D.2 Electric and magnetic fields flashcards: IB Physics SL and HL

All 76 D.2 Electric and magnetic fields cards

1. DefinitionSL & HL

   Define electric charge and state the law of conservation of charge.

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   Electric charge is the property of matter that causes it to experience a force in an electric field; there are two kinds, positive and negative, and like charges repel while unlike charges attract. SI unit coulomb (C); charge is a scalar. Conservation of charge: the net (algebraic) charge of an isolated system is constant — charge can be transferred from one body to another but cannot be created or destroyed. Exam tip: students write "charge cannot be created or destroyed" and stop; the mark scheme wants "net charge of an isolated/closed system remains constant". In every nuclear or particle reaction the total charge before must equal the total charge after.

2. DefinitionSL & HL

   State what is meant by the quantisation of electric charge and define the elementary charge.

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   Charge is quantised: any observed charge is an integer multiple of the elementary charge e, so Q = ne with n an integer. The elementary charge e = 1.60 × 10⁻¹⁹ C is the magnitude of the charge on a proton (+e) or an electron (−e). SI unit C; scalar. Exam tip: state "integer multiple of e", not merely "charge comes in packets". Quarks carry ±⅓e and ±⅔e but are never observed in isolation (confinement), so free particles still carry whole multiples of e. Millikan's oil-drop experiment is the standard experimental evidence quoted for quantisation.

3. DefinitionSL & HL

   State Coulomb's law in words.

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   The magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the separation of their centres; the force acts along the line joining them and is repulsive for like charges, attractive for unlike. SI unit newton (N); force is a vector. Exam tip: three marking points — proportional to product of charges, inverse square of separation, and "point charges" (or spheres treated as point charges at their centres). Omitting "point" or writing "inversely proportional to distance" loses marks. The law applies strictly in a vacuum (or air, to good approximation) and obeys Newton's third law: the forces on the two charges are equal and opposite.

4. DefinitionSL & HL

   Define electric field strength at a point.

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   Electric field strength E is the force per unit charge experienced by a small positive test charge placed at that point: E = F/q. SI unit N C⁻¹ (equivalently V m⁻¹); E is a vector with the same direction as the force on a positive charge. Exam tip: the two words that earn the marks are "force per unit charge" AND "positive test charge" — omitting "positive" loses the direction mark, and omitting "small"/"test" ignores the fact that a large charge would distort the field being measured. Do not define it as "force on a charge"; that is not per unit charge. A region has an electric field if a stationary charge placed there experiences a force.

5. DefinitionSL & HL

   State the conventions and rules for drawing electric field lines.

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   Field lines show the direction of the force on a small positive test charge: they start on positive charge and end on negative charge (or run to/from infinity). The tangent to a line gives the field direction; the density (number per unit cross-sectional area) is proportional to the field strength, so closely spaced lines mean a strong field. Lines never cross, because the field has a unique direction at each point, and they meet a conductor's surface at 90°. Exam tip: arrows are compulsory — an unarrowed diagram scores zero. For an isolated point charge the lines are radial and evenly spaced in angle; equal numbers should leave/enter equal charges.

6. DefinitionSL & HL

   Define a uniform electric field and state where one is produced in the laboratory.

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   A uniform electric field is one in which the electric field strength has the same magnitude and the same direction at every point, represented by straight, parallel, equally spaced field lines. It is produced in the region between two parallel, oppositely charged conducting plates, where E = V/d directed from the positive to the negative plate. SI unit N C⁻¹ or V m⁻¹. Exam tip: "same magnitude" alone is not enough — the direction must also be constant. In sketches, show edge (fringing) effects: the lines bulge outwards near the ends of the plates, so the field is uniform only well inside the plates. A charge moving in this field feels a constant force qE, hence constant acceleration.

7. DefinitionSL & HL

   Define the permittivity of free space and state how it relates to the Coulomb constant.

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   The permittivity of free space ε₀ is the constant appearing in Coulomb's law that characterises the ability of a vacuum to permit electric field lines; it fixes the strength of the electrostatic interaction. ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻² (equivalently F m⁻¹), a scalar. The Coulomb constant is k = 1/(4πε₀) = 8.99 × 10⁹ N m² C⁻². Exam tip: the 4π arises because the field of a point charge spreads over a sphere of area 4πr². A medium other than vacuum has permittivity ε = ε_r ε₀ with relative permittivity ε_r > 1, which reduces the force. Quoting k as 9 × 10⁹ is acceptable, but keep 3 significant figures in final answers.

8. DefinitionSL & HL

   Describe the state of a conductor in electrostatic equilibrium.

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   In electrostatic equilibrium the charges in a conductor are at rest, so: the electric field inside the conducting material is zero; any excess charge resides entirely on the outer surface; the whole conductor (surface and interior) is at the same electric potential, so its surface is an equipotential; and the field just outside is perpendicular to the surface. Charge concentrates where the surface curvature is greatest (points), giving the strongest external field there. Exam tip: the reasoning mark is "if E inside were not zero, free electrons would move, contradicting equilibrium". This is the basis of the Faraday cage: a hollow conductor shields its interior from external electric fields.

