Electric and magnetic fields HL Paper 2

D.2 Electric and Magnetic Fields: HL Question Bank

15 multiple-choice questions • 15 extended-response questions

Paper 1 style and Paper 2 style • Expandable worked answers

Practise the core and additional HL content of D.2. The questions cover charge, electric forces, electric field strength, potential, energy, equipotentials, conducting spheres and magnetic field patterns. Attempt each question before opening its answer. These are original IB-style practice questions; the suggested marks are for self-assessment.

Instructions, data and notation

Select one answer from A–D for each MCQ.
Show working, units and physical reasoning in extended responses.
Give numerical answers to 2–3 significant figures unless stated otherwise.
Treat the stated charges as point charges in vacuum unless another model or medium is specified.
For isolated charge systems, take electric potential and potential energy to be zero at infinite separation.
All derivations use algebra and potential gradients without calculus.

$k=8.99\times10^9\,\mathrm{N\,m^2\,C^{-2}},\qquad e=1.60\times10^{-19}\,\mathrm{C}$

$1\,\mathrm{eV}=1.60\times10^{-19}\,\mathrm{J},\qquad 1\,\mu\mathrm{C}=10^{-6}\,\mathrm{C}$

IB notation: $E$ is electric field strength, $V_e$ is electric potential and $E_p$ is electric potential energy. $V$ in the parallel-plate formula denotes the magnitude of the potential difference. $W_{\mathrm{external}}=q\Delta V_e$ applies to a slow transfer with no change in kinetic energy; work done by the electric field has the opposite sign.

Section A — Multiple-choice questions: Paper 1 style

MCQ 1 — Quantization of charge [1 mark]

An initially neutral object acquires a charge of $-3.2\times10^{-19}\,\mathrm{C}$ . What has happened?

It has lost two electrons.
It has gained two electrons.
It has gained one electron.
It has lost one electron.

Show answer and explanation

Answer: B. The negative sign indicates excess electrons. Divide the charge magnitude by the elementary charge.

$n=\frac{|Q|}{e}=\frac{3.2\times10^{-19}}{1.60\times10^{-19}}=2$

MCQ 2 — Coulomb force [1 mark]

Charges $+3.0\,\mu\mathrm{C}$ and $-2.0\,\mu\mathrm{C}$ are separated by $0.40\,\mathrm{m}$ . Which gives the force magnitude and interaction?

0.34 N, repulsive.
0.13 N, attractive.
0.034 N, repulsive.
0.34 N, attractive.

Show answer and explanation

Answer: D. Opposite charges attract. Calculate the magnitude using the absolute value of the charge product.

$F=k\frac{|q_1q_2|}{r^2}=0.337\,\mathrm{N}\approx0.34\,\mathrm{N}$

MCQ 3 — Force scaling [1 mark]

Both point charges are doubled and their separation is doubled. By what factor does their force magnitude change?

$1$
$2$
$4$
$1/4$

Show answer and explanation

Answer: A. The charge product increases by a factor of four and the squared separation also increases by four.

$\frac{F_{\mathrm{new}}}{F_{\mathrm{old}}}=\frac{2\times2}{2^2}=1$

MCQ 4 — Field of a point charge [1 mark]

What is the electric field magnitude $0.20\,\mathrm{m}$ from a $+5.0\,\mu\mathrm{C}$ point charge, to 3 significant figures?

$2.25\times10^5\,\mathrm{N\,C^{-1}}$
$1.12\times10^5\,\mathrm{N\,C^{-1}}$
$1.12\times10^6\,\mathrm{N\,C^{-1}}$
$2.25\times10^6\,\mathrm{N\,C^{-1}}$

Show answer and explanation

Answer: C. Use the inverse-square field relationship, converting microcoulombs to coulombs.

$E=\frac{k|Q|}{r^2}=\frac{(8.99\times10^9)(5.0\times10^{-6})}{(0.20)^2}=1.12\times10^6\,\mathrm{N\,C^{-1}}$

MCQ 5 — Equal positive charges [1 mark]

At the midpoint between two equal positive point charges, which statement is correct?

Field and potential are both zero.
Field is zero and potential is positive.
Field is non-zero and potential is zero.
Field is zero and potential is negative.

