Summary to follow. 18 syllabus statements (8 HL) · 47 questions · about twenty minutes.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank ·
Specialist review in progress
· How these pages are made
Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Physics guide (first assessment 2025, updated November 2023).
Learn
In preparation: 0 of 18 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
Electric charge
Electric charge
A property of matter that gives rise to electric forces. There are two types, positive and negative: like charges repel and unlike charges attract, and the forces the two charges exert on each other are equal in size and opposite in direction (Newton's third law). A charged object also attracts an uncharged one, by inducing charge separation in it. SI unit: coulomb (C).
Students often think Of two unequal charges, the larger charge exerts the larger force on the other. In fact No. The two forces form a Newton's third law pair, so they are equal in size and opposite in direction. F = kq₁q₂/r² contains the product of the charges, so it is the same number for each body.
Students often think A charged object exerts no force at all on an uncharged object; attraction requires two opposite charges. In fact Yes. A charged object attracts a nearby uncharged object: it induces charge separation (in a conductor) or polarizes the molecules (in an insulator), so the nearer side carries opposite charge and is attracted more strongly than the further side is repelled.
Coulomb's law
Coulomb's law
The electric force between two point charges q₁ and q₂ a distance r apart is F = k q₁q₂/r², directed along the line joining them: repulsive for like charges, attractive for unlike charges. For charged spheres with a uniform (spherically symmetric) charge distribution, r is measured between their centres.
Coulomb constant and permittivity of free space
k = 1/4πε₀, where ε₀ = 8.85 × 10⁻¹² F m⁻¹ (C² N⁻¹ m⁻²) is the permittivity of free space; k = 8.99 × 10⁹ N m² C⁻². These values apply to charges in a vacuum (and very nearly in air).
Permittivity of a medium
When the charges are in a medium, ε₀ is replaced by the permittivity ε of that medium: F = q₁q₂/4πεr². Insulating media have ε > ε₀ (for example ε ≈ 80ε₀ for water), so the force between the same charges at the same separation is smaller than in a vacuum. SI unit: F m⁻¹.
Point charge
A charged body whose size is negligible compared with the distances involved, so that all its charge can be taken to be at one point. A sphere with a uniform charge distribution behaves, at points outside it, as a point charge at its centre.
Students often think Electric force and field strength are inversely proportional to distance, so tripling the distance divides them by three. In fact No. Coulomb's law is an inverse-square law: tripling r reduces the force and the field strength to one-ninth.
Students often think The Coulomb constant k = 8.99 × 10⁹ N m² C⁻² (equivalently ε₀) applies whatever medium the charges are in. In fact No. k = 1/4πε₀ is for a vacuum (and very nearly for air). In a medium, ε₀ is replaced by the permittivity ε of the medium: F = q₁q₂/4πεr².
Conservation of electric charge
Conservation of electric charge
The total electric charge of an isolated system is constant. Charge is never created or destroyed: charging an object moves existing charge (in solids, electrons) from one body to another, so one body gains exactly the charge the other loses.
Students often think Charges combine as amounts without sign, so +5.0 nC and −3.0 nC make 8.0 nC in total. In fact No. Charge is a signed quantity: +5.0 nC and −3.0 nC add to +2.0 nC, not 8.0 nC.
Students often think When a positive and a negative charge meet, they neutralize each other completely, so both bodies are left uncharged. In fact Only if their charges were equal in size. Otherwise the total, e.g. +5.0 nC + (−3.0 nC) = +2.0 nC, remains and is shared between them.
Quantization of charge and the elementary charge
Quantization of charge and the elementary charge
Electric charge exists only in whole-number multiples of the elementary charge e = 1.60 × 10⁻¹⁹ C, the size of the charge on an electron or a proton: q = ne, where n is an integer.
Millikan's oil-drop experiment
Charged oil drops are observed between two horizontal parallel plates. Adjusting the potential difference until a drop is held stationary gives qE = mg, so q = mgd/V for a drop of known weight. Millikan found that every drop's charge was a whole-number multiple of one smallest value, about 1.6 × 10⁻¹⁹ C: evidence that charge is quantized.
Students often think The smallest charge found on any drop is the basic unit of charge. In fact Not necessarily. The smallest measured charge may itself be a multiple of the elementary charge (e.g. 3e). The basic unit is the largest value of which every measured charge is a whole-number multiple.
Students often think The charges measured on different drops are repeated measurements of one quantity, so their mean gives the elementary charge. In fact No. The drops carry different numbers of elementary charges, so their mean has no special meaning; it is not a unit of which each charge is a multiple.
Conductors and insulators
Conductors and insulators
In a metal (a conductor) some electrons are free to move through the material, so charge spreads over it and can flow to or from it. In an insulator the electrons are bound to their atoms, so charge placed on it stays where it was put, although the molecules can still be polarized.
Charging by friction
When two different insulators are rubbed together, electrons are transferred from one surface to the other. The body that gains electrons becomes negatively charged; the other is left with an equal positive charge.
Charging by contact
When a charged conductor touches another conductor, charge flows between them until there is no potential difference between them. For two identical conducting spheres the total charge is shared equally.
Electrostatic induction
A charged object brought near a conductor, without touching it, causes the conductor's free electrons to redistribute: charge opposite in sign to the inducing charge gathers on the near side and like charge on the far side. No charge passes between the two objects.
Grounding (earthing)
Connecting a conductor to the Earth, a very large reservoir of charge. Electrons can then flow between the conductor and the Earth. If the conductor is earthed while a charged object is held nearby and the earth connection is broken before the object is removed, the conductor is left with a charge opposite in sign to that of the object (charging by induction).
Students often think Rubbing (friction) creates electric charge on the surface that is rubbed. In fact No. Rubbing transfers existing electrons from one surface to the other; the charge gained by one object equals the charge lost by the other, so the total is unchanged.
Students often think Charging a solid object involves the movement of positive charge (protons) as well as, or instead of, electrons. In fact No. In solids the protons are fixed in the nuclei of the atoms. Only electrons move: a positive object has lost electrons and a negative object has gained them.
Electric field strength
Electric field strength
The electric force per unit charge experienced by a small positive test charge placed at a point: E = F/q. It is a vector in the direction of the force on a positive charge; a negative charge experiences a force opposite to E. E at a point does not depend on the test charge used. SI unit: N C⁻¹ (equivalent to V m⁻¹).
Radial field of a point charge or charged sphere
Combining E = F/q with Coulomb's law gives E = kQ/r² at a distance r from a point charge Q, directed away from a positive charge and towards a negative one. Outside a charged conducting sphere the field is the same as that of a point charge at its centre; inside the conductor the electric field is zero.
Students often think Electric field strength is the electric force at a point, measured in newtons: the force measured on a charge placed there is the field strength. In fact No. E is the force per unit charge, E = F/q, measured with a small test charge. At a given point it has the same value whatever test charge is used; only the force F = qE changes with q.
Students often think The force on any charge in a field acts in the direction of the field, whatever the sign of the charge (or: the field direction is the direction of the force on whatever charge is considered). In fact No. The field direction is defined as the direction of the force on a positive charge. A negative charge, such as an electron, experiences a force opposite to the field.
Electric field line
Electric field line
A line drawn so that its direction at every point is the direction of the electric field (the force on a positive test charge). Electric field lines start on positive charges and end on negative charges (or at infinity), never cross, and meet the surface of a conductor at right angles.
Neutral point
A point at which the resultant electric field of two or more charges is zero. For two equal like charges it is at the midpoint between them; for unequal like charges it is closer to the smaller charge. No neutral point exists between two unlike charges.
Students often think Field lines from different charges can cross where the fields of the charges overlap. In fact No. The field has one direction at each point, and a field line shows that direction; two lines crossing would give two directions at one point.
Students often think In the field of two charges, the field lines run from one charge to the other, joining them, whatever the signs of the charges. In fact No. Lines join two charges only when the charges are of opposite sign. For two like charges, lines from each charge bend away from the other charge and go off to infinity.
Field-line density
Field-line density
The number of field lines crossing unit area at right angles to the field. The field strength is proportional to this density: the field is strongest where the lines are closest together. For a point charge the same number of lines crosses every sphere of area 4πr², so the density, and E, falls as 1/r².
Students often think The field strength is the same at every point along a field line, because the same line passes through all of them. In fact No. The field strength is shown by how close the lines are, not by which line a point is on. Along a radial line from a point charge the field falls as 1/r².
Students often think The number of field lines drawn is arbitrary, so the density of the lines carries no information about the field strength. In fact No. The number drawn is arbitrary, but once chosen it is used consistently across the diagram, so the relative density of the lines shows where the field is stronger or weaker.
Uniform field between parallel plates
Uniform field between parallel plates
Between two oppositely charged parallel plates, away from their edges, the field has the same magnitude and direction at every point: E = V/d, where V is the potential difference between the plates and d their separation. The field lines are parallel, equally spaced and run from the positive to the negative plate. SI unit: V m⁻¹.
Edge effects
Near the edges of a pair of parallel plates the field is not uniform: the field lines bulge outwards beyond the edges of the plates, so E = V/d applies only in the central region.
Students often think Distances quoted in millimetres or centimetres can be substituted directly into E = V/d (or other equations) without converting to metres. In fact No. d must be in metres to give E in V m⁻¹: 5.0 mm = 5.0 × 10⁻³ m.
Students often think The field strength between parallel plates is E = Vd, so it increases in proportion to the separation. In fact No. E = V/d: the field is the potential difference per unit distance, so it is smaller when the plates are further apart.
Magnetic field line
Magnetic field line
A line whose direction at every point is the direction in which the north pole of a small compass would point. Magnetic field lines form closed loops (outside a magnet they run from its north pole to its south pole, and inside it from south to north), never cross, and are closest together where the field is strongest.
Right-hand grip rule
For a straight current-carrying wire: grip the wire with the right hand, thumb in the direction of the conventional current; the fingers curl in the direction of the magnetic field lines, which are concentric circles centred on the wire. For a coil or solenoid, fingers curled in the direction of the current give the thumb pointing along the field inside it (towards the north-pole end).