9. DefinitionSL & HL

   Define potential difference between two points in terms of work and charge.

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   The potential difference ΔV between two points is the work done per unit charge in moving a small positive charge between those points: ΔV = W/q, so W = qΔV. SI unit volt (V), where 1 V = 1 J C⁻¹; potential difference is a scalar. Exam tip: "work done per unit charge" earns the mark; "energy used by a charge" does not. One volt is the potential difference between two points when 1 J of work is done moving 1 C between them. Because ΔV is a scalar, potentials from several charges add algebraically with sign — a frequent contrast with fields, which must be added as vectors.

10. DefinitionSL & HL

    Describe how a magnetic field is produced and state what magnetic field lines represent.

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    A magnetic field is a region in which a moving charge, a current-carrying conductor or a magnetic material experiences a force; it is produced by moving charges (currents) and by the intrinsic magnetic moments of electrons in permanent magnets. Magnetic field lines run from the north-seeking pole to the south-seeking pole outside a magnet and from S to N inside it, so they always form closed loops; the tangent gives the direction of the field (the direction a free N pole would be pushed), and line density is proportional to field strength. Exam tip: lines never cross and never start or stop — there are no magnetic monopoles, which is the key structural difference from electric field lines.

11. DefinitionSL & HL

    Define magnetic flux density B and state its unit.

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    Magnetic flux density B is defined from the force on a current: B is the force per unit current per unit length on a straight conductor placed perpendicular to the field, B = F/(IL). SI unit tesla (T), where 1 T = 1 N A⁻¹ m⁻¹; B is a vector. Equivalently, from F = qvB sinθ, 1 T gives 1 N of force on a charge of 1 C moving at 1 m s⁻¹ perpendicular to the field. Exam tip: the mark is lost by omitting "perpendicular" — at any other angle F = BIL sinθ. The tesla is a large unit: the Earth's field is about 5 × 10⁻⁵ T, a school bar magnet about 10⁻² T, an MRI magnet 1–3 T.

12. DefinitionSL & HL

    State the right-hand grip rule and use it to describe the field pattern of a long straight current-carrying wire.

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    Right-hand grip rule: point the right thumb along the conventional current and the curled fingers give the direction of the magnetic field. For a long straight wire the field lines are concentric circles centred on the wire, lying in planes perpendicular to it, with the spacing increasing with distance because B ∝ 1/r. Exam tip: use conventional current (positive to negative), not electron flow — reversing this is the most common single error. On a plane diagram show ⊗ for current into the page and ⊙ for current out of the page, and mark arrows on the circles: clockwise for current into the page.

13. DefinitionSL & HL

    Describe the magnetic field of a solenoid and identify its poles.

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    A solenoid is a long coil of N closely wound turns. Inside, the field lines are straight, parallel and equally spaced along the axis, so the field is uniform, strong and given by B = μ₀NI/L; outside, the pattern is that of a bar magnet, weak and spreading. The end where conventional current flows anticlockwise as viewed is the north pole (right-hand grip rule applied to the turns: fingers follow the current, thumb points to N). Exam tip: field lines must be drawn continuous through the solenoid and closed outside — leaving them stopping at the ends loses the mark. Adding a soft-iron core greatly increases B; this is the electromagnet.

14. DefinitionSL & HL

    Define the permeability of free space and describe the Earth's magnetic field.

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    The permeability of free space μ₀ = 4π × 10⁻⁷ T m A⁻¹ (H m⁻¹) is the constant relating the magnetic field produced to the current producing it in a vacuum; it is a scalar. The Earth's magnetic field resembles that of a bar magnet tilted about 11° to the rotation axis, with the magnetic pole in the northern hemisphere behaving as a magnetic south pole (which is why a compass N-pole points north). Its magnitude is roughly 5 × 10⁻⁵ T and it has both horizontal and vertical (dip) components, becoming vertical near the magnetic poles. Exam tip: quote the field as arising from convection currents in the liquid outer core; state that it deflects charged particles from the solar wind, producing aurorae.

15. DefinitionSL & HL

    Explain what is meant by charging by friction and by induction, and by earthing.

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    Charging by friction transfers electrons between two insulators in contact; the one that gains electrons becomes negative and the other equally positive, conserving charge. Charging by induction requires no contact with the charging body: a charged rod brought near an isolated conductor causes charge separation (polarisation), the conductor is then earthed so that electrons flow to or from the ground, the earth connection is removed and finally the rod is removed, leaving the conductor with a charge opposite to the rod. Earthing means connecting a body by a conductor to a very large reservoir of charge (the Earth) held at zero potential. Exam tip: the order — earth first, then remove the earth, then remove the rod — carries the marks.

16. EquationSL & HLData booklet: Yes

    State Coulomb's law as an equation, defining every symbol and its unit.