Show answer and explanation

Answer: B. The equal and opposite field vectors cancel, while the two positive scalar potentials add.

MCQ 6 — Equal and opposite charges [1 mark]

At the midpoint between $+Q$ and $-Q$ , with $Q$ positive, which statement is correct?

Field and potential are both zero.
Field is zero and potential is positive.
Field is non-zero and potential is negative.
Potential is zero and the field points from the positive charge towards the negative charge.

Show answer and explanation

Answer: D. The scalar potentials cancel, but both electric-field contributions point from the positive charge towards the negative charge.

MCQ 7 — Inside a conductor [1 mark]

A positively charged isolated solid conducting sphere is in electrostatic equilibrium. Which describes a point inside the metal?

The electric field is zero and the potential equals the positive surface potential.
The electric field and potential are both zero.
The electric field is radial and the potential is zero.
The electric field is radial and the potential decreases towards the centre.

Show answer and explanation

Answer: A. Free charges redistribute until the internal field is zero. The conductor is an equipotential; zero field does not require zero potential.

MCQ 8 — Parallel-plate field [1 mark]

Two large parallel plates have a fixed potential difference. Their separation doubles. Neglecting edge effects, what happens to the field magnitude between them?

It doubles.
It quadruples.
It halves.
It is unchanged.

Show answer and explanation

Answer: C. For a fixed potential difference, the field magnitude is inversely proportional to plate separation.

$E=\frac{V}{d}$

MCQ 9 — Negative-source potential [1 mark]

A negative point charge is isolated in vacuum. What happens to its electric potential as distance from it increases?

It becomes more negative.
It increases towards zero.
It remains constant.
It becomes positive.

Show answer and explanation

Answer: B. Its potential is negative at finite distance and approaches zero from below as distance increases.

$V_e=\frac{kQ}{r},\qquad Q\text{ negative}$

MCQ 10 — Electric potential energy [1 mark]

Charges $+4.0\,\mu\mathrm{C}$ and $+6.0\,\mu\mathrm{C}$ are separated by $0.30\,\mathrm{m}$ . What is their electric potential energy?

$+0.72\,\mathrm{J}$
$-0.72\,\mathrm{J}$
$+2.4\,\mathrm{J}$
$-2.4\,\mathrm{J}$

Show answer and explanation

Answer: A. The product of two positive charges is positive. Use one power of separation for potential energy.

$E_p=k\frac{q_1q_2}{r}=0.719\,\mathrm{J}\approx+0.72\,\mathrm{J}$

MCQ 11 — Potential and a negative charge [1 mark]

A negative test charge moves to a point of higher electric potential. What happens to its electric potential energy?

It increases.
It is unchanged.
It decreases.
It becomes zero in every case.

Show answer and explanation

Answer: C. The potential change is positive and the charge is negative, so the potential-energy change is negative.

$\Delta E_p=q\Delta V_e$

MCQ 12 — Potential gradient [1 mark]

Over a small region, potential falls linearly from $300\,\mathrm{V}$ to $100\,\mathrm{V}$ as $x$ increases by $0.020\,\mathrm{m}$ . What is the signed field component along increasing $x$ ?

$-1.0\times10^4\,\mathrm{N\,C^{-1}}$
$+4.0\,\mathrm{N\,C^{-1}}$
$-4.0\,\mathrm{N\,C^{-1}}$
$+1.0\times10^4\,\mathrm{N\,C^{-1}}$

Show answer and explanation

Answer: D. The potential gradient is negative, so the electric-field component is positive.

$E_x=-\frac{\Delta V_e}{\Delta x}=-\frac{100-300}{0.020}=+1.0\times10^4\,\mathrm{N\,C^{-1}}$

MCQ 13 — Magnetic field around a wire [1 mark]

A straight wire passes perpendicular to the page. Conventional current is directed out of the page towards you. Viewed from your side, what is the magnetic field pattern?

Anticlockwise circles around the wire.
Clockwise circles around the wire.
Radial lines pointing away from the wire.
Radial lines pointing towards the wire.