Magnetic field of a coil and of a solenoid
At the centre of a flat circular coil the field is along the coil's axis, perpendicular to its plane. Inside a long air-core solenoid the field lines are parallel and evenly spaced along the axis (a strong, nearly uniform field); outside, the pattern is like that of a bar magnet, with the lines forming closed loops through the inside.
Students often think The magnetic field lines around a straight wire are radial, pointing directly away from (or towards) the wire, like the electric field of a charged wire. In fact No. They are concentric circles centred on the wire, in planes perpendicular to it.
Students often think The magnetic field produced by a current points in the same direction as the current. In fact No. The magnetic field of a current circulates around it: around a straight wire it is perpendicular to the wire, and at the centre of a coil it is along the coil's axis, perpendicular to the current.
Electric potential energy HL
Electric potential energy
The electric potential energy E_p of a system of charges is the work done to assemble the system from infinite separation of its components (where E_p = 0). It belongs to the system, not to any one charge. It is positive for like charges (work must be done to push them together) and negative for unlike charges. SI unit: joule (J).
Students often think Electric potential energy, like gravitational potential energy, is negative for any system of bodies brought together from infinity. In fact No. For two like charges E_p = kq₁q₂/r is positive, because they repel and positive work must be done to push them together from infinity. Only attractive (unlike) pairs have negative E_p.
Students often think The electric potential energy is stored in the charge that was moved (or in the smaller charge), not in the system of charges. In fact No. E_p belongs to the system of charges; it depends on their separation and is shared by the pair, whichever charge moved.
Electric potential energy of two point charges HL
Electric potential energy of two point charges
For two point charges q₁ and q₂ a distance r apart, E_p = k q₁q₂/r, with the signs of the charges included. E_p varies as 1/r, not 1/r², and is a scalar.
Students often think Each charge of a pair has its own potential energy, its charge times the potential due to the other charge, so the potential energy of the system is the sum of the two, 2kq₁q₂/r. In fact No. E_p = kq₁q₂/r is the potential energy of the pair (the system) as a whole. It is the work done to assemble the pair once, so it is counted once.
Students often think The potential energy of a bound pair is half of kq₁q₂/r, as in the total energy of an orbiting satellite. In fact No. E_p = kq₁q₂/r. The expression with 2r is the total energy (E_p + E_k) of a charge in a circular orbit, which is half of E_p.
Electric potential as a scalar HL
Electric potential as a scalar
Electric potential is a scalar, defined as zero at infinity. The potential at a point due to several charges is the algebraic sum of the potentials due to each charge, V = Σ kQᵢ/rᵢ, with the sign of each charge included and no directions involved.
Students often think Potentials due to several charges are added as magnitudes, ignoring the signs of the charges. In fact No. Potential is a scalar with a sign: a negative charge contributes a negative potential, which reduces the total.
Students often think The potential of a point charge is kQ/r², falling with the square of the distance like the field strength. In fact No. V = kQ/r falls as 1/r; it is the field strength E = kQ/r² that falls as 1/r².
Electric potential HL
Electric potential
The electric potential V_e at a point is the work done per unit charge to bring a small positive test charge from infinity to that point. For a point charge Q, V_e = kQ/r: positive near a positive charge, negative near a negative charge. SI unit: volt (V = J C⁻¹).
Students often think Electric potential is a vector, pointing away from positive charges and towards negative charges, so potentials are added like fields. In fact No. Potential is a scalar: it has a size and a sign but no direction. Potentials from several charges are added algebraically, not as vectors.
Students often think Electric potential and electric potential energy are the same thing, so the potential at a point is the energy of any charge there, measured in joules. In fact No. The potential is the work done per unit charge (in J C⁻¹ = V). The potential energy of a charge q at that point is qV.
Electric potential gradient HL
Electric potential gradient
The electric field strength equals minus the potential gradient: E = −ΔV_e/Δr. The field points in the direction in which the potential decreases most rapidly, and its magnitude is the rate of change of potential with distance. In a uniform field the potential changes linearly with distance along the field.
Students often think The electric field points in the direction of increasing potential, so the minus sign in E = −ΔV/Δr can be ignored. In fact No. E = −ΔV/Δr: the field points in the direction in which the potential decreases, from high potential to low.
Students often think In a uniform field the potential is uniform too, so it has the same value at every point. In fact No. A uniform field means a constant potential gradient: the potential changes steadily with distance along the field, by E × Δx.
Work done moving a charge HL
Work done moving a charge
The work done by an external force in moving a charge q between two points at constant kinetic energy is W = qΔV_e, where ΔV_e is the change in potential. It depends only on the start and end points, not on the path taken. The work done by the field is −qΔV_e.
Electronvolt
The energy transferred when a charge of e moves through a potential difference of 1 V: 1 eV = 1.60 × 10⁻¹⁹ J. A particle of charge q = ne moving through a potential difference ΔV has work done on it of n × ΔV electronvolts.
Students often think Any charged particle gains one electronvolt of energy for each volt of potential difference it moves through, whatever its charge. In fact No. 1 eV is the energy for a charge of e moving through 1 V. A particle of charge 2e, such as an alpha particle, gains 2 eV per volt.
Students often think The work done on a charge by the external force and the work done by the field are the same thing, so the sign of qΔV can be taken either way. In fact No. When a charge is moved at constant speed, the external force and the electric force are equal and opposite, so the work done by the external force is W = qΔV_e and the work done by the field is −qΔV_e.
Equipotential surface HL
Equipotential surface
A surface on which every point is at the same electric potential. Around a point charge they are concentric spheres, closer together near the charge for equal steps of potential; between parallel plates they are equally spaced planes parallel to the plates; the whole of a charged conductor (solid or hollow), including the space inside a hollow one, is at a single potential equal to that at its surface. Around a collection of point charges, the equipotentials close to each charge are nearly spherical surfaces around that charge; further out they merge into surfaces that enclose several charges, and far away they approach spheres around the whole collection. Outside a charged conducting sphere, solid or hollow, they are concentric spheres, as for a point charge at its centre.
Students often think Equipotential surfaces drawn at equal steps of potential are equally spaced in every field, as they are between parallel plates. In fact No. V = kQ/r changes rapidly near the charge and slowly far away, so equal potential steps give spheres that are close together near the charge and further apart further out.
Students often think Equipotential surfaces run along the field lines (or are another name for them), so the field acts along an equipotential. In fact No. Equipotentials are everywhere perpendicular to the field lines. Along an equipotential the potential is constant, so there is no component of the field along it.
Equipotentials and field lines HL
Equipotentials and field lines
Field lines cross equipotential surfaces at right angles and point from higher to lower potential. Because the field is perpendicular to an equipotential, no work is done in moving a charge along one. Where equipotentials (at equal potential steps) are closer together, the field is stronger.
Students often think Because the potential does not change on an equipotential surface, there is no electric field there, so no force acts on a charge on it. In fact No. The field is not zero on an equipotential; it is perpendicular to the surface. What is zero is the component of the field along the surface.
Students often think Whenever a force acts on a moving object it does work on it, whatever the angle between the force and the motion. In fact No. Work is done only by the component of the force along the displacement: W = Fs cos θ. A force at right angles to the motion does no work.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 Which statement about electric charges and the forces they exert is correct?
Answer and reasoning
Unlike charges attract each other, but the larger charge exerts the larger force. — A student who thinks the 'stronger' charge pulls harder picks this. F = kq₁q₂/r² depends on the product of the charges, which is the same for both; by Newton's third law the forces are equal and opposite.
Unlike charges attract, and the two forces on them are equal in size. — Two types of charge: like repel, unlike attract. The force of each charge on the other forms a Newton's third law pair, so the forces are equal in size and opposite in direction, whatever the sizes of the charges.
A charged object exerts no force at all on any uncharged object near it. — A student who thinks attraction needs two opposite charges picks this. A charged object induces charge separation in a nearby uncharged object; the nearer, opposite charge is attracted more strongly than the further one is repelled, so there is a net attraction.
A stationary positive charge attracts a magnet's north pole and repels its south pole. — A student who treats charges and magnetic poles as the same kind of thing picks this. A stationary charge exerts no magnetic force on either pole; magnetic forces act on moving charges and currents. Any small attraction of the magnet is by induction and acts on both ends alike.
2 Two small spheres, treated as point charges, carry charges of +4.0 nC and +6.0 nC. They are 2.0 cm apart, immersed in an oil of permittivity ε = 4.2 × 10⁻¹¹ F m⁻¹. Take ε₀ = 8.85 × 10⁻¹² F m⁻¹. What is the magnitude of the electric force between them?
Answer and reasoning
5.4 × 10⁻⁴ N — A student who uses the vacuum value ε₀ (k = 8.99 × 10⁹ N m² C⁻²) whatever the medium gets 5.4 × 10⁻⁴ N, the force in a vacuum. In the oil ε is 4.7 times larger, so the force is 4.7 times smaller.
2.6 × 10⁻³ N — A student who thinks a higher permittivity 'permits' a larger force multiplies the vacuum force by ε/ε₀ = 4.75 and gets 2.6 × 10⁻³ N. Permittivity is in the denominator of Coulomb's law: the force is divided by 4.75, not multiplied.
1.1 × 10⁻⁴ N — In the oil, k = 1/4πε with ε = 4.2 × 10⁻¹¹ F m⁻¹: F = q₁q₂/4πεr² = (4.0 × 10⁻⁹)(6.0 × 10⁻⁹)/(4π × 4.2 × 10⁻¹¹ × (0.020)²) = 1.1 × 10⁻⁴ N.
1.4 × 10⁻³ N — A student who writes k = 1/ε, dropping the 4π, gets q₁q₂/εr² = 1.4 × 10⁻³ N. The Coulomb constant in a medium is k = 1/4πε, so this answer is 4π times too large.