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    F = kq₁q₂/r² with k = 1/(4πε₀). F is the magnitude of the electrostatic force in newtons (N); q₁ and q₂ are the point charges in coulombs (C); r is the separation of their centres in metres (m); k = 8.99 × 10⁹ N m² C⁻² and ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻². Valid for point charges (or uniformly charged spheres, using centre-to-centre r) at rest in a vacuum. Data booklet: Yes, both forms. Common misuse: forgetting to square r, or using diameters instead of centre separation for spheres. Sanity check: two 1.0 μC charges 1.0 m apart give F = 8.99 × 10⁹ × (10⁻⁶)²/1² = 9.0 × 10⁻³ N — small, so μC-scale charges give mN-scale forces.

17. EquationSL & HLData booklet: Yes

    State the defining equation for electric field strength and its rearrangements.

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    E = F/q, so F = qE and q = F/E. E is electric field strength in N C⁻¹ (= V m⁻¹); F is the force in newtons on the charge q in coulombs. The field direction is the direction of F on a positive charge; for a negative charge F is antiparallel to E. Valid for any electric field, uniform or not, provided q is small enough not to disturb the field. Data booklet: Yes. Common misuse: treating E as the force itself, or ignoring the sign of q so the force is drawn the wrong way for an electron. Sanity check: an electron in E = 1.0 × 10³ N C⁻¹ feels F = 1.6 × 10⁻¹⁹ × 10³ = 1.6 × 10⁻¹⁶ N, giving a ≈ 1.8 × 10¹⁴ m s⁻²; gravity is utterly negligible here.

18. EquationSL & HLData booklet: Yes

    State the equation for the electric field strength due to a point charge and how fields combine.

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    E = kq/r² = q/(4πε₀r²). E is field strength in N C⁻¹; q is the source charge in C; r is the distance from the charge in m; k = 8.99 × 10⁹ N m² C⁻². The field is radial, directed away from a positive q and towards a negative q, and the expression also applies outside a uniformly charged conducting sphere (measuring r from the centre); inside such a conductor E = 0. Data booklet: Yes. Fields from several charges add by vector superposition, not algebraically. Common misuse: adding magnitudes when the two fields are not collinear, or using the charge being tested rather than the source charge. Sanity check: 2.0 nC at 3.0 cm gives E = 8.99 × 10⁹ × 2.0 × 10⁻⁹/(0.030)² = 2.0 × 10⁴ N C⁻¹.

19. EquationSL & HLData booklet: Yes

    State the equation for the field between parallel plates and the relation between the units of E.

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    E = V/d for a uniform field. E is the field strength in V m⁻¹ (= N C⁻¹); V is the potential difference between the plates in volts; d is the plate separation in metres. The field is directed from the positive plate to the negative plate and is valid only where the field is uniform, i.e. away from the edges of large, parallel plates. Data booklet: Yes. Rearrangements: V = Ed and d = V/E; the force on a charge is F = qV/d. Common misuse: using the distance of the charge from one plate instead of the full plate separation d. Sanity check: 200 V across 5.0 mm gives E = 200/0.0050 = 4.0 × 10⁴ V m⁻¹, and a proton there feels 6.4 × 10⁻¹⁵ N.

20. EquationSL & HLData booklet: Yes

    State the equation linking work, charge and potential difference, and give its rearrangements.

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    W = qΔV, so ΔV = W/q and q = W/ΔV. W is the work done (energy transferred) in joules; q is the charge moved in coulombs; ΔV is the potential difference in volts (J C⁻¹). Valid for moving a charge between two points at a stated potential difference; the work is independent of the path taken because the electrostatic field is conservative. Data booklet: Yes. In a uniform field this becomes W = qEd. Common misuse: using the potential at one point rather than the difference, or losing the sign — moving a positive charge to a higher potential requires positive work done against the field. Sanity check: an electron accelerated through 500 V gains 1.6 × 10⁻¹⁹ × 500 = 8.0 × 10⁻¹⁷ J = 500 eV.

21. EquationSL & HLData booklet: Yes

    State the equation for the magnetic flux density around a long straight current-carrying wire.

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    B = μ₀I/(2πr). B is the magnetic flux density in tesla (T); μ₀ = 4π × 10⁻⁷ T m A⁻¹; I is the current in amperes (A); r is the perpendicular distance from the axis of the wire in metres. Valid for a long straight wire in vacuum/air, at distances small compared with the wire's length and outside the wire itself. Data booklet: Yes. So B ∝ I and B ∝ 1/r — an inverse first power, not inverse square. Common misuse: dropping the 2π, or measuring r from the surface rather than from the centre of the wire. Sanity check: 5.0 A at 2.0 cm gives B = (4π × 10⁻⁷ × 5.0)/(2π × 0.020) = 5.0 × 10⁻⁵ T, comparable to the Earth's field.

22. EquationSL & HLData booklet: Yes

    State the equation for the magnetic flux density inside a solenoid and explain each symbol.

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    B = μ₀NI/L. B is the flux density on the axis inside the solenoid in tesla; μ₀ = 4π × 10⁻⁷ T m A⁻¹; N is the total number of turns; I is the current in amperes; L is the length of the solenoid in metres. N/L is the number of turns per metre (m⁻¹), so B is independent of the solenoid's radius and of position along the axis, provided L ≫ diameter. Data booklet: Yes. Common misuse: substituting the length of wire used rather than the length of the coil, or using turns per centimetre without converting. Sanity check: 400 turns over 0.20 m carrying 2.0 A gives B = 4π × 10⁻⁷ × (400/0.20) × 2.0 = 5.0 × 10⁻³ T.