Show answer and explanation

Answer: A. Point your right thumb out of the page. Your curled fingers indicate the anticlockwise magnetic-field direction.

MCQ 14 — Polarity of a circular coil [1 mark]

Viewed face-on, conventional current flows anticlockwise around a circular coil. Which describes the face you are viewing?

It is a south face and the field at the centre points away from you.
It is a north face and the field at the centre points towards you.
It is a north face and the field at the centre points away from you.
It is a south face and the field at the centre points towards you.

Show answer and explanation

Answer: B. Curl the fingers of your right hand with the current. Your thumb points towards you, indicating the field through the centre and the north face.

MCQ 15 — Magnetic field lines [1 mark]

Which statement correctly describes the magnetic field lines of a bar magnet?

They begin at the north pole and end at the south pole without continuing inside.
They point south to north outside and north to south inside.
They cross where the magnetic field is strongest.
They form closed loops, pointing north to south outside and south to north inside.

Show answer and explanation

Answer: D. Magnetic field lines form continuous closed loops and do not intersect.

Section B — Extended-response questions: Paper 2 style

Suggested marks are given for each part. The worked answers distinguish force direction, signed potential and signed work. Magnetic questions assess field patterns and current direction.

Extended response 1 — Charge quantization and experimental evidence [7 marks]

(a) Explain what is meant by quantization of electric charge for an ordinary isolated body. [2]

(b) An initially neutral body acquires charge $-3.2\times10^{-19}\,\mathrm{C}$ . Determine the number of excess electrons. [2]

(c) Describe how Millikan’s oil-drop experiment provided evidence for charge quantization. [2]

(d) State the principle of conservation of electric charge. [1]

Show answer and working

(a) The net charge of an ordinary isolated body occurs in integer multiples of the elementary charge. The integer can be positive, negative or zero:

$Q=ne,\qquad e=1.60\times10^{-19}\,\mathrm{C}$

$\text{(b)}\quad n_{\mathrm{excess}}=\frac{|Q|}{e}=2$

The body has gained two electrons.

(c) Charges inferred for different oil drops, and changes in a drop’s charge, were consistent with integer multiples of a common elementary amount. Balancing electric and gravitational forces allowed the drop charges to be determined.

(d) The total electric charge of an isolated system remains constant. Charging transfers charge between bodies.

Extended response 2 — Charging by electrostatic induction [7 marks]

A negatively charged rod is brought near an initially neutral metal sphere on an insulating stand. The rod does not touch the sphere.

(a) Describe the redistribution of charge on the sphere. [2]

(b) Explain how to use grounding to leave the sphere positively charged, including the order in which the ground connection and rod are removed. [3]

(c) Explain how this process is consistent with conservation of charge. [2]

Show answer and working

(a) Electrons in the metal are repelled towards the side farther from the rod. The near side has a deficit of electrons and the far side an excess. Before grounding, the sphere’s net charge remains zero.

(b) With the rod still nearby, connect the sphere to Earth. Electrons leave the sphere through the ground connection. Disconnect the ground first, then remove the rod. The remaining positive charge redistributes over the sphere’s surface.

(c) Electrons have moved from the sphere to Earth. The sphere’s positive net charge is balanced by negative charge transferred to Earth; no net charge is created.

Extended response 3 — Coulomb’s law and the effect of a medium [7 marks]

(a) State Coulomb’s law for two point charges and explain how their signs determine the interaction. [2]

(b) Calculate the force magnitude between $+3.0\,\mu\mathrm{C}$ and $-2.0\,\mu\mathrm{C}$ separated by $0.40\,\mathrm{m}$ in vacuum. State the direction of each force. [3]

(c) The charges are placed in a uniform medium of relative permittivity $\varepsilon_r=4.0$ . Using $k_{\mathrm{medium}}=k/\varepsilon_r$ , determine the new force magnitude. [2]

Show answer and working

(a) The force magnitude is proportional to the product of the charge magnitudes and inversely proportional to separation squared. Like charges repel; unlike charges attract.