Working In the oil, F = q₁q₂/(4πεr²) with ε = 4.2 × 10⁻¹¹ F m⁻¹ and r = 2.0 cm = 0.020 m. F = (4.0 × 10⁻⁹ C)(6.0 × 10⁻⁹ C)/(4π × 4.2 × 10⁻¹¹ F m⁻¹ × (0.020 m)²) = 2.4 × 10⁻¹⁷/2.11 × 10⁻¹³ = 1.1 × 10⁻⁴ N (repulsive). For comparison, in a vacuum (ε₀ = 8.85 × 10⁻¹² F m⁻¹) the force would be 5.4 × 10⁻⁴ N.
3 In an experiment of Millikan's type, an oil drop of weight 4.8 × 10⁻¹⁴ N is held stationary between two horizontal parallel plates 5.0 mm apart, with a potential difference of 500 V across them. The field between the plates is uniform. Take e = 1.60 × 10⁻¹⁹ C. What is the charge on the drop?
Answer and reasoning
9.6 × 10⁻¹⁷ C — Using F = qV, q = W/V = 4.8 × 10⁻¹⁴/500 = 9.6 × 10⁻¹⁷ C. The electric force is F = qE, and E = V/d, not V: the drop's charge is 4.8 × 10⁻¹⁹ C.
4.8 × 10⁻¹⁹ C — E = V/d = 500/0.0050 = 1.0 × 10⁵ V m⁻¹. For the drop to be stationary, qE = W, so q = 4.8 × 10⁻¹⁴/1.0 × 10⁵ = 4.8 × 10⁻¹⁹ C. This is exactly 3 × 1.60 × 10⁻¹⁹ C: a whole-number multiple of e, as quantization of charge requires.
4.8 × 10⁻¹⁶ C — Using d = 5.0 (millimetres not converted), E = 100 V m⁻¹ and q = 4.8 × 10⁻¹⁶ C. Convert to metres: d = 5.0 × 10⁻³ m gives E = 1.0 × 10⁵ V m⁻¹ and q = 4.8 × 10⁻¹⁹ C.
1.9 × 10⁻¹⁴ C — Using E = Vd = 500 × 0.0050 = 2.5 V m⁻¹ gives q = 1.9 × 10⁻¹⁴ C. The field between plates is E = V/d, larger when the plates are closer, so q = 4.8 × 10⁻¹⁹ C.
Working E = V/d = 500 V / (5.0 × 10⁻³ m) = 1.0 × 10⁵ V m⁻¹. Stationary drop: qE = W, so q = W/E = (4.8 × 10⁻¹⁴ N)/(1.0 × 10⁵ V m⁻¹) = 4.8 × 10⁻¹⁹ C. Check against quantization: q/e = 4.8 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 3.0, a whole number, so the drop carries 3 excess electrons.
4 Which statement defines the electric field strength at a point?
Answer and reasoning
The force on any charged object that is placed at that point, measured in newtons. — A student who equates field strength with the force picks this. The force depends on the charge placed there; the field strength is the force per unit charge, which does not.
The force per unit charge on any small charge, in the direction of the force on it. — A student who thinks the field points along the force on any charge picks this. For a negative charge the force is opposite to the field; the field direction is defined by the force on a positive charge.
The energy per unit charge transferred to a small charge placed at that point. — A student who confuses field strength with potential difference picks this. Energy per unit charge is measured in volts; field strength is force per unit charge, in N C⁻¹.
The force per unit charge exerted on a small positive test charge placed at that point. — E = F/q, defined using a small positive test charge, so that the test charge does not disturb the source charges and the direction of E is the direction of the force on a positive charge. Unit: N C⁻¹.
5 The field lines of an isolated negative point charge spread out in three dimensions, and the same number of lines passes through every sphere centred on the charge. Point P is 2.0 cm from the charge and point Q is 6.0 cm from it. How does the electric field strength at P compare with that at Q?
Answer and reasoning
E at P is 3 times E at Q, as P is 3 times closer, so the lines are 3 times as close. — A student who reads spacing from a flat sketch, or assumes E ∝ 1/r, picks this. In three dimensions the lines spread over an area ∝ r², so the density changes by 3² = 9.
E is the same at P and Q, as each line that passes P's sphere also passes Q's sphere. — A student who thinks a field line carries the same strength along its length picks this. The same lines spread over a larger area further out, so they are less dense and the field is weaker at Q.
E at P is 9 times E at Q, as there are 9 times as many lines per unit area at P. — The same lines pass through spheres of area 4πr². Q's sphere has 3² = 9 times the area of P's, so the line density, and hence E, is 9 times greater at P, in agreement with E = kQ/r².
E cannot be compared, as the number of field lines drawn is merely an arbitrary choice. — A student who thinks line density is a mere drawing convention picks this. Whatever number is chosen, it is fixed within the picture, so the relative density at P and Q compares the field strengths.
Working The same number N of lines crosses spheres of area 4πr². Density at P ∝ N/(4π × 2.0²), at Q ∝ N/(4π × 6.0²). Ratio P : Q = 6.0²/2.0² = 9. Since E ∝ line density, E_P = 9E_Q, consistent with E = kQ/r².
6 A long air-core solenoid carries a steady current. Which description of its magnetic field lines is correct?
Answer and reasoning
Almost no field inside; outside, lines loop from one end to the other, as for a bar magnet. — A student who knows only the outside pattern of a bar magnet picks this. The lines continue through the inside of the solenoid, where the field is strongest.
Strongest inside at the two ends, where the lines crowd together, and weakest at the centre. — A student who thinks the field is strongest at the poles picks this. Inside a long solenoid the field is nearly uniform over most of the length and falls near the ends, where the lines spread out.
Lines start at the north-pole end and stop at the south-pole end, so none of them form closed loops. — A student who treats magnetic field lines like electric ones, beginning and ending on something, picks this. There are no isolated poles: every magnetic field line is a closed loop.
Parallel, evenly spaced lines inside; outside, like a bar magnet's; each is a closed loop. — Inside a long solenoid the field is strong and nearly uniform along the axis. Outside, the pattern resembles a bar magnet's. Every line is a closed loop, returning through the inside of the solenoid.
7 Two small spheres, each carrying a positive charge, are brought from infinite separation to a separation r. Which statement about the electric potential energy E_p of the system is correct? HL
Answer and reasoning
It is positive, since positive work must be done to push the like charges together. — E_p is the work done to assemble the system from infinite separation, where E_p = 0. Like charges repel, so an external force must do positive work to bring them together: E_p = kq₁q₂/r > 0.
It is negative, as it is for two masses, since E_p is zero at infinite separation. — A student who carries over the result for gravity picks this. Gravitational E_p is negative because masses attract; like charges repel, so the work to assemble them, and E_p, is positive.
It is stored in the sphere that was moved, not in the system of the two spheres. — A student who assigns potential energy to one object picks this. E_p = kq₁q₂/r depends on both charges and their separation; it belongs to the system.
It is positive, and it becomes larger if the spheres are moved further apart. — A student who carries over 'higher means more E_p' from mgh picks this. For like charges E_p = kq₁q₂/r falls as r increases: they repel, so moving apart releases energy.
8 Four point charges are fixed at the corners of a square of side 0.20 m. Three of the charges are +2.0 nC and the fourth is −2.0 nC. Electric potential is zero at infinity. Take k = 8.99 × 10⁹ N m² C⁻². What is the electric potential at the centre of the square? HL
Answer and reasoning
5.1 × 10² V — A student who adds the potentials as magnitudes, ignoring the negative charge's sign, uses 8.0 nC and gets 5.1 × 10² V. A scalar still has a sign: the −2.0 nC charge contributes a negative potential.
1.3 × 10² V — A student who uses the whole diagonal, 0.283 m, as the distance gets 1.3 × 10² V. The centre is half a diagonal from each corner, 0.141 m.
1.8 × 10³ V — A student who uses kQ/r², as for field strength, gets 1.80 × 10³ (which is not in volts). Potential falls as 1/r: V = kQ/r.
2.5 × 10² V — Each corner is a/√2 = 0.141 m from the centre. Potential is a scalar: V = kΣq/r = 8.99 × 10⁹ × (2.0 + 2.0 + 2.0 − 2.0) × 10⁻⁹/0.141 = 2.5 × 10² V.
Working Distance from each corner to the centre r = (0.20 m)/√2 = 0.141 m. Net charge Σq = 3(+2.0 nC) + (−2.0 nC) = +4.0 nC. V = kΣq/r = (8.99 × 10⁹ N m² C⁻²)(4.0 × 10⁻⁹ C)/(0.141 m) = 35.96/0.1414 = 2.5 × 10² V.
9 In a region of uniform electric field of strength 4.0 × 10³ V m⁻¹ directed in the +x direction, point A at x = 0 has an electric potential of +500 V. What is the electric potential at point B, at x = 0.050 m? HL
Answer and reasoning
+300 V — E = −ΔV/Δx, so ΔV = −EΔx = −(4.0 × 10³)(0.050) = −200 V: potential falls along the field. V_B = 500 V − 200 V = +300 V.
+700 V — A student who thinks the field points towards higher potential adds 200 V. The field points from high to low potential, so moving along it lowers V.
−200 V — A student who takes the change in potential, ΔV = −EΔx = −200 V, as the potential at B picks this. That is only the change from A to B; the starting value must be included: V_B = 500 V − 200 V = +300 V.
+500 V — A student who thinks a uniform field means a uniform potential picks this. A uniform field is a constant potential gradient: V changes by 4.0 × 10³ V for each metre along the field.
Working E = −ΔV/Δx ⇒ ΔV = −EΔx = −(4.0 × 10³ V m⁻¹)(0.050 m) = −200 V. V_B = V_A + ΔV = +500 V − 200 V = +300 V.
10 Equipotential surfaces are drawn at equal steps of potential. Which description is correct for the arrangement named? HL
Answer and reasoning
Between oppositely charged parallel plates: evenly spaced planes parallel to the plates. — Between the plates (away from the edges) E is uniform, so V changes by equal amounts in equal distances: the equipotentials are equally spaced planes parallel to the plates, perpendicular to the field lines.
Around an isolated point charge: concentric spheres, evenly spaced from the charge outwards. — A student who carries over the parallel-plate pattern picks this. V = kQ/r changes fastest near the charge, so equal steps of V give spheres that are close together near the charge and further apart further out.