23. EquationSL & HLData booklet: No – memorise

    State the equation for the quantisation of charge and use it in a calculation.

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    Q = ne, where Q is the total charge in coulombs, n is an integer (the number of elementary charges gained or lost) and e = 1.60 × 10⁻¹⁹ C. Valid for any observed free charge; n must be a whole number, which is the essential physical content. Data booklet: No – memorise (e is given in the constants list). Common misuse: obtaining a non-integer n from experimental data and rounding without comment; in Millikan-style questions the point is that the drop charges are common multiples of e. Sanity check: a sphere carrying −3.2 × 10⁻⁹ C has n = 3.2 × 10⁻⁹/1.60 × 10⁻¹⁹ = 2.0 × 10¹⁰ excess electrons.

24. EquationSL & HL

    State how electric fields and forces from several charges are combined, and give the equation for the field on the axis between two point charges.

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    Superposition: the resultant field is the vector sum of the individual fields, E_res = ΣE_i, each E_i = kq_i/r_i² directed radially. For two charges on a line, add signed components along that line; off-axis, resolve into perpendicular components and combine with Pythagoras and tan θ. Units N C⁻¹; E is a vector. Data booklet: N/A (principle). Common misuse: adding magnitudes of fields that point in different directions. Between two equal positive charges the field is zero at the midpoint; between a +q and a −q the fields add there, giving a maximum. For unequal like charges the null point lies nearer the smaller charge, found from kq₁/x² = kq₂/(d − x)².

25. Graph/diagramSL & HL

    Sketch and interpret a graph of electric field strength E against distance r from an isolated point charge, and state how it is linearised.

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    Axes: E/N C⁻¹ (y) against r/m (x). Shape: a decreasing curve, E ∝ 1/r², asymptotic to both axes — it never touches either, and E → ∞ as r → 0 (an idealisation of the point model). Halving r quadruples E. To find q, linearise by plotting E against 1/r² (x-axis units m⁻²): the result is a straight line through the origin with gradient kq = q/(4πε₀), so q = gradient/k. For a charged conducting sphere of radius R, E = 0 for r < R, jumps to kq/R² at the surface and then follows the 1/r² curve. Changing q scales the whole curve vertically without changing its shape; a negative q simply reverses the field direction.

26. Graph/diagramSL & HL

    Describe the electric field line patterns for a single point charge, two like charges, two unlike charges and parallel plates.

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    Single positive charge: straight radial lines pointing outwards, evenly spaced in angle (inwards for negative). Two equal positive charges: lines curve away from each other with a neutral point (E = 0) at the midpoint where no line passes. Two unlike charges (dipole): lines leave the + and curve round to enter the −, densest on the line joining them; no null point between them. Parallel plates: straight, parallel, equally spaced lines from + plate to − plate, meeting the plates at 90°, with outward bulging (fringing) at the edges. Exam tip: arrows, correct start/end on charges, no crossing lines, and symmetry are the marking points. Denser lines = stronger field, so field strength can be compared by eye at two labelled points.

27. Graph/diagramSL & HL

    Sketch and interpret a graph of magnetic flux density B against distance r from a long straight current-carrying wire.

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    Axes: B/T (y) against r/m (x). Shape: a decreasing curve with B ∝ 1/r (a rectangular hyperbola), falling more slowly than the 1/r² electric field of a point charge. Linearisation: plot B against 1/r (x-axis m⁻¹) to obtain a straight line through the origin of gradient μ₀I/(2π); hence I = 2π × gradient/μ₀. Alternatively plot ln B against ln r: the gradient is −1, confirming the inverse first power, with intercept ln(μ₀I/2π). Doubling the current doubles B at every r, scaling the curve vertically. A non-zero intercept in the experiment usually signals a background field such as the Earth's, which is a systematic error.

28. Graph/diagramSL & HL

    Describe a graph of B against I for a solenoid and against position along its axis.

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    B against I: axes B/T (y) against I/A (x); a straight line through the origin of gradient μ₀N/L, from which the turns per unit length N/L = gradient/μ₀ can be found. The line curves and saturates only if a ferromagnetic core is present. B against position along the axis: B is essentially constant (a flat plateau) over the central region and falls off towards each end, reaching about half the central value exactly at each open end, then decaying rapidly outside. Increasing L for fixed N/L lengthens the plateau; increasing N/L raises it. Exam tip: state explicitly that the flat region is what justifies calling the interior field uniform.

29. Graph/diagramSL & HL

    Explain how a Coulomb's law investigation is linearised graphically.

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    Measure the force F between two small charged spheres (for example from the deflection of a suspended sphere or an electronic balance reading) at several separations r. A plot of F against r gives a curve, which cannot be used to test the power law. Plot F against 1/r² (x-axis m⁻²): a straight line through the origin of gradient kq₁q₂ confirms the inverse-square law and gives the product of the charges as gradient/k. Alternatively plot ln F against ln r: a straight line of gradient −2.0 confirms the exponent, and the intercept gives ln(kq₁q₂). Error bars on 1/r² are strongly asymmetric at small r; a non-zero F-intercept suggests a systematic error such as induced charge or air currents.