$F=k\frac{|q_1q_2|}{r^2}$

$\text{(b)}\quad F=(8.99\times10^9)\frac{|(3.0\times10^{-6})(-2.0\times10^{-6})|}{(0.40)^2}=0.337\,\mathrm{N}$

The magnitude is $0.34\,\mathrm{N}$ . Each charge experiences a force towards the other along their joining line. A negative charge product identifies attraction; it does not by itself specify a Cartesian force direction.

$\text{(c)}\quad F_{\mathrm{medium}}=\frac{F_{\mathrm{vacuum}}}{4.0}=0.0843\,\mathrm{N}\approx0.084\,\mathrm{N}$

Extended response 4 — Electric field strength and force [7 marks]

(a) Use Coulomb’s law and the definition of field strength to obtain the field magnitude around a point charge. [2]

(b) Calculate the field magnitude $0.20\,\mathrm{m}$ from a $+5.0\,\mu\mathrm{C}$ point charge. State its direction. [3]

(c) In a separate uniform field of magnitude $3.0\times10^4\,\mathrm{N\,C^{-1}}$ , determine the force on a $+2.0\,\mu\mathrm{C}$ charge. [2]

Show answer and working

$\text{(a)}\quad E=\frac{F}{q_{\mathrm{test}}}=\frac{k|Q|q_{\mathrm{test}}/r^2}{q_{\mathrm{test}}}=\frac{k|Q|}{r^2}$

Here the test charge is positive. The field direction is the force direction on a positive test charge: away from a positive source and towards a negative source.

$\text{(b)}\quad E=\frac{(8.99\times10^9)(5.0\times10^{-6})}{(0.20)^2}=1.12\times10^6\,\mathrm{N\,C^{-1}}$

The field points radially outwards.

$\text{(c)}\quad F=qE=(2.0\times10^{-6})(3.0\times10^4)=0.060\,\mathrm{N}$

The force is in the electric-field direction. For a negative charge it would be opposite to that direction.

Extended response 5 — Electric potential and potential energy [8 marks]

(a) Define electric potential at a point, with zero at infinity. [2]

(b) Given $E_p=kQq/r$ , obtain the electric potential of a point charge using $V_e=E_p/q$ . [2]

(c) Determine the potential $0.30\,\mathrm{m}$ from a $+2.0\,\mu\mathrm{C}$ point charge. [2]

(d) Determine the potential energy of a $-3.0\,\mathrm{nC}$ test charge at that point. [2]

Show answer and working

(a) Electric potential is the external work done per unit positive test charge in bringing it slowly from infinity to the point, with no change in kinetic energy. It is a scalar with unit $\mathrm{J\,C^{-1}}$ , equivalent to the volt.

$\text{(b)}\quad V_e=\frac{E_p}{q}=\frac{kQq/r}{q}=\frac{kQ}{r}$

$\text{(c)}\quad V_e=\frac{(8.99\times10^9)(2.0\times10^{-6})}{0.30}=+5.99\times10^4\,\mathrm{V}$

$\text{(d)}\quad E_p=qV_e=(-3.0\times10^{-9})(5.99\times10^4)=-1.80\times10^{-4}\,\mathrm{J}$

The potential is positive because of the source charge. The test charge’s potential energy is negative because its charge is negative.

Extended response 6 — Work and the energy of two charges [8 marks]

Two charges $+4.0\,\mu\mathrm{C}$ and $+6.0\,\mu\mathrm{C}$ are initially separated by $0.30\,\mathrm{m}$ . One is held fixed while the other is moved slowly to a separation of $0.60\,\mathrm{m}$ .

(a) Calculate the initial electric potential energy. [2]

(b) Calculate the change in electric potential energy. [2]

(c) Determine the work done by the external agent and the work done by the electric force. [2]

(d) Explain the signs of the two work values. [2]

Show answer and working

$\text{(a)}\quad E_{p,i}=k\frac{q_1q_2}{r_i}=0.719\,\mathrm{J}\approx+0.72\,\mathrm{J}$

$\text{(b)}\quad E_{p,f}=k\frac{q_1q_2}{r_f}=0.360\,\mathrm{J}$

$\Delta E_p=E_{p,f}-E_{p,i}=-0.360\,\mathrm{J}$

$\text{(c)}\quad W_{\mathrm{external}}=\Delta E_p=-0.36\,\mathrm{J},\qquad W_{\mathrm{electric}}=-\Delta E_p=+0.36\,\mathrm{J}$

(d) Repulsion acts in the direction of outward displacement, so the electric force does positive work. To maintain slow motion with no kinetic-energy change, the external agent opposes that motion and removes energy, doing negative work.