Inside a charged hollow metal sphere: the potential is zero everywhere, as the field is zero. — A student who links zero field to zero potential picks this. With no field inside, no work is done moving a charge inside, so the whole interior is at the same potential as the surface, kQ/R, which is not zero.
Near two equal positive point charges: curves that follow the field lines out from the charges. — A student who confuses equipotentials with field lines picks this. Equipotentials cross the field lines at right angles; near each charge they are closed surfaces around it.
Read the ones marked not yet in Learn, then Verify.
Verify confirm before you go
37 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 Two small insulating spheres, each of diameter 2.0 cm, carry charges of +10 nC and +15 nC spread uniformly over them. They are in a vacuum with a gap of 3.0 cm between their surfaces. Take k = 8.99 × 10⁹ N m² C⁻². What is the magnitude of the electric force between them?
Answer and reasoning
5.4 × 10⁻⁴ N — Each uniformly charged sphere acts as a point charge at its centre, so r = 1.0 cm + 3.0 cm + 1.0 cm = 5.0 cm = 0.050 m. F = kq₁q₂/r² = 8.99 × 10⁹ × 10 × 10⁻⁹ × 15 × 10⁻⁹/(0.050)² = 5.4 × 10⁻⁴ N.
1.5 × 10⁻³ N — A student who uses the gap between the surfaces, 0.030 m, as r gets 1.5 × 10⁻³ N. The charges act as if at the centres, which are 0.050 m apart.
2.8 × 10⁻⁴ N — A student who adds the full diameters to the gap, r = 2.0 + 3.0 + 2.0 = 7.0 cm, gets 2.8 × 10⁻⁴ N. From each surface to its centre is the radius, 1.0 cm, not the diameter.
2.7 × 10⁻⁵ N — A student who divides by r instead of r² gets kq₁q₂/r = 2.70 × 10⁻⁵ (which is not even in newtons). Coulomb's law is an inverse-square law: divide by (0.050 m)².
Working Uniformly charged spheres act as point charges at their centres: r = gap + two radii = 3.0 cm + 1.0 cm + 1.0 cm = 5.0 cm = 0.050 m. F = kq₁q₂/r² = (8.99 × 10⁹ N m² C⁻²)(10 × 10⁻⁹ C)(15 × 10⁻⁹ C)/(0.050 m)² = 1.349 × 10⁻⁶/2.5 × 10⁻³ = 5.4 × 10⁻⁴ N (repulsive).
2 Two identical small metal spheres, X and Y, stand on insulating bases. X carries a charge of +5.0 nC and Y carries −3.0 nC. The spheres are touched together and then separated. What is the charge on X after they are separated?
Answer and reasoning
4.0 nC — A student who adds the sizes of the charges without their signs gets 8.0 nC in total and 4.0 nC each. Charge is signed: −3.0 nC cancels 3.0 nC of the +5.0 nC, leaving +2.0 nC to share.
0.0 nC — A student who thinks opposite charges neutralize each other completely picks this. Only 3.0 nC of each sign can cancel; the remaining +2.0 nC is conserved and shared, so neither sphere is left uncharged.
5.0 nC — A student who thinks static charge stays where it was put picks this. The spheres are metal: free electrons flow from Y to X when they touch, until the charge is shared equally.
1.0 nC — Charge is conserved: the total is +5.0 nC + (−3.0 nC) = +2.0 nC. Electrons flow between the touching conductors until the identical spheres share this equally, so X is left with +1.0 nC (and Y with +1.0 nC).
Working Total charge before = +5.0 nC + (−3.0 nC) = +2.0 nC. Charge is conserved, and identical conducting spheres share it equally on contact: each is left with +2.0 nC/2 = +1.0 nC. (4.0 nC of electrons, i.e. 2.5 × 10¹⁰ electrons, flow from Y to X.)
3 In an experiment of Millikan's type, the charges on four oil drops were measured in arbitrary units as 7.50, 12.5, 17.5 and 22.5 units. Assuming that electric charge is quantized, what is the largest value of the basic unit of charge that is consistent with all four results?
Answer and reasoning
7.50 units — A student who takes the smallest measured charge as the unit picks this. 12.5 units is 1.67 times 7.50 units, not a whole number, so the smallest drop must itself carry several units.
2.50 units — The basic unit must divide every charge a whole number of times. 7.50, 12.5, 17.5 and 22.5 are 3, 5, 7 and 9 times 2.50, and no larger value divides all four: 5.00 fails for 7.50 (1.5 units) and 7.50 fails for 12.5.
15.0 units — A student who averages the four charges gets 15.0 units. The drops carry different numbers of units, so their mean is not a unit of charge; 7.50 units is less than it, which a unit could not allow.
5.00 units — A student who takes the constant step between the sorted values as the unit gets 5.00 units. But 7.50 units is 1.5 times 5.00 units, so 5.00 cannot be the unit; every charge must be a whole-number multiple.
Working Test candidate units by dividing each charge by them. Unit 2.50: 7.50/2.50 = 3, 12.5/2.50 = 5, 17.5/2.50 = 7, 22.5/2.50 = 9, all whole numbers. Unit 5.00: 7.50/5.00 = 1.5, not whole. Unit 7.50: 12.5/7.50 = 1.67, not whole. The largest consistent unit is 2.50 units.
4 A polythene rod becomes negatively charged when it is rubbed with a woollen cloth. Which statement explains how the rod becomes charged?
Answer and reasoning
Rubbing creates new negative charge in the surface layer of the rod. — A student who thinks friction generates charge, as it generates heat, picks this. Charge is never created: the cloth ends up with a positive charge exactly equal in size to the rod's negative charge.
Protons move from the rod to the cloth, leaving the rod with a negative charge. — A student who thinks positive charges move during charging picks this. In a solid the protons are locked in the nuclei; only electrons are transferred.
Electrons move from the cloth to the rod, so the rod gains extra electrons. — Rubbing transfers electrons between the surfaces. The rod gains electrons (an excess makes it negative) and the cloth loses the same number, becoming equally positive: charge is conserved.
Electrons move from the rod to the cloth, as negative means a lack of electrons. — A student who reads 'negative' as 'a shortage' picks this. Electrons carry negative charge, so a negative rod has more electrons than normal: they moved onto it, not off it.
5 An uncharged metal sphere stands on an insulating base. A negatively charged rod is held near the sphere, without touching it. While the rod is held there, the sphere is connected to earth by a wire. The wire is then removed, and after that the rod is taken away. Which statement describes the electron flow in the wire and the final charge on the sphere?
Answer and reasoning
Electrons flowed from the sphere to earth, and the sphere is left positively charged. — The negative rod repels free electrons in the sphere; with the earth wire connected they flow away to earth. Breaking the connection before removing the rod traps the deficit, so the sphere is left positive (charging by induction).
Electrons flowed from earth to the sphere, and it is left negatively charged. — A student who thinks the Earth always supplies electrons picks this. The nearby negative rod repels electrons, so they are pushed out of the sphere into the earth, not drawn in.
Electrons flowed from the sphere to earth, and the sphere is left with no charge. — A student who thinks earthing always leaves an object uncharged picks this. Electrons left the sphere and could not return once the wire was removed; by conservation of charge the sphere must now be positive.
Electrons crossed from the rod to the sphere, and the sphere is left negatively charged. — A student who thinks induction transfers charge from the rod picks this. The rod never touches the sphere and no charge crosses the gap; induction only redistributes the sphere's own electrons.
6 At point P, a charge of +2.0 µC experiences an electric force of 5.0 × 10⁻³ N directed east. This charge is removed and a charge of −6.0 µC is placed at P; the charges that produce the field do not move. Taking east as positive, what is the electric force on the −6.0 µC charge?
Answer and reasoning
+1.5 × 10⁻² N — A student who thinks every charge is pushed along the field picks this. The field is east, but the charge is negative, so F = qE points west: the sign is negative.
−5.0 × 10⁻³ N — A student who thinks the force at a point is fixed reverses its direction for the negative charge but keeps its size. The field is fixed; the force is proportional to the charge, so it is three times larger.
−1.5 × 10⁻² N — E at P = F/q = 5.0 × 10⁻³ N/2.0 × 10⁻⁶ C = 2.5 × 10³ N C⁻¹ east, and it does not change when the test charge is replaced. F = qE = (−6.0 × 10⁻⁶)(+2.5 × 10³) = −1.5 × 10⁻² N, i.e. 1.5 × 10⁻² N west.
−3.0 × 10⁻⁸ N — A student who treats the field strength as the force takes E = 5.0 × 10⁻³ (the force on the first charge) and then uses F = qE = (−6.0 × 10⁻⁶)(5.0 × 10⁻³) = −3.0 × 10⁻⁸ N. The field strength is force per unit charge: E = 5.0 × 10⁻³ N/2.0 × 10⁻⁶ C = 2.5 × 10³ N C⁻¹.
Working E = F/q = (5.0 × 10⁻³ N)/(2.0 × 10⁻⁶ C) = 2.5 × 10³ N C⁻¹, directed east (the direction of the force on the positive charge). E at P is unchanged when the test charge is replaced. F = qE = (−6.0 × 10⁻⁶ C)(+2.5 × 10³ N C⁻¹) = −1.5 × 10⁻² N: 1.5 × 10⁻² N directed west.
7 Two point charges of +4.0 nC and +1.0 nC are fixed 0.30 m apart. At one point on the line between them the resultant electric field strength is zero. How far is this point from the +4.0 nC charge?
Answer and reasoning
0.15 m — A student who puts the neutral point of any two like charges at the midpoint picks this. At 0.15 m the +4.0 nC charge's field is four times the other's; the fields balance only nearer the smaller charge.
0.24 m — A student who takes the fields as proportional to 1/r solves 4/x = 1/(0.30 − x) and gets 0.24 m. The fields vary as 1/r², so the distance ratio is √4 = 2, not 4.
0.06 m — A student who treats the neutral point like a centre of mass, nearer the larger charge, divides 0.30 m in the ratio 1 : 4 from the +4.0 nC charge. The larger charge's field is stronger, so the balance point must be further from it, not nearer.