30. Concept/explainSL & HL

    Explain why the electric field inside a conductor in electrostatic equilibrium is zero and where the excess charge sits.

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    • A conductor contains free electrons that respond to any internal field
    • If a field existed inside, the free charges would accelerate, so the situation would not be static
    • Charges therefore redistribute until their own field exactly cancels the applied field, giving E = 0 everywhere inside
    • Since E = 0 inside, no work is done moving charge within the metal, so the whole conductor is at one potential (an equipotential volume)
    • Excess charge is forced to the outer surface by mutual repulsion, and sits most densely where the surface curves most sharply
    • This is the basis of electrostatic shielding by a Faraday cage
    • Exam tip: say the redistribution stops when the internal field is cancelled, not that metal blocks fields.
31. Concept/explainSL & HLData booklet: Yes

    Explain why the field between two parallel charged plates is uniform, and describe what happens near the edges.

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    • Each plate carries a uniform surface charge, and every element of charge contributes to the field at a point between them
    • The contributions parallel to the plates cancel by symmetry, leaving a resultant perpendicular to the plates
    • Because the plates are large compared with their separation, the resultant has the same magnitude and direction everywhere in the central region
    • This uniform field is drawn as equally spaced, parallel, straight lines running from the positive to the negative plate
    • Its magnitude is E = V/d, so E depends on the potential difference and separation, not on where the charge sits
    • Near the edges the lines bulge outward and the field weakens — edge or fringing effects
    • Exam tip: quote the large-plates condition when asked why the field is uniform.
32. Concept/explainSL & HLData booklet: Yes

    Explain why the electric field of an isolated point charge obeys an inverse square law, and what this means physically.

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    • The field of a point charge is radial and spherically symmetric, so the flux leaving the charge spreads over a sphere of area 4πr²
    • The same total flux crossing an ever larger area means the field strength falls as 1/r², giving E = kq/r² with k = 1/(4πε₀)
    • Doubling r therefore quarters E, and trebling r reduces E to one ninth
    • The direction is radially outward for a positive charge and radially inward for a negative charge
    • E is a vector with unit N C⁻¹, equivalently V m⁻¹
    • The same geometry gives Newton's law of gravitation its 1/r² form
    • Exam tip: an inverse square law is not the same as inverse proportionality; halving E needs r to increase by √2, not by 2.
33. Concept/explainSL & HLData booklet: Yes

    Explain how the resultant electric field at a point due to several charges is found.

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    • Electric fields obey the principle of superposition, so each charge contributes as if the others were absent
    • Calculate the magnitude of each contribution from E = kq/r² using the magnitude of each charge
    • Draw each contribution as an arrow at the point: away from a positive charge, towards a negative charge
    • Add the contributions as vectors, either by resolving into perpendicular components or by scale drawing
    • On a line joining two charges the arrows are collinear, so add or subtract magnitudes according to direction
    • A null point exists where the contributions are equal in magnitude and opposite in direction; it lies between two like charges and outside the pair for unlike charges
    • Exam tip: never add fields by adding the charges first.
34. Concept/explainSL & HLData booklet: Yes

    Describe the magnetic field pattern of a bar magnet and of a long straight current-carrying wire.

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    • Bar magnet: lines emerge from the north pole and enter the south pole outside the magnet, and continue from south to north inside it, so every line is a closed loop
    • The lines are most concentrated at the poles, where the field is strongest, and they never cross
    • Straight wire: the lines are concentric circles in the plane perpendicular to the wire, centred on it
    • Their direction is given by the right-hand grip rule — thumb along the conventional current, fingers curl in the sense of B
    • The circles are spaced further apart with distance because B = μ₀I/(2πr), an inverse first-power law
    • Exam tip: magnetic field lines have no start or end because isolated magnetic poles do not exist.
35. Concept/explainSL & HLData booklet: Yes

    Explain the magnetic field of a solenoid and how it can be increased.

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    • Each turn produces circular field lines that reinforce along the axis and largely cancel between adjacent turns
    • The result inside a long solenoid is a nearly uniform axial field, drawn as parallel equally spaced lines
    • Outside, the field is weak and spreads out like that of a bar magnet, so the solenoid has an effective north and south end
    • The magnitude is B = μ₀NI/L, so B depends on the current and on the number of turns per unit length, not on the coil radius
    • Increase B by raising I, winding more turns per metre, or inserting a soft-iron core to raise the permeability
    • Exam tip: L is the length of the solenoid, not the length of wire used.
36. Concept/explainSL & HL

    Describe the Earth's magnetic field and explain why it matters in school magnetism experiments.