Extended response 7 — Zero resultant field between unequal charges [10 marks]

Charges $+3.0\,\mu\mathrm{C}$ and $+12\,\mu\mathrm{C}$ are fixed $0.90\,\mathrm{m}$ apart. A point between them is at distance $x$ from the smaller charge.

(a) Write the condition for zero resultant electric field. [3]

(b) Determine $x$ . [3]

(c) Calculate the electric potential at that point. [3]

(d) Explain why zero field does not mean zero potential. [1]

Show answer and working

(a) Between like charges, the field contributions point in opposite directions. Set their magnitudes equal:

$\frac{k(3.0\times10^{-6})}{x^2}=\frac{k(12\times10^{-6})}{(0.90-x)^2}$

$\text{(b)}\quad (0.90-x)^2=4x^2\quad\Rightarrow\quad 0.90-x=2x$

$x=0.30\,\mathrm{m},\qquad 0.90-x=0.60\,\mathrm{m}$

$\text{(c)}\quad V_e=k\left(\frac{3.0\times10^{-6}}{0.30}+\frac{12\times10^{-6}}{0.60}\right)=+2.70\times10^5\,\mathrm{V}$

(d) Electric fields add as vectors and cancel here. Potentials add as scalars; both contributions are positive.

Extended response 8 — Equal and opposite charges: field and potential [8 marks]

A $+2.0\,\mu\mathrm{C}$ charge is fixed on the left and a $-2.0\,\mu\mathrm{C}$ charge on the right, separated by $0.40\,\mathrm{m}$ . Consider their midpoint.

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

(b) Determine the electric field magnitude and direction there. [3]

(c) Explain how a zero potential can coexist with a non-zero field. [2]

(d) State the direction of force on a negative test charge at the midpoint. [1]

Show answer and working

$\text{(a)}\quad V_e=k\left(\frac{2.0\times10^{-6}}{0.20}-\frac{2.0\times10^{-6}}{0.20}\right)=0$

(b) The field from the positive charge points right, away from it. The field from the negative charge also points right, towards it. Add their magnitudes:

$E=2\frac{k(2.0\times10^{-6})}{(0.20)^2}=8.99\times10^5\,\mathrm{N\,C^{-1}}$

The resultant field points to the right.

(c) Zero potential is a scalar value at a point. Field depends on the potential gradient, which need not be zero at a point of zero potential.

(d) The force on a negative test charge points left, opposite to the electric field.

Extended response 9 — Parallel plates and the uniform-field relationship [9 marks]

Two large parallel plates have a potential difference of $600\,\mathrm{V}$ and separation $0.030\,\mathrm{m}$ . The positive plate is on the left and the negative plate on the right.

(a) Derive the field-magnitude relationship $E=V/d$ using work and potential difference. [3]

(b) Calculate the field magnitude and state its direction well away from the edges. [2]

(c) Determine the force magnitude and direction on an electron between the plates. [2]

(d) Describe the field near the plate edges and explain the limitation of the uniform-field model. [2]

Show answer and working

(a) Consider a positive test charge moving a distance $d$ along a uniform electric field. The field does work $qEd$ . The potential falls by a magnitude $V$ , so the loss in potential energy is $qV$ .

$qEd=qV\quad\Rightarrow\quad E=\frac{V}{d}$

$\text{(b)}\quad E=\frac{600}{0.030}=2.0\times10^4\,\mathrm{N\,C^{-1}}$

The field points right, from the positive plate to the negative plate.

$\text{(c)}\quad F=eE=(1.60\times10^{-19})(2.0\times10^4)=3.2\times10^{-15}\,\mathrm{N}$

The electron’s force points left, towards the positive plate.

(d) Near the edges, field lines curve outwards and their spacing varies. These fringing fields are non-uniform. $E=V/d$ is a good approximation in the central region when the plate dimensions are large compared with their separation.