0.20 m — At the neutral point the two fields are equal and opposite: k(4.0 nC)/x² = k(1.0 nC)/(0.30 − x)². Taking square roots, x/(0.30 − x) = 2, so x = 0.20 m from the +4.0 nC charge (0.10 m from the +1.0 nC charge).
Working Between two like charges the fields are opposite in direction, so E = 0 where kQ₁/x² = kQ₂/(d − x)². With Q₁ = 4.0 nC, Q₂ = 1.0 nC and d = 0.30 m: x/(0.30 m − x) = √(Q₁/Q₂) = √4 = 2 ⇒ x = 2(0.30 m − x) ⇒ 3x = 0.60 m ⇒ x = 0.20 m from the +4.0 nC charge.
8 A student sketches the electric field lines around two equal positive point charges a short distance apart. Which description of a correct sketch is right?
Answer and reasoning
Lines leave each charge and curve away from the other, with a neutral point at the midpoint. — Like charges: every line starts on a positive charge and goes out to infinity. Lines between the charges curve away from the other charge, and the midpoint is a neutral point where the two fields cancel (E = 0), so no line crosses it.
Lines leave each charge, and some of them cross one another in the region midway between them. — A student who overlays each charge's field picks this. The resultant field has one direction at each point, so field lines can never cross.
Lines run from one charge to the other, joining the two charges across the gap. — A student who generalizes the pattern for opposite charges picks this. Lines end only on negative charges; with two positive charges no line can end on the other charge.
Straight radial lines leave each charge, exactly as if the other charge were absent. — A student who does not add the two fields picks this. At each point the field is the vector sum of both, so the lines near the charges are curved, not straight and radial.
9 A hollow metal sphere on an insulating stand is given a positive charge. Which description of the electric field lines is correct?
Answer and reasoning
Lines inside as well as outside, since the charge is spread through the whole sphere. — A student who pictures the charge spread through the sphere picks this. In a conductor the charge moves to the outer surface until the field inside is zero.
None inside; outside, radial lines pointing away, as if from a point charge at the centre. — The charge sits on the outer surface of the conductor and the field inside is zero. Outside, the field is that of a point charge at the centre: radial lines leaving the surface at right angles, spreading out with distance.
Lines leave the surface both ways, pointing inwards and outwards from the charge. — A student who treats each bit of surface charge as an isolated point charge picks this. Inside, the fields of all the surface charges cancel exactly, so no lines point inwards.
None inside; outside, parallel and equally spaced lines, as the charge is spread evenly. — A student who links an evenly spread charge with a uniform field picks this. Outside a sphere the lines are radial and spread out, so the field weakens as 1/r²; parallel lines belong to flat plates.
10 Which description of the electric field lines of two oppositely charged parallel plates is correct?
Answer and reasoning
Parallel and evenly spaced from the positive to the negative plate, stopping sharply at the plate edges. — A student who ignores edge effects picks this. The lines cannot simply stop at the edges: they curve outwards there, and the field is non-uniform near the edges.
From the positive to the negative plate, closest near each plate and widest apart midway. — A student who applies point-charge thinking to the plates picks this. Away from the edges the field between large plates is uniform: the lines are equally spaced right across the gap.
Closed loops that leave the positive plate and curve round to return to it, as magnetic field lines do. — A student who carries over the closed loops of magnetic fields picks this. Electric field lines start on positive charge and end on negative charge; they do not return to where they began.
Parallel and evenly spaced from the positive to the negative plate, bulging outwards at the edges. — In the central region the field is uniform: parallel, equally spaced lines from + to −. Near the edges the lines curve outwards beyond the plates (the edge effect), so the field there is not uniform.
11 Two parallel metal plates 5.0 mm apart in a vacuum are connected to a 1500 V supply. An electron is between the plates, well away from their edges. Take e = 1.60 × 10⁻¹⁹ C. What is the magnitude of the electric force on the electron?
Answer and reasoning
4.8 × 10⁻¹⁴ N — E = V/d = 1500 V/(5.0 × 10⁻³ m) = 3.0 × 10⁵ V m⁻¹. F = eE = 1.60 × 10⁻¹⁹ C × 3.0 × 10⁵ V m⁻¹ = 4.8 × 10⁻¹⁴ N.
4.8 × 10⁻¹⁷ N — A student who substitutes d = 5.0 without converting millimetres to metres gets E = 300 V m⁻¹ and F = 4.8 × 10⁻¹⁷ N. d must be 5.0 × 10⁻³ m.
1.2 × 10⁻¹⁸ N — A student who multiplies V by d gets E = 7.5 and F = 1.2 × 10⁻¹⁸ N. The field is the potential difference per unit distance, E = V/d, in V m⁻¹.
2.4 × 10⁻¹⁶ N — A student who uses F = qV gets 2.4 × 10⁻¹⁶, but qV is the energy (in J) an electron gains crossing the gap, not the force on it. F = qE = qV/d.
Working E = V/d = 1500 V/(5.0 × 10⁻³ m) = 3.0 × 10⁵ V m⁻¹, uniform between the plates away from the edges. F = eE = (1.60 × 10⁻¹⁹ C)(3.0 × 10⁵ V m⁻¹) = 4.8 × 10⁻¹⁴ N, directed towards the positive plate.
12 Two parallel plates remain connected to the same supply while their separation is doubled. A small charged dust particle is held between the plates, well away from their edges. How does the electric force on the particle change?
Answer and reasoning
It is unchanged, since E depends only on V, which the supply fixes. — A student who thinks the field is set by the potential difference alone picks this. E = V/d: the same potential difference across twice the separation gives half the field, so E and F halve.
It halves, since E = V/d and d has doubled while V is unchanged. — The supply keeps V fixed. Between the plates E = V/d, so doubling d halves E, and F = qE halves.
It falls to a quarter, since E ∝ 1/d², like the field of a point charge. — A student who applies the inverse-square law to every field picks this. Only radial fields fall as 1/r²; between plates E = V/d, so doubling d halves E.
It doubles, since E = Vd and d has doubled at constant V. — A student who multiplies V by d picks this. E = V/d (volts per metre): a wider gap at the same potential difference gives a weaker field.
13 A long, straight, vertical wire carries a steady current upwards. Point P is 5.0 cm due east of the wire, level with it. What is the direction of the magnetic field of the wire at P?
Answer and reasoning
Due south, at a tangent to a circle around the wire — A student who uses the left hand, or curls the fingers the wrong way, gets the reversed direction. With the right thumb upwards the field circulates anticlockwise seen from above, which is north at P.
Due east, directly away from the wire along a radius — A student who expects a radial pattern, like the electric field of a charged wire, picks this. The magnetic field lines of a straight wire are circles around it, so the field at P is tangential, not radial.
Due north, at a tangent to a circle centred on the wire — Right-hand grip rule: thumb upwards along the current, fingers curl anticlockwise when seen from above. The field lines are circles round the wire; at a point east of the wire, anticlockwise means due north.
Vertically upwards, parallel to the direction of the current — A student who thinks the field points along the current picks this. The field of a straight current circulates around the wire, perpendicular to it.
14 A flat horizontal circular coil carries a steady current that flows anticlockwise when viewed from above. What is the direction of the magnetic field at the centre of the coil?
Answer and reasoning
There is no field: the fields of opposite sides cancel out. — A student who thinks opposite currents give cancelling fields picks this. The centre is on opposite sides of the two currents, so the grip rule gives the same direction for both: the fields add, upwards along the axis.
Vertically upwards, along the central axis of the coil. — Apply the right-hand grip rule to any part of the coil: with the thumb along the anticlockwise current, the curled fingers pass upwards through the inside of the loop. Every part gives an upward field at the centre, so the field there is vertically upwards, along the axis.
Horizontal, in the plane of the coil, circling like the current. — A student who thinks the field points along the current picks this. The field is perpendicular to the current in each part of the coil, so at the centre it is along the axis, perpendicular to the plane of the coil.
Vertically downwards, along the central axis of the coil. — A student who uses the left hand, or curls the fingers the wrong way, gets the reversed direction. With the right thumb along the anticlockwise current, the curled fingers pass upwards through the inside of the loop: the field at the centre is upwards.
15 In a simple model of the hydrogen atom, the electron is 5.3 × 10⁻¹¹ m from the proton. Take e = 1.60 × 10⁻¹⁹ C and ε₀ = 8.85 × 10⁻¹² F m⁻¹. What is the electric potential energy of the electron–proton system? HL
Answer and reasoning
−8.7 × 10⁻¹⁸ J — A student who gives each charge its own potential energy and adds the two gets twice E_p. The potential energy belongs to the pair and is the work done to assemble it once: E_p = kq₁q₂/r.
−5.5 × 10⁻¹⁷ J — A student who writes k = 1/ε₀, leaving out the 4π, gets an answer 4π times too large. k = 1/4πε₀ = 8.99 × 10⁹ N m² C⁻².
−4.3 × 10⁻¹⁸ J — E_p = kq₁q₂/r with k = 1/4πε₀ = 8.99 × 10⁹ N m² C⁻²: E_p = 8.99 × 10⁹ × (+1.60 × 10⁻¹⁹)(−1.60 × 10⁻¹⁹)/5.3 × 10⁻¹¹ = −4.3 × 10⁻¹⁸ J (about −27 eV). Negative, because the charges attract.
−2.2 × 10⁻¹⁸ J — A student who uses kq₁q₂/2r, the total energy of a charge in a circular orbit, gets half of E_p. The potential energy alone is kq₁q₂/r.
Working k = 1/4πε₀ = 1/(4π × 8.85 × 10⁻¹² F m⁻¹) = 8.99 × 10⁹ N m² C⁻². E_p = kq₁q₂/r = (8.99 × 10⁹)(+1.60 × 10⁻¹⁹ C)(−1.60 × 10⁻¹⁹ C)/(5.3 × 10⁻¹¹ m) = −2.30 × 10⁻²⁸/5.3 × 10⁻¹¹ = −4.3 × 10⁻¹⁸ J. In electronvolts: −4.3 × 10⁻¹⁸ J/1.60 × 10⁻¹⁹ J eV⁻¹ = −27.1 eV.
