) P: left; Q: up the page; (b) weaker (half); (c) north; (d) uniform: parallel, equally spaced lines along the axis
Examiner tip: Use the right-hand grip rule: thumb along the current (out of the page), fingers curl anticlockwise; the compass points along the field, tangent to the circle.

6Standard5 marks

Two small spheres A and B carry charges of +4.0 μC and +1.0 μC. Their centres are 0.30 m apart. M is the midpoint of AB.

(a) Define electric field strength. [1]

(b) Calculate the magnitude of the electric field strength at M and state its direction. [3]

(a) force per unit charge on a small/test positive charge (placed at the point) ; accept $E=F/q$ with symbols defined and $q$ positive(1)
(b) field from each charge at 0.15 m: $E_A=1.6\times10^{6}$ N C$^{-1}$ and $E_B=4.0\times10^{5}$ N C$^{-1}$ (both seen)(1)
(b) fields are opposite, so subtract: $E=1.2\times10^{6}$ N C$^{-1}$ ; accept $1.20\times10^6$; allow V m$^{-1}$(1)
(b) directed from A towards B / away from A(1)
(c) $\dfrac{4.0}{x^2}=\dfrac{1.0}{(0.30-x)^2}$ gives $x=0.20$ m from A ; ignore the rejected root 0.60 m if stated as outside AB(1)
Answer: (b) $1.2\times10^{6}$ N C$^{-1}$ towards B; (c) 0.20 m from A
Examiner tip: Field strengths are vectors: between two like charges they point opposite ways, so subtract them; adding gives $2.0\times10^6$ and loses two marks.

7HardHL only6 marks

A point charge A of +6.0 nC and a point charge B of −2.0 nC are fixed 0.40 m apart in a vacuum. P is the midpoint of AB.

(a) Calculate the electric potential at P. [2]

(b) Z is the point on line AB, between A and B, where the electric potential is zero. Show that Z is 0.30 m from A. [1]

(d) A student claims that because the potential at Z is zero, the electric field strength at Z must also be zero. Explain why the student is wrong. [2]

(a) potentials are scalars and add: $V=k\left(\dfrac{6.0\times10^{-9}}{0.20}+\dfrac{-2.0\times10^{-9}}{0.20}\right)$(1)
(a) $V_P=180$ V ; accept 179.8 V(1)
(b) $\dfrac{k(6.0\times10^{-9})}{x}=\dfrac{k(2.0\times10^{-9})}{0.40-x}$ leading to $x=0.30$ m (or check: $6.0/0.30-2.0/0.10=0$)(1)
(c) $W=q(V_P-V_Z)=3.0\times10^{-9}\times180=5.4\times10^{-7}$ J ; allow $-5.4\times10^{-7}$ J if clearly described as work done against the field / by an external agent(1)
(d) field strength is (minus) the potential gradient, $E=-\Delta V/\Delta r$, which is not zero at Z (V is still changing with position) ; OR field is a vector sum, potential a scalar sum(1)
(d) between A and B both fields point from A towards B, so they add / do not cancel (E at Z $\approx2.4\times10^{3}$ N C$^{-1}$)(1)
Answer: (a) 180 V; (b) 0.30 m; (c) $5.4\times10^{-7}$ J; (d) $E=-\Delta V/\Delta r\neq0$; fields from A and B are in the same direction at Z
Examiner tip: Zero potential and zero field are different conditions: potential depends on the value of V, field on its gradient.

8StandardHL only5 marks

The graph shows how the electric potential $V$ varies with distance $r$ from the centre of an isolated charged conducting sphere of radius 0.10 m, for points outside the sphere.

(a) Use the graph to determine the charge on the sphere. [2]

(b) Use the graph to estimate the magnitude of the electric field strength at $r=0.20$ m. [2]

(a) a correct reading from the graph, e.g. $V=900$ V at $r=0.20$ m, used in $V=\dfrac{kQ}{r}$(1)
(a) $Q=\dfrac{Vr}{k}=2.0\times10^{-8}$ C ; accept 1.9 to $2.1\times10^{-8}$ C(1)
(b) tangent drawn at $r=0.20$ m and its gradient found / $E=-\dfrac{\Delta V}{\Delta r}$ used ; accept $E=\dfrac{kQ}{r^2}$ or $E=V/r$ using the value from (a)(1)
(b) $E\approx4.5\times10^{3}$ V m$^{-1}$ ; accept 4.0 to $5.0\times10^{3}$ V m$^{-1}$ (N C$^{-1}$); ignore negative sign(1)
(c) $W=q\Delta V=1.60\times10^{-19}\times(900-450)=7.2\times10^{-17}$ J ; accept 7.0 to $7.4\times10^{-17}$ J(1)
Answer: (a) $2.0\times10^{-8}$ C; (b) $\approx4.5\times10^{3}$ V m$^{-1}$; (c) $7.2\times10^{-17}$ J
Examiner tip: Field strength is the gradient of the V–r graph, not V at that point; draw a long tangent so the gradient reading is accurate.