Extended response 10 — Potential gradient, signed work and equipotentials [8 marks]

Electric potential varies approximately linearly with position along a small region:

Position $x$ / m	Potential $V_e$ / V
0.000	160
0.010	140
0.020	120
0.030	100

(a) Determine the signed electric-field component along increasing $x$ . [2]

(b) A $-2.0\,\mathrm{nC}$ charge moves slowly from $x=0.010\,\mathrm{m}$ to $x=0.030\,\mathrm{m}$ . Determine its potential-energy change and the external work. [3]

(c) Determine the work done by the electric field during this move. [1]

(d) State the work done by the electric field when a charge moves along one equipotential surface, and give the reason. [2]

Show answer and working

$\text{(a)}\quad E_x=-\frac{\Delta V_e}{\Delta x}=-\frac{100-160}{0.030}=+2.0\times10^3\,\mathrm{N\,C^{-1}}$

The electric field points in the direction of increasing $x$ .

$\text{(b)}\quad \Delta V_e=100-140=-40\,\mathrm{V}$

$\Delta E_p=q\Delta V_e=(-2.0\times10^{-9})(-40)=+8.0\times10^{-8}\,\mathrm{J}$

The external work is $+8.0\times10^{-8}\,\mathrm{J}$ because the kinetic energy is unchanged.

$\text{(c)}\quad W_{\mathrm{electric}}=-\Delta E_p=-8.0\times10^{-8}\,\mathrm{J}$

(d) The work is zero along the same equipotential because $\Delta V_e=0$ . Moving between different equipotentials generally involves non-zero work. Field lines are perpendicular to equipotential surfaces wherever the field is non-zero.

Extended response 11 — A charged conducting sphere [8 marks]

An isolated solid conducting sphere of radius $0.10\,\mathrm{m}$ carries charge $+4.0\,\mathrm{nC}$ and is in electrostatic equilibrium. Let $r$ be distance from its centre.

(a) State where its excess charge resides and the electric field inside the metal. [2]

(b) Calculate its surface potential and state the potential at its centre. [2]

(c) Calculate the field magnitude and potential at $r=0.20\,\mathrm{m}$ . [2]

(d) Describe the electric field and potential inside an otherwise identical hollow conducting shell with no charge in its closed cavity. [2]

Show answer and working

(a) Excess charge resides on the outer surface. The field inside the metal is zero; otherwise free charges would continue to redistribute.

$\text{(b)}\quad V_{e,\mathrm{surface}}=\frac{kQ}{R}=\frac{(8.99\times10^9)(4.0\times10^{-9})}{0.10}=+360\,\mathrm{V}$

The centre has the same potential as the surface because the entire conductor is an equipotential.

$\text{(c)}\quad E=\frac{kQ}{r^2}=899\,\mathrm{N\,C^{-1}},\qquad V_e=\frac{kQ}{r}=+180\,\mathrm{V}$

The exterior field points radially outwards. Unrounded calculated values are $899\,\mathrm{N\,C^{-1}}$ and $179.8\,\mathrm{V}$ .

(d) With no charge inside the closed cavity, the field in both the metal and cavity is zero. Their potential is constant and equals the shell’s surface potential, which is generally non-zero. Outside, the spherical shell has the same field and potential as a point charge at its centre.

Extended response 12 — Comparing electric and gravitational potential [6 marks]

(a) Write the potential-energy expressions for two point charges and two point masses, using zero at infinite separation. [2]

(b) Explain the difference between the possible signs of these two potential energies. [2]

(c) Describe how electric potential changes with increasing distance for a positive source charge and for a negative source charge. [2]

Show answer and working

$\text{(a)}\quad E_{p,\mathrm{electric}}=k\frac{q_1q_2}{r},\qquad E_{p,\mathrm{gravitational}}=-G\frac{m_1m_2}{r}$

(b) Electric potential energy is positive for like charges and negative for unlike charges. Gravitational potential energy is negative for positive masses because gravity is attractive. Both expressions vary with inverse separation.

(c) For a positive source, $V_e=kQ/r$ is positive and decreases towards zero as distance increases. For a negative source, it is negative and increases towards zero. Therefore, “potential always decreases with distance” is not correct.