16 Point P is a distance r from an isolated point charge −Q. Which statement about the electric potential at P is correct? HL
Answer and reasoning
It is directed towards the charge, with a size equal to the work done per unit charge from infinity to P. — A student who treats potential as a vector picks this. Potential is a scalar: it has a sign (negative here) but no direction.
It is the work done per unit charge to bring a small positive test charge from infinity to P. — V_e is defined as the work done per unit charge bringing a small positive test charge from infinity (V = 0) to P. The charge is attracted towards −Q, so this work is negative: V_e = −kQ/r.
It is the potential energy, measured in joules, of any small charge that happens to be placed at P. — A student who confuses potential with potential energy picks this. V_e is energy per unit charge, in volts; a charge q at P has potential energy qV_e, which depends on q.
It is inversely proportional to r², falling off in the same way as the field strength. — A student who attaches the inverse-square law to potential picks this. V_e = −kQ/r varies as 1/r; it is the field strength that varies as 1/r².
17 Point charges +Q and −Q are fixed a short distance apart, and M is the midpoint between them. A student claims: 'The potential at M is zero, so no work is needed to bring a small test charge from infinity to M, and the test charge feels no electric force at M.' Which evaluation of the claim is correct? HL
Answer and reasoning
The whole claim is right, as the field strength is zero wherever the potential is zero. — A student who links a zero potential to a zero field picks this. E is the potential gradient; at M the potential is zero but still changing with position, so the field (towards −Q) is not zero.
The first part is wrong, as the potentials of the two charges add as vectors at M. — A student who treats potential as a vector picks this. Potential is a scalar: +kQ/r and −kQ/r add algebraically to zero at M.
The second part is wrong, as the work needed depends on the path from infinity to M. — A student who thinks work depends on the route picks this. The electric force is conservative: W = qΔV_e depends only on the potentials at the ends of the path, both zero here.
The first two parts are right, but the field at M is not zero, as V changes with position there. — V_M = kQ/r − kQ/r = 0, and W = qΔV_e = q(0 − 0) = 0, so the first two parts are right. But E = −ΔV/Δr depends on the gradient of V, not its value: along the line V falls from +Q to −Q, so E ≠ 0 at M and a force acts.
18 An alpha particle, of charge +2e, is moved at constant speed from a point where the electric potential is +2000 V to a point where it is +500 V. What is the work done on the alpha particle by the external force that moves it? HL
Answer and reasoning
+3.0 keV — A student who uses the work done by the field gets +3.0 keV. The field does −qΔV_e = +3.0 keV; the external force, acting against the field at constant speed, does +qΔV_e = −3.0 keV.
−3.0 keV — W = qΔV_e = (2e)(500 V − 2000 V) = 2e × (−1500 V) = −3000 eV = −3.0 keV (−4.8 × 10⁻¹⁶ J). The field pushes the positive particle towards lower potential, so the external force must hold it back to keep its speed constant: its work is negative.
−1.5 keV — A student who takes 1 eV per volt for any particle gets −1500 eV. The alpha particle's charge is 2e, so each volt corresponds to 2 eV: −3000 eV.
+1.0 keV — A student who uses the final potential alone, 2e × 500 V, gets +1.0 keV. Work depends on the change in potential, ΔV_e = 500 V − 2000 V = −1500 V.
Working ΔV_e = V_final − V_initial = +500 V − (+2000 V) = −1500 V. W = qΔV_e = (+2e)(−1500 V) = −3000 eV = −3.0 keV. In joules: −3000 × 1.60 × 10⁻¹⁹ J = −4.8 × 10⁻¹⁶ J. (The field does +3.0 keV of work on the particle; its kinetic energy is unchanged.)
19 A point charge of +5.0 nC is fixed in place. A second charge of +2.0 nC is moved slowly from a point 0.30 m from the fixed charge to a point 0.10 m from it. Take k = 8.99 × 10⁹ N m² C⁻². What is the work done on the +2.0 nC charge by the external force? HL
Answer and reasoning
+9.0 × 10⁻⁷ J — A student who uses the potential at the final point only, 450 V, gets 9.0 × 10⁻⁷ J. The charge starts at 150 V, not at infinity, so ΔV_e = 300 V.
+8.0 × 10⁻⁶ J — A student who calculates 'potentials' with kQ/r² gets 8.0 × 10⁻⁶ J. Potential falls as 1/r, so V = kQ/r at each point.
+6.0 × 10⁻⁷ J — V at 0.30 m = kQ/r = 150 V; V at 0.10 m = 450 V. W = qΔV_e = 2.0 × 10⁻⁹ C × 300 V = +6.0 × 10⁻⁷ J.
+2.0 × 10⁻⁷ J — A student who multiplies the force at the start by the 0.20 m moved gets 2.0 × 10⁻⁷ J. The force grows as the charges approach (∝ 1/r²), so W = Fd with one value of F is wrong; use W = qΔV_e.
Working V₁ = kQ/r₁ = (8.99 × 10⁹)(5.0 × 10⁻⁹)/0.30 = 149.8 V. V₂ = kQ/r₂ = 44.95/0.10 = 449.5 V. W = qΔV_e = (2.0 × 10⁻⁹ C)(449.5 V − 149.8 V) = (2.0 × 10⁻⁹)(299.7) = 5.99 × 10⁻⁷ J ≈ +6.0 × 10⁻⁷ J.
20 A hollow metal sphere of radius 0.12 m carries a charge of +4.0 nC. Electric potential is zero at infinity. Take k = 8.99 × 10⁹ N m² C⁻². What is the electric potential at the centre of the sphere? HL
Answer and reasoning
Zero, as the field inside the sphere is zero — A student who links a zero field to a zero potential picks this. E = 0 means V does not change inside; it stays at the surface value, kQ/R, not zero.
3.00 × 10² V, the same as on its surface — At the surface V = kQ/R = 8.99 × 10⁹ × 4.0 × 10⁻⁹/0.12 = 3.00 × 10² V. Inside, E = 0, so no work is done moving a charge from the surface to the centre: the whole interior is an equipotential at 3.00 × 10² V.
2.50 × 10³ V, from kQ/R² at its surface — A student who uses kQ/R², the expression for field strength, gets 2.50 × 10³ (which is in N C⁻¹, not V). Potential is kQ/R.
Infinite, since r = 0 at the centre in V = kQ/r — A student who applies the point-charge formula inside the sphere picks this. The charge is on the surface, a distance R from the centre; V = kQ/r holds only outside, and inside V is constant at kQ/R.
Working Surface potential V = kQ/R = (8.99 × 10⁹ N m² C⁻²)(4.0 × 10⁻⁹ C)/(0.12 m) = 299.7 V = 3.00 × 10² V. Inside a conductor E = 0, so ΔV = 0 between the surface and the centre: V(centre) = 3.00 × 10² V.
21 A small positive charge is moved along an equipotential surface in the electric field of a fixed point charge. Which statement is correct? HL
Answer and reasoning
The field is perpendicular to its path, so the field does no work on the charge. — Field lines cross equipotentials at right angles, so the electric force is perpendicular to the displacement: W = Fs cos 90° = 0. Equivalently, W = qΔV_e = 0 because ΔV_e = 0 along the surface.
There is no electric field on the surface, so no force acts on the charge at all. — A student who thinks a constant potential means no field picks this. The field is not zero on an equipotential; it points across the surface, so a force acts, but at right angles to the motion.
The field does work on the charge, as the electric force acts on it throughout. — A student who thinks any force on a moving object does work picks this. Work is Fs cos θ; with the force perpendicular to the motion, θ = 90° and no work is done.
The field acts along the surface, in the same direction as the charge moves. — A student who thinks field lines run along equipotentials picks this. Along an equipotential ΔV = 0, so the field has no component along it: the field is perpendicular to the surface.
22 Point charges +Q and −Q are fixed a distance 2d apart. A small positively charged ball is released from rest at the midpoint between them. Which statement about the electric force on the ball at the moment of release is correct?
Answer and reasoning
It is zero, since the ball is the same distance from the two fixed charges and so their forces on it cancel out. — Equal distances give equal sizes, but the directions are fixed by the signs. Away from +Q and towards −Q are the same direction here, so the forces add rather than cancel.
It is zero, since the fixed charges add up to no net charge and so together they exert no force at all on the ball. — Each fixed charge exerts its own force on the ball. Net charge zero does not mean net force zero; the two forces here point the same way and add.
It is directed towards −Q, since it is pushed away from +Q and pulled towards −Q, both forces acting one way. — Like charges repel, so +Q pushes the ball away from itself, towards −Q. Unlike charges attract, so −Q pulls the ball towards itself. The two forces are equal in size and in the same direction, so the resultant is towards −Q.
It is directed towards +Q, since it is pulled towards +Q and pushed away from −Q, both forces acting the same way in that direction. — The rule is the other way round. Like charges repel and unlike charges attract, so the ball is pushed away from +Q and pulled towards −Q.
23 A polythene rod is rubbed with a dry woollen cloth and gains a charge of −8.0 nC. Which statement about the cloth after the rubbing is correct?
Answer and reasoning
Its charge is −8.0 nC, because rubbing charges both surfaces in the same way. — Rubbing does not generate charge on each surface; it transfers electrons from one to the other. The total charge of rod plus cloth is still zero, so the cloth must be +8.0 nC.
Its charge is +8.0 nC, as the electrons the rod gained came from the cloth. — Charge is conserved. The rod became negative by gaining electrons, and the only place they came from is the cloth, which is left short of the same number of electrons: +8.0 nC.
It is uncharged, because the friction created the charge on the rod alone. — Friction creates no charge. The rod's −8.0 nC is made of electrons that existed before, on the cloth; conservation of charge requires the cloth to carry +8.0 nC.
Its charge is +8.0 nC, because protons moved from the rod to the cloth. — Protons are bound in nuclei and do not move between solids. The cloth is +8.0 nC because it lost electrons to the rod, not because it gained protons.