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    • The Earth behaves approximately as a bar magnet tilted from the rotation axis, with the field thought to arise from convection currents in the liquid iron outer core
    • Field lines run from the geographic south region to the geographic north region outside the Earth, so the magnet's south pole lies near the geographic North Pole and a compass north-seeking pole points there
    • The field has a horizontal component of order 2 × 10⁻⁵ T and a dip angle that varies with latitude
    • In experiments this component is comparable with the field of a modest current, so it adds vectorially to the field being measured
    • Exam tip: control for it by aligning apparatus east–west, or by reversing the current and averaging.
37. Concept/explainSL & HLData booklet: Yes

    Explain the meaning of potential difference and how it relates to the work done moving a charge.

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    • The potential difference between two points is the work done per unit charge in moving charge between them, so ΔV = W/q and W = qΔV
    • One volt is one joule per coulomb
    • Moving a positive charge from low to high potential requires external work, and the charge gains electric potential energy
    • Released, the charge is accelerated by the field and that energy becomes kinetic energy, giving qΔV = ½mv² for a charge starting from rest
    • In a uniform field the work can also be written W = qEd, and comparing the two gives E = V/d
    • Exam tip: potential difference is a scalar; use magnitudes and decide the sign from whether the field does the work or work is done against it.
38. Worked problemSL & HLData booklet: Yes

    Two point charges of +3.0 μC and −5.0 μC are 0.20 m apart in vacuum. Determine the magnitude and nature of the force between them.

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    Use Coulomb's law F = kq₁q₂/r² with k = 8.99 × 10⁹ N m² C⁻². Substitute magnitudes: F = (8.99 × 10⁹ × 3.0 × 10⁻⁶ × 5.0 × 10⁻⁶)/(0.20)². Numerator = 8.99 × 10⁹ × 1.5 × 10⁻¹¹ = 0.13485 N m². Denominator = 0.040 m². F = 0.13485/0.040 = 3.37 N, so F = 3.4 N to 2 s.f. The charges have opposite signs, so the force is attractive, and by Newton's third law each charge experiences 3.4 N directed towards the other. Check/Trap: convert μC to C before squaring anything, and square the whole separation — using 0.20 instead of 0.040 makes the answer five times too small. Do not put the signs into the formula; get the magnitude, then state attractive or repulsive by inspection.

39. Worked problemSL & HLData booklet: Yes

    Two point charges exert a force of 36 N on each other. One charge is doubled and the separation is trebled. Determine the new force.

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    Coulomb's law gives F ∝ q₁q₂/r². Doubling one charge multiplies the numerator by 2. Trebling r multiplies the denominator by 3² = 9. So F_new = F_old × 2/9 = 36 × 2/9 = 8.0 N. The force remains along the line joining the charges and its nature (attractive or repulsive) is unchanged, because neither sign changed. Check/Trap: this is a proportionality question, so no values of k, q or r are needed — students who try to solve for the individual charges waste time and usually run out of information. The commonest error is dividing by 3 rather than by 9; always apply the square to the whole length factor. A useful sanity check: the answer must be smaller than 36 N because the separation increase dominates the charge increase.

40. Worked problemSL & HLData booklet: No – memorise

    A metal sphere carries a charge of −4.8 × 10⁻¹⁷ C. Determine the number of excess electrons on the sphere.

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    Charge is quantised, so q = ne with e = 1.60 × 10⁻¹⁹ C. Rearranging, n = q/e = (4.8 × 10⁻¹⁷)/(1.60 × 10⁻¹⁹) = 3.0 × 10². So the sphere carries 300 excess electrons. The sign is negative because electrons have been added; the sphere has not gained protons, and its mass rises by only 300 × 9.11 × 10⁻³¹ kg ≈ 2.7 × 10⁻²⁸ kg, which is undetectable. Check/Trap: n must be a whole number — a non-integer answer means an arithmetic slip. Use the magnitude of the charge; a negative n is meaningless. Watch the exponent subtraction: −17 − (−19) = +2, so the answer is of order 10², not 10⁻². Quote n as a pure number with no unit.

41. Worked problemSL & HLData booklet: Yes

    Determine the electric field strength 5.0 cm from an isolated point charge of +2.0 nC, and state its direction.

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    For a point charge E = kq/r². Convert first: q = 2.0 × 10⁻⁹ C and r = 5.0 × 10⁻² m, so r² = 2.5 × 10⁻³ m². E = (8.99 × 10⁹ × 2.0 × 10⁻⁹)/(2.5 × 10⁻³) = 17.98/2.5 × 10⁻³ = 7.19 × 10³. So E = 7.2 × 10³ N C⁻¹ (equivalently 7.2 × 10³ V m⁻¹), directed radially away from the charge, since the charge is positive. Check/Trap: the two most frequent errors are leaving r in centimetres, which makes E 10⁴ times too small, and forgetting to square r. Note that N C⁻¹ and V m⁻¹ are the same unit — either is accepted. If asked for the force on a −1.0 nC charge placed there, use F = qE = 7.2 × 10⁻⁶ N directed towards the source charge.

42. Worked problemSL & HLData booklet: Yes

    Charges of +9.0 μC and +4.0 μC are fixed 0.50 m apart. Determine the position on the line between them where the resultant electric field is zero.