Extended response 13 — Magnetic field around a straight wire [6 marks]

A long straight wire passes perpendicular to a sheet of paper. Conventional current flows out of the paper towards you.

(a) Describe the magnetic field pattern in the plane of the paper. [2]

(b) Determine its direction as viewed from your side, explaining the rule used. [2]

(c) State how the pattern’s direction changes if the current reverses. [1]

(d) State what happens to magnetic field strength as distance from the wire increases. [1]

Show answer and working

(a) Field lines are concentric circles centred on the wire. They are closer together near the wire, representing a stronger field.

(b) The direction is anticlockwise. Point the thumb of the right hand out of the paper with conventional current; the curled fingers indicate the magnetic field direction.

(c) Reversing the current reverses the field direction to clockwise.

(d) The field strength decreases with increasing distance from the wire.

Extended response 14 — Circular coils, solenoids and bar magnets [6 marks]

(a) A circular coil carries anticlockwise conventional current when viewed from one face. State that face’s magnetic polarity and the field direction through the centre. [2]

(b) Describe the magnetic field pattern inside and outside a long air-core solenoid carrying a steady current. [2]

(c) Explain the effect of reversing the solenoid current. [1]

(d) State one similarity between a solenoid’s external field and a bar magnet’s external field. [1]

Show answer and working

(a) The viewed face is north and the field at the centre points towards the viewer. Curl the right-hand fingers with the current; the thumb gives the axial field direction.

(b) Well inside a long solenoid, field lines are approximately straight, parallel and evenly spaced, indicating an approximately uniform field. Outside they curve round to form closed loops; the external field is comparatively weak away from the ends.

(c) Reversing current reverses the field and swaps the north and south ends.

(d) Both have a dipole-like external pattern, with field lines directed from the north end towards the south end outside.

Extended response 15 — Lab-style investigation: mapping potential and field [10 marks]

A student uses two long parallel electrodes on conducting paper, connected to a low-voltage DC supply. A voltmeter probe measures potential along a line perpendicular to the electrodes in the central region. The zero reference is one electrode, rather than infinity. The measurements are:

Position $x$ / m	Potential $V_e$ / V
0.000	0.0
0.020	10.0
0.040	20.0
0.060	30.0

(a) Identify a graph that tests whether the field component along this line is uniform. [2]

(b) Use the data to determine the field magnitude and direction along the line. [2]

(c) Describe how to map an equipotential line using the probe. [2]

(d) The total potential difference is $(30.0\pm0.5)\,\mathrm{V}$ over $(0.060\pm0.002)\,\mathrm{m}$ . Estimate the percentage uncertainty in the field magnitude. [2]

(e) Calculate the external work needed to move a $+3.0\,\mu\mathrm{C}$ test charge slowly from the zero-potential electrode to the $30.0\,\mathrm{V}$ electrode in the corresponding field model. [2]

Show answer and working

(a) Plot potential $V_e$ vertically against position $x$ horizontally. A straight line has constant gradient and therefore a constant field component along that line.

$\text{(b)}\quad E_x=-\frac{\Delta V_e}{\Delta x}=-\frac{30.0}{0.060}=-500\,\mathrm{N\,C^{-1}}$

The magnitude is $5.0\times10^2\,\mathrm{N\,C^{-1}}$ and the direction is towards decreasing $x$ , from higher potential towards lower potential.

(c) Keep the reference lead connected to the same electrode. Move the probe to find several positions giving the same voltage, mark those positions and join them smoothly. Repeat for other voltage values. Field lines meet these equipotential lines at right angles.

$\text{(d)}\quad \frac{\Delta E}{E}\approx\frac{\Delta V}{V}+\frac{\Delta d}{d}=\frac{0.5}{30.0}+\frac{0.002}{0.060}=0.050$

The estimated percentage uncertainty is $5.0\%$ . This uses the IB approximation of adding fractional uncertainties for division.