Working Conservation of charge: total charge of rod + cloth before rubbing = 0, and rubbing only transfers electrons. Rod: −8.0 nC (gained electrons), so cloth: 0 − (−8.0 nC) = +8.0 nC (lost the same electrons).
24 A sketch shows the electric field lines near a charged object. A small positive test charge is placed at a point P that lies midway between two neighbouring field lines, in a region where the lines are closely spaced. Which statement is correct?
Answer and reasoning
No force acts on it at P, since the electric field exists only along the lines that are drawn. — The lines are a representation, not the field itself. Infinitely many lines could be drawn; the field, and the force on the charge, exist at P just as on a line.
A force acts on it at P, but its size cannot be judged, as the number of lines drawn is a matter of choice. — The number drawn is a choice, but once chosen, the lines are spaced in proportion to the field strength. Close spacing at P means a relatively strong field there.
A force acts on it at P, of the same size as anywhere between those two lines, as the field is constant along a line. — The field is not constant along a line. Where the two lines spread apart further along, the field between them is weaker; the local spacing gives the local strength.
A force acts on it at P, and the close spacing of the lines shows that the field there is relatively strong. — The field fills the whole region; the drawn lines are a sample of it. At P the force is along the local field direction, and its size is indicated by how closely the lines are spaced there.
25 Three point charges, each of +2.0 nC, are brought from infinite separation and fixed at the corners of an equilateral triangle of side 0.10 m. Take k = 8.99 × 10⁹ N m² C⁻². What is the total work done by the external forces to assemble the system? HL
Answer and reasoning
2.2 × 10⁻⁶ J — Counting each pair twice (an energy 'for each charge' of the pair) doubles the answer. Each pair shares one potential energy kq₁q₂/r; there are three pairs, giving 1.1 × 10⁻⁶ J.
7.2 × 10⁻⁷ J — This is only the work done to bring in the third charge (two pairs). The energy belongs to the whole system: the second charge also needed work against the first, making three pairs in all.
1.1 × 10⁻⁶ J — Bringing in the first charge needs no work. The second needs kq²/a = 3.6 × 10⁻⁷ J against the first. The third needs kq²/a against each of the two already there, 7.2 × 10⁻⁷ J. Total: 3 pairs × 3.6 × 10⁻⁷ J = 1.1 × 10⁻⁶ J.
1.1 × 10⁻⁵ J — Using kq²/a² for each pair gives 1.1 × 10⁻⁵ J. Potential energy falls as 1/r, not 1/r²: E_p = kq₁q₂/r for each pair.
Working The work done to assemble the system is the electric potential energy of the system: the sum over every pair, E_p = Σ kq₁q₂/r. Three pairs, each with kq²/a = (8.99 × 10⁹)(2.0 × 10⁻⁹)²/0.10 = 3.60 × 10⁻⁷ J. Total W = 3 × 3.60 × 10⁻⁷ = 1.08 × 10⁻⁶ J ≈ 1.1 × 10⁻⁶ J (positive: work must be done to push like charges together).
26 Two point charges, +q and −q, are held a distance r apart, with electric potential energy zero at infinite separation. The separation is then increased to 2r. Which statement about the electric potential energy E_p of the system is correct? HL
Answer and reasoning
E_p rises from −kq²/r to −kq²/2r, so an external force does positive work on it. — E_p = kq₁q₂/r = −kq²/r at separation r and −kq²/2r at 2r. Since −kq²/2r is greater (less negative) than −kq²/r, E_p increases; the increase is supplied by positive work from the external force pulling the unlike charges apart.
E_p falls from −kq²/r to −kq²/2r, because its size has been halved. — Halving the size of a negative number makes it larger, not smaller: −kq²/2r > −kq²/r. E_p increases as unlike charges are separated, which is why work must be done to separate them.
E_p changes from −kq²/r to −kq²/4r, because it obeys an inverse-square law. — The force obeys an inverse-square law; the potential energy is E_p = kq₁q₂/r, proportional to 1/r. Doubling r halves the size of E_p, giving −kq²/2r.
E_p falls from +kq²/r to +kq²/2r, because potential energy is a positive quantity. — With zero at infinity, the potential energy of unlike charges is negative: the field does positive work as they come together, so E_p = −kq²/r at separation r. Keeping the signs of the charges in kq₁q₂/r gives this automatically.
Working E_p = kq₁q₂/r with q₁ = +q, q₂ = −q: at separation r, E_p = −kq²/r; at 2r, E_p = −kq²/(2r). Change ΔE_p = −kq²/2r − (−kq²/r) = +kq²/2r > 0: E_p increases (becomes less negative). Work done by the external force = ΔE_p = +kq²/2r, positive.
27 Electric potential is defined to be zero at infinity. A student instead takes the potential to be zero at the surface of a charged conducting sphere, and repeats a calculation of the work needed to move a small charge between two points P and Q outside the sphere. Which statement is correct? HL
Answer and reasoning
The field strength at the surface of the sphere changes, because the field must be zero wherever the potential is zero. — Potential and field strength are different quantities: E is the gradient of V, not its value. Adding a constant to every potential leaves every gradient, and so every field, unchanged.
The work needed is unchanged: every potential shifts by the same constant, and only V_Q − V_P matters. — Moving the zero adds the same constant to the potential of every point. W = qΔV_e depends only on the difference between the potentials at Q and P, which is unaffected. The zero at infinity is a convention, not a law.
The work needed changes, because the potential energy of the moved charge at P and at Q has changed. — The potential energy at each point does change, but by the same amount at P and at Q. Work depends on the change in potential energy between the two points, which is the same as before.
The calculation is invalid, because the potential at the surface of a charged sphere is not zero. — The potential at any point has no absolute value; only differences are measurable. Zero at infinity is the IB convention, but any other reference point gives the same fields and the same work done.
28 A metal sphere of diameter 0.20 m carries a charge of +6.0 nC, spread uniformly over its surface. Electric potential is zero at infinity. Take k = 8.99 × 10⁹ N m² C⁻². What is the electric potential at a point 0.20 m from the surface of the sphere? HL
Answer and reasoning
1.8 × 10² V — The distance in V_e = kQ/r is measured from the centre of the sphere: r = radius + 0.20 m = 0.10 + 0.20 = 0.30 m. V_e = (8.99 × 10⁹ × 6.0 × 10⁻⁹)/0.30 = 1.8 × 10² V.
2.7 × 10² V — Using r = 0.20 m, the distance from the surface, gives 2.7 × 10² V. Outside the sphere the charge acts as if at the centre, so r is measured from the centre: 0.30 m.
1.3 × 10² V — Adding the full diameter, 0.20 + 0.20 = 0.40 m, gives 1.3 × 10² V. The distance from the centre to the surface is the radius, 0.10 m, so r = 0.30 m.
6.0 × 10² V — Using kQ/r² with r = 0.30 m gives 6.0 × 10² V. That expression is the field strength (in V m⁻¹); potential is V_e = kQ/r.
Working Outside the sphere the field and potential are those of a point charge at its centre. Radius = 0.10 m, so r = 0.10 + 0.20 = 0.30 m from the centre. V_e = kQ/r = (8.99 × 10⁹)(6.0 × 10⁻⁹)/0.30 = 53.94/0.30 = 179.8 V ≈ 1.8 × 10² V.
29 Three equipotential surfaces in an electric field are drawn at +30 V, +20 V and +10 V. A single electric field line passes through all three surfaces. Which statement about this field line is correct? HL
Answer and reasoning
It crosses each surface at right angles and is directed from the +10 V surface towards the +30 V surface. — The field points down the potential gradient, from high to low potential: from +30 V towards +10 V. A positive charge released on the +30 V surface would move towards +10 V.
It runs along the surfaces rather than across them, as the electric field acts along an equipotential. — Along an equipotential the potential does not change, so the field has no component along it. The field, and the field line, are perpendicular to each equipotential surface.
It crosses each surface at right angles, directed from the +30 V surface to the +10 V surface. — Field lines are always perpendicular to equipotential surfaces (otherwise the field would have a component along the surface and V would change along it), and the field points from high to low potential, E = −ΔV_e/Δr.
It is broken at each surface, since there is no electric field where the potential is constant. — The potential is constant along the surface, not across it. There is a field at every point of the surface, directed perpendicular to it, so the line passes straight through.
Working Along a field line the potential falls: E = −ΔV_e/Δr, so the field is directed from +30 V through +20 V to +10 V. On an equipotential surface V is constant, so E has no component along the surface; the field line therefore meets each surface at 90°.
30 The diagram shows the electric field lines of two point charges, X and Y. Which statement about the charges is correct?
Answer and reasoning
X is positive and Y negative, and the two charges have exactly the same size. — The signs are right, but the line count is not arbitrary: it is drawn in proportion to the charge. Only half the lines leaving X end on Y, so Y's charge is half the size of X's.
X is positive and Y negative, and the size of X's charge is twice that of Y's. — Lines leave positive charges and end on negative ones, so X is positive and Y negative. The number of lines drawn to a charge is proportional to its size: ten leave X but only five end on Y, so |Q_X| = 2|Q_Y|.
X and Y carry charges of the same sign, because the field lines join them together. — Lines join two charges only when they have opposite signs; lines from two like charges curve away from each other. Lines leaving X and ending on Y show that X is positive and Y negative.
Y is positive and X negative, since the arrows show how a negative charge is pushed. — Field-line arrows show the direction of the force on a positive test charge. Arrows point away from X, so X is positive; they point into Y, so Y is negative.
31 The diagram shows the electric field lines of a positively charged metal sphere and four points A, B, C and D. Which statement about the electric field strength E at these points is correct?
Answer and reasoning
E is greatest at A and falls steadily through B, C and D, as the charge is inside the sphere. — The charge on a conductor sits on its surface, and there are no field lines inside: E = 0 at A and B. The field is greatest just outside the surface, at C, and falls off with distance.
E is zero at A and at B, and E is greater at C than at D, where the lines are wider apart. — No field lines are drawn inside the sphere: the field inside a charged conductor is zero, so E = 0 at both A and B. Outside, the lines spread apart with distance, so the line density, and E, is greater at C than at D.