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    Between two positive charges the two field contributions point in opposite directions, so a null point exists there. Let the point be a distance x from the 9.0 μC charge, so it is (0.50 − x) from the 4.0 μC charge. Set the magnitudes equal: k(9.0 × 10⁻⁶)/x² = k(4.0 × 10⁻⁶)/(0.50 − x)². Cancel k and 10⁻⁶: 9/x² = 4/(0.50 − x)². Take the square root of both sides: 3/x = 2/(0.50 − x). Cross-multiply: 3(0.50 − x) = 2x, so 1.5 = 5x and x = 0.30 m from the 9.0 μC charge. Check/Trap: the null point must be nearer the smaller charge, and 0.30 m > 0.20 m confirms this. Taking the square root early avoids a quadratic; keep only the root that lies between the charges.

43. Worked problemSL & HLData booklet: Yes

    Two parallel plates 4.0 mm apart have a potential difference of 120 V. Determine the field strength between them and the acceleration of an electron released there.

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    The field between parallel plates is uniform: E = V/d = 120/(4.0 × 10⁻³) = 3.0 × 10⁴ V m⁻¹, directed from the positive to the negative plate. Force on the electron: F = qE = 1.60 × 10⁻¹⁹ × 3.0 × 10⁴ = 4.8 × 10⁻¹⁵ N, directed towards the positive plate because the electron is negative. Acceleration: a = F/m_e = (4.8 × 10⁻¹⁵)/(9.11 × 10⁻³¹) = 5.3 × 10¹⁵ m s⁻². Check/Trap: d must be in metres — using 4.0 mm as 4.0 gives an answer 1000 times too small. The field is uniform, so E, F and a are the same everywhere between the plates and do not depend on the electron's position. Gravity here is about 10⁻²⁹ N and is entirely negligible compared with the electric force.

44. Worked problemSL & HLData booklet: Yes

    An electron starting from rest is accelerated through a potential difference of 2.5 kV. Determine its final speed.

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    Energy approach: the work done by the field equals the gain in kinetic energy, so qΔV = ½m_ev². Work: W = 1.60 × 10⁻¹⁹ × 2.5 × 10³ = 4.0 × 10⁻¹⁶ J. Then v = √(2W/m_e) = √((2 × 4.0 × 10⁻¹⁶)/(9.11 × 10⁻³¹)) = √(8.78 × 10¹⁴) = 2.96 × 10⁷, so v = 3.0 × 10⁷ m s⁻¹ to 2 s.f. Check/Trap: this is 0.10c, so a non-relativistic treatment is acceptable — above roughly 0.1c the IB expects you to note the limitation. The answer does not depend on the plate separation or on the field strength, only on ΔV, so do not look for d. Remember to take the square root: a very common slip is quoting 8.8 × 10¹⁴ m s⁻¹, which exceeds c and should be spotted instantly.

45. Worked problemSL & HLData booklet: Yes

    An electron enters midway between horizontal plates of length 6.0 cm and separation 2.0 cm, moving at 3.0 × 10⁷ m s⁻¹ parallel to them. The plates are at 200 V. Determine the vertical deflection on exit.

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    Treat this as projectile-style motion: constant velocity along the plates, uniform acceleration perpendicular to them. Field: E = V/d = 200/(2.0 × 10⁻²) = 1.0 × 10⁴ V m⁻¹. Acceleration: a = eE/m_e = (1.60 × 10⁻¹⁹ × 1.0 × 10⁴)/(9.11 × 10⁻³¹) = 1.76 × 10¹⁵ m s⁻². Time inside the plates: t = L/v = (6.0 × 10⁻²)/(3.0 × 10⁷) = 2.0 × 10⁻⁹ s. Deflection: y = ½at² = 0.5 × 1.76 × 10¹⁵ × (2.0 × 10⁻⁹)² = 3.5 × 10⁻³ m = 3.5 mm. Check/Trap: y must be less than d/2 = 10 mm or the electron would hit a plate — 3.5 mm confirms it exits. Use the horizontal length for t, never the plate separation, and do not use v in the vertical equation since the initial vertical velocity is zero.

46. Worked problemSL & HLData booklet: Yes

    A long straight wire carries 5.0 A. Determine the magnetic flux density 2.0 cm from it, and the resultant field there if the Earth's horizontal component of 1.8 × 10⁻⁵ T is perpendicular to it.

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    Wire field: B = μ₀I/(2πr) = (4π × 10⁻⁷ × 5.0)/(2π × 2.0 × 10⁻²). The π cancels, leaving B = (2 × 10⁻⁷ × 5.0)/(2.0 × 10⁻²) = (1.0 × 10⁻⁶)/(2.0 × 10⁻²) = 5.0 × 10⁻⁵ T, in circles around the wire given by the right-hand grip rule. The two fields are perpendicular, so add as vectors: B_R = √((5.0 × 10⁻⁵)² + (1.8 × 10⁻⁵)²) = √(28.2 × 10⁻¹⁰) = 5.3 × 10⁻⁵ T. Angle from the wire field: θ = tan⁻¹(1.8/5.0) = 20°. Check/Trap: cancel 2π against 4π × 10⁻⁷ to get 2 × 10⁻⁷ and the arithmetic becomes trivial. This is a 1/r law, not 1/r², so doubling r halves B. Magnetic fields add as vectors, never by simple addition of magnitudes.