9HardHL only6 marks

Compare and explain the equipotential surfaces and electric field lines for (i) the region between two oppositely charged parallel plates and (ii) the region around an isolated positive point charge. In your answer, explain why no work is done when a charge moves along an equipotential and why equipotentials drawn at equal potential intervals are spaced as they are in each case. [6]

field lines are everywhere perpendicular to equipotential surfaces(1)
moving along an equipotential, $\Delta V=0$ so $W=q\Delta V=0$ / force (along field line) is perpendicular to the displacement(1)
parallel plates: field lines parallel, equally spaced, from + plate to − plate (uniform field); ignore edge effects(1)
parallel plates: equipotentials are planes parallel to the plates, equally spaced for equal $\Delta V$(1)
equal spacing because $E$ is constant, so $\Delta r=\Delta V/E$ is constant(1)
point charge: field lines radial, pointing outwards, getting further apart with distance(1)
point charge: equipotentials are concentric spheres (circles in 2D) centred on the charge(1)
spacing increases with distance because $E=kQ/r^2$ decreases, so a larger $\Delta r$ is needed for the same $\Delta V$ / because $V\propto1/r$(1)
potential decreases along the direction of the field lines (in both cases)(1)

Level	Marks	What it looks like
Level 3	5–6	Both arrangements described correctly with field lines perpendicular to equipotentials; zero work explained using ΔV = 0 or force perpendicular to motion; spacing in both cases explained by linking it to field strength or the potential gradient. Logical, well-organised and uses correct terminology.
Level 2	3–4	Correct shapes of field lines and equipotentials for both cases, with some explanation of either the zero work or the spacing, but the explanation of spacing is incomplete or given for only one case. Reasoning is mostly clear.
Level 1	1–2	Some correct description of field lines or equipotentials for at least one arrangement; little or no explanation; ideas may be disjointed or contain errors.
Level 0	0	No relevant physics.

Answer: Equipotentials are perpendicular to field lines; no work since ΔV = 0. Plates: parallel, equally spaced planes (uniform E). Point charge: concentric spheres, spacing increasing outwards as E falls with r.
Examiner tip: Many answers describe the shapes but never explain the spacing; link it to $E=-\Delta V/\Delta r$ for the top band.

10StretchHL only5 marks

Two protons are held at rest a distance $1.0\times10^{-12}$ m apart in a vacuum and then released. Assume no other forces act.

(a) Calculate the initial electric potential energy of the system. [1]

(b) Determine the speed of each proton when they are a very large distance apart. [2]

(a) $E_p=\dfrac{ke^2}{r}=\dfrac{8.99\times10^{9}\times(1.60\times10^{-19})^2}{1.0\times10^{-12}}=2.3\times10^{-16}$ J(1)
(b) momentum conserved so protons have equal speeds and share $E_p$ equally: each $E_k=1.15\times10^{-16}$ J(1)
(b) $v=\sqrt{\dfrac{2\times1.15\times10^{-16}}{1.67\times10^{-27}}}=3.7\times10^{5}$ m s$^{-1}$ ; award 1 of 2 marks for $5.2\times10^{5}$ m s$^{-1}$ (all energy given to one proton)(1)
(c) energy released is $E_p(r)-E_p(2r)=$ half of initial $E_p$ $=1.15\times10^{-16}$ J, so each proton has $E_k=5.8\times10^{-17}$ J(1)
(c) $v=2.6\times10^{5}$ m s$^{-1}$ ; allow ECF from (a); accept $v_{(b)}/\sqrt2$(1)
Answer: (a) $2.3\times10^{-16}$ J; (b) $3.7\times10^{5}$ m s$^{-1}$; (c) $2.6\times10^{5}$ m s$^{-1}$
Examiner tip: Both protons move: momentum conservation means the energy is shared equally, so each gets half of the change in potential energy.

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