E is zero at A and at B, and E at C equals E at D, since the same lines pass both points. — The same lines pass C and D, but at D they are further apart. The field strength is shown by the spacing of the lines, not by which lines pass a point, so E is smaller at D.
E is zero at A only, and E at B is greater than E at C, as B is nearest the surface charge. — The surface charge produces no field anywhere inside the conductor, not only at the centre: the contributions of all parts of the surface cancel at every interior point, so E = 0 at B as well as at A.
32 The diagram shows a long straight wire viewed end-on; the cross indicates that the current is directed into the page. What is the direction of the magnetic field due to the wire at point P?
Answer and reasoning
Towards the top of the page, along the tangent to the dashed circle at P — This is the anticlockwise sense, which the grip rule gives for a current out of the page. With the thumb pointing into the page the fingers curl clockwise, so at P the field is downwards.
Directly away from the wire, along the radius of the dashed circle through P — Magnetic field lines around a straight wire are closed circles centred on the wire, not radial lines. The field at P is tangential to the circle through P.
Towards the bottom of the page, along the tangent to the dashed circle at P — The field lines are circles centred on the wire. Right-hand grip rule: thumb into the page (with the current), fingers curl clockwise as seen on the page. At P, to the right of the wire, clockwise means downwards.
Into the page, parallel to the current, and perpendicular to the dashed circles — The magnetic field of a current is not along the current: it circles the wire in the plane perpendicular to the current. At P it lies in the page, along the tangent to the circle.
33 The diagram shows a section through an air-core solenoid. Current flows out of the page in the wires above the axis and into the page in the wires below it. What is the direction of the magnetic field on the axis, inside the solenoid?
Answer and reasoning
Along the axis, in the direction from the end labelled Q to the end labelled P — This is the reversed sense. Curl the fingers of the right hand out of the page at the top and into the page at the bottom; the thumb then points from P towards Q.
Out of the page above the axis and into the page below it, along the currents — The field is not along the current. Each turn's current circulates around the axis, and the field it produces inside the solenoid is along the axis, perpendicular to the plane of the turn.
Zero on the axis, because the fields of the upper and lower rows of wires cancel — The two rows are the two sides of the same turns. Their contributions on the axis are in the same direction, along the axis, and add to give a strong uniform field inside the solenoid.
Along the axis, directed from the end labelled P towards the end labelled Q — Take one turn: out of the page at the top, into the page at the bottom. Right-hand grip rule with the fingers following the current around the turn: the thumb points along the axis from P to Q. Every turn gives the same direction, so the field inside is from P towards Q.
34 The diagram shows two point charges, Q₁ = +4.0 nC and Q₂ = −2.0 nC, fixed on a line, and a point P on the same line. Take k = 8.99 × 10⁹ N m² C⁻². What is the magnitude of the resultant electric field strength at P?
Answer and reasoning
9.8 × 10² N C⁻¹ — Adding the two magnitudes gives 9.8 × 10² N C⁻¹, but the fields at P point in opposite directions: away from the positive Q₁ and towards the negative Q₂. Subtract them.
4.0 × 10¹ N C⁻¹ — Using kQ/r instead of kQ/r² gives 120 − 80 = 40 N C⁻¹. Field strength follows an inverse-square law: E = kQ/r².
6.2 × 10² N C⁻¹ — E₁ = kQ₁/r₁² with r₁ = 0.45 m: 1.8 × 10² N C⁻¹ away from Q₁ (rightwards). E₂ = kQ₂/r₂² with r₂ = 0.15 m: 8.0 × 10² N C⁻¹ towards Q₂ (leftwards). Opposite directions, so the resultant is 8.0 × 10² − 1.8 × 10² = 6.2 × 10² N C⁻¹, towards Q₂.
4.0 × 10² N C⁻¹ — Using the marked 0.30 m as the distance from Q₁ to P gives 8.0 × 10² − 4.0 × 10² = 4.0 × 10² N C⁻¹. The 0.30 m is the distance from Q₁ to Q₂; P is 0.30 + 0.15 = 0.45 m from Q₁.
Working Distance of P from Q₁ = 0.30 + 0.15 = 0.45 m; from Q₂ = 0.15 m. E₁ = kQ₁/r² = (8.99 × 10⁹)(4.0 × 10⁻⁹)/0.45² = 35.96/0.2025 = 177.6 N C⁻¹, directed away from Q₁ (to the right). E₂ = (8.99 × 10⁹)(2.0 × 10⁻⁹)/0.15² = 17.98/0.0225 = 799.1 N C⁻¹, directed towards Q₂ (to the left). Resultant = 799.1 − 177.6 = 621.5 N C⁻¹ ≈ 6.2 × 10² N C⁻¹, directed towards Q₂.
35 The diagram shows three point charges, A, B and C, fixed on a straight line with the distances marked. Take k = 8.99 × 10⁹ N m² C⁻². What is the magnitude of the resultant electric force on charge A?
Answer and reasoning
2.5 × 10⁻⁶ N — Adding the two sizes gives 2.5 × 10⁻⁶ N, but B pushes A to the left while C pulls it to the right. The forces act in opposite directions and must be subtracted.
9.0 × 10⁻⁸ N — Using kq₁q₂/r instead of kq₁q₂/r² gives 3.6 × 10⁻⁷ − 2.7 × 10⁻⁷ = 9.0 × 10⁻⁸ N. Coulomb's law is an inverse-square law.
1.3 × 10⁻⁶ N — Using the marked 0.20 m as the distance from A to C gives 2.7 × 10⁻⁶ − 1.35 × 10⁻⁶ = 1.3 × 10⁻⁶ N. The 0.20 m marker joins A and B; A is 0.20 + 0.10 = 0.30 m from C.
1.5 × 10⁻⁷ N — B repels A, pushing it to the left (away from B): F_B = k(3.0 nC)(2.0 nC)/0.20² = 1.35 × 10⁻⁶ N. C attracts A, pulling it to the right, and A is 0.20 + 0.10 = 0.30 m from C: F_C = k(3.0 nC)(4.0 nC)/0.30² = 1.20 × 10⁻⁶ N. Opposite directions, so the resultant is 1.35 × 10⁻⁶ − 1.20 × 10⁻⁶ = 1.5 × 10⁻⁷ N, directed away from B.
Working Force on A due to B (both positive, repel; B is to the right, so the force on A is to the left): F_B = k(3.0 × 10⁻⁹)(2.0 × 10⁻⁹)/0.20² = 5.394 × 10⁻⁸/0.040 = 1.35 × 10⁻⁶ N. A is 0.20 + 0.10 = 0.30 m from C. Force on A due to C (unlike, attract; C is to the right, so the force on A is to the right): F_C = k(3.0 × 10⁻⁹)(4.0 × 10⁻⁹)/0.30² = 1.0788 × 10⁻⁷/0.090 = 1.20 × 10⁻⁶ N. Opposite directions: resultant = 1.35 × 10⁻⁶ − 1.20 × 10⁻⁶ = 1.5 × 10⁻⁷ N, directed away from B (to the left).
36 The diagram shows equipotential lines around a positive point charge, with their potentials and their distances from the charge marked. Estimate the electric field strength at point M. HL
Answer and reasoning
5.0 × 10⁰ V m⁻¹ — Dividing 50 V by 10 (centimetres, not converted) gives 5.0 V m⁻¹. The gap between the two equipotentials is 30 − 20 = 10 cm = 0.10 m; E = 50/0.10 = 5.0 × 10² V m⁻¹.
1.0 × 10³ V m⁻¹ — Dividing the potential of one equipotential, 100 V, by the gap gives 1.0 × 10³ V m⁻¹. The gradient uses the change in potential across the gap, 150 − 100 = 50 V.
5.0 × 10² V m⁻¹ — M is between the 150 V and 100 V equipotentials, which are 0.10 m apart (r = 20 cm and 30 cm). The field strength is the potential gradient: E ≈ ΔV/Δr = 50 V/0.10 m = 5.0 × 10² V m⁻¹, directed outwards from the charge.
7.5 × 10² V m⁻¹ — Dividing the 150 V potential by its radius, 0.20 m, applies E = V/d, which holds only for the uniform field between parallel plates. In any field, E is the potential gradient ΔV/Δr between neighbouring equipotentials.
Working M lies between the 150 V equipotential (r = 20 cm = 0.20 m) and the 100 V equipotential (r = 30 cm = 0.30 m). E = −ΔV_e/Δr; in magnitude E ≈ ΔV/Δr = (150 − 100) V/(0.30 − 0.20) m = 50/0.10 = 5.0 × 10² V m⁻¹, directed away from the charge (down the potential gradient). Check: at r = 0.25 m, kQ/r² with kQ = 30 V m gives 30/0.0625 = 4.8 × 10² V m⁻¹, consistent with the estimate.
37 The diagram shows equipotential lines in an electric field and three points X, Y and Z on them. A proton is moved slowly from X to Z. What is the work done on the proton by the external force? HL
Answer and reasoning
−30 eV — The field does −30 eV of work on the proton; the external force does the opposite, +30 eV. Moving a positive charge to a higher potential needs positive work from the external force.
+30 eV — ΔV_e = V_Z − V_X = +20 − (−10) = +30 V. For a charge of +e moved through +30 V, W = qΔV_e = +30 eV: positive work is done by the external force to move the proton up the potential.
+20 eV — Using the potential at Z alone ignores where the proton started. X is at −10 V, not at 0 V, so ΔV_e = 20 − (−10) = 30 V and W = +30 eV.
+10 eV — Subtracting the sizes, 20 − 10, drops the sign of the potential at X. ΔV_e = V_Z − V_X = 20 − (−10) = 30 V, so W = +30 eV.
Working X is on the −10 V equipotential and Z on the +20 V equipotential. W = qΔV_e = e × (V_Z − V_X) = e × (+20 − (−10)) V = e × 30 V = +30 eV (positive: the proton is moved to a higher potential, against the field). In joules, 30 × 1.60 × 10⁻¹⁹ = 4.8 × 10⁻¹⁸ J.
That was your twenty minutes. Real practice on D.2 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
DP Science Cafe