$\text{(e)}\quad W_{\mathrm{external}}=q\Delta V_e=(3.0\times10^{-6})(30.0)=+9.0\times10^{-5}\,\mathrm{J}$

The electric field does $-9.0\times10^{-5}\,\mathrm{J}$ of work. Measurements should be taken away from electrode ends to reduce edge effects; regular electrode spacing and a high-resistance voltmeter help the field mapping.

Exam skills and mastery — Common misconceptions

Field lines, potential and work

Field lines are not automatically particle paths. They show the instantaneous force direction for a positive test charge. Motion also depends on initial velocity, mass and other forces.

Potential is not potential energy. $V_e=E_p/q$ is energy per unit charge. Potential is determined by the source charges; the test charge’s potential energy also depends on its sign and magnitude.

Electric force can attract or repel. Like charges repel and unlike charges attract. Calculate force magnitude using $|q_1q_2|$ and state its direction separately.

Zero field does not imply zero potential. A charged conductor can have zero internal field and non-zero constant potential. A dipole midpoint can have zero potential and non-zero field.

Zero work applies along one equipotential. Between different equipotentials, $\Delta V_e$ is generally non-zero. For a slow transfer, $W_{\mathrm{external}}=q\Delta V_e$ .

Command terms and mastery goal

Command term	Expected response
Define	Give the precise meaning, including per-unit-charge wording where needed.
State	Give the requested law, fact, direction or result.
Calculate / determine	Use the data, show working and give units and direction where relevant.
Explain	Connect the result to a physical principle.
Show / derive	Display algebraic steps from an established relationship.
Describe / sketch	Give the field pattern, arrows and relative line spacing.

Mastery goal: distinguish electric field, electric potential and potential energy; handle charge signs and work correctly; use algebraic field relationships; and describe magnetic field patterns with the right-hand rule.

Open the quick MCQ answer key

Question	Answer	Question	Answer	Question	Answer
1	B	6	D	11	C
2	D	7	A	12	D
3	A	8	C	13	A
4	C	9	B	14	B
5	B	10	A	15	D

IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

Curriculum

Electric and magnetic fields HL Paper 2

D.2 Electric and Magnetic Fields: HL Question Bank

Instructions, data and notation

Section A — Multiple-choice questions: Paper 1 style

MCQ 1 — Quantization of charge [1 mark]

MCQ 2 — Coulomb force [1 mark]

MCQ 3 — Force scaling [1 mark]

MCQ 4 — Field of a point charge [1 mark]

MCQ 5 — Equal positive charges [1 mark]

MCQ 6 — Equal and opposite charges [1 mark]

MCQ 7 — Inside a conductor [1 mark]

MCQ 8 — Parallel-plate field [1 mark]

MCQ 9 — Negative-source potential [1 mark]

MCQ 10 — Electric potential energy [1 mark]

MCQ 11 — Potential and a negative charge [1 mark]

MCQ 12 — Potential gradient [1 mark]

MCQ 13 — Magnetic field around a wire [1 mark]

MCQ 14 — Polarity of a circular coil [1 mark]

MCQ 15 — Magnetic field lines [1 mark]

Section B — Extended-response questions: Paper 2 style

Extended response 1 — Charge quantization and experimental evidence [7 marks]

Extended response 2 — Charging by electrostatic induction [7 marks]

Extended response 3 — Coulomb’s law and the effect of a medium [7 marks]

Extended response 4 — Electric field strength and force [7 marks]

Extended response 5 — Electric potential and potential energy [8 marks]

Extended response 6 — Work and the energy of two charges [8 marks]

Extended response 7 — Zero resultant field between unequal charges [10 marks]

Extended response 8 — Equal and opposite charges: field and potential [8 marks]

Extended response 9 — Parallel plates and the uniform-field relationship [9 marks]

Extended response 10 — Potential gradient, signed work and equipotentials [8 marks]

Extended response 11 — A charged conducting sphere [8 marks]

Extended response 12 — Comparing electric and gravitational potential [6 marks]

Extended response 13 — Magnetic field around a straight wire [6 marks]

Extended response 14 — Circular coils, solenoids and bar magnets [6 marks]

Extended response 15 — Lab-style investigation: mapping potential and field [10 marks]

Exam skills and mastery — Common misconceptions

Field lines, potential and work

Command terms and mastery goal

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