47. Worked problemSL & HLData booklet: Yes

    A solenoid of length 25 cm has 500 turns and carries 2.0 A. Determine the flux density at its centre.

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    For a long solenoid B = μ₀NI/L, where L is the length of the solenoid. Substitute: B = (4π × 10⁻⁷ × 500 × 2.0)/(0.25). Numerator = 1.257 × 10⁻⁶ × 1000 = 1.257 × 10⁻³ T m. Then B = (1.257 × 10⁻³)/0.25 = 5.03 × 10⁻³, so B = 5.0 × 10⁻³ T along the axis, its direction given by curling the right hand round the turns in the sense of the conventional current with the thumb pointing to the north end. Check/Trap: N/L is turns per metre, here 2000 m⁻¹ — do not use the length of wire wound on the former. The radius does not appear, so a wider solenoid of the same length and turns gives the same central field. Doubling the current doubles B: the relationship is linear, not inverse square.

48. Worked problemSL & HLData booklet: Yes

    Determine the ratio of the electric force to the gravitational force between two protons, and comment on the result.

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    The separation r cancels because both forces obey an inverse square law. F_E/F_G = (ke²)/(Gm_p²). Numerator: 8.99 × 10⁹ × (1.60 × 10⁻¹⁹)² = 8.99 × 10⁹ × 2.56 × 10⁻³⁸ = 2.30 × 10⁻²⁸ N m². Denominator: 6.67 × 10⁻¹¹ × (1.67 × 10⁻²⁷)² = 6.67 × 10⁻¹¹ × 2.79 × 10⁻⁵⁴ = 1.86 × 10⁻⁶⁴ N m². Ratio = (2.30 × 10⁻²⁸)/(1.86 × 10⁻⁶⁴) = 1.2 × 10³⁶. Comment: the electric repulsion is about 10³⁶ times the gravitational attraction, so gravity is completely negligible inside atoms and nuclei, and the strong nuclear force, not gravity, is what holds nuclei together. Check/Trap: r must cancel — quoting a separation shows the structure was missed. The ratio is dimensionless, so no unit. Squaring 1.60 × 10⁻¹⁹ gives 2.56 × 10⁻³⁸, not 2.56 × 10⁻³⁷.

49. Exam technique/trapSL & HLData booklet: Yes

    Identify the substitution traps in Coulomb's law questions and state the safe method.

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    Trap: students substitute microcoulombs, nanocoulombs or centimetres directly. Convert every charge to coulombs and every distance to metres first, and write the converted values down — the mark scheme awards the substitution mark for the correct numbers. Trap: squaring only the number and not the power of ten, so (2.0 × 10⁻²)² becomes 4.0 × 10⁻² instead of 4.0 × 10⁻⁴. Trap: putting the signs into F = kq₁q₂/r² and then arguing about the sign of the answer. The safe method is to substitute magnitudes, obtain a positive force, then state attractive for unlike charges and repulsive for like charges. Trap: r is the centre-to-centre separation of the spheres, not the gap between their surfaces and not a radius. Final answer to 2 or 3 s.f. with the unit N.

50. Exam technique/trapSL & HLData booklet: Yes

    Explain the traps in using E = V/d and in quoting units of electric field strength.

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    Trap: applying E = V/d to a point charge or a curved field. It is valid only for the uniform field between parallel plates, well away from the edges, where fringing distorts the lines. For a point charge use E = kq/r². Trap: leaving d in millimetres, the single commonest arithmetic error, which inflates E by 10³. Trap: thinking the field is stronger nearer the positive plate — in a uniform field E is the same everywhere between the plates. Trap: unit confusion. N C⁻¹ and V m⁻¹ are identical and both are accepted; V alone or N alone scores zero. Direction: the field points from the positive plate to the negative plate, which is the direction of force on a positive charge, so an electron is pushed the opposite way.

51. Exam technique/trapSL & HL

    State what the mark scheme looks for when a question says sketch or draw the electric field pattern.

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    Sketch means a qualitatively correct diagram with labelled key features; draw usually implies more care with spacing and scale. Marks are given for: arrowheads on every line, pointing away from positive and towards negative charge; lines that never cross or touch; lines meeting a conductor surface at 90°; symmetry about the axis; and correct relative spacing, with lines closer where the field is stronger. For parallel plates draw straight, parallel, equally spaced lines in the central region and show the lines bulging outward at the edges. For two equal like charges show a null point where no line passes. For magnetic patterns, every line must be a closed loop with arrows from north to south outside the magnet. Traps: drawing too few lines to show a pattern, and drawing lines that stop in empty space.

52. Exam technique/trapSL & HLData booklet: Yes

    Explain the direction traps in magnetic field questions and how to avoid them.

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    Trap: using electron flow instead of conventional current. The right-hand grip rule uses conventional current, from + to − outside the source; if the question describes electron drift, reverse it first. Trap: applying a left-hand rule learned elsewhere. In the IB course use the right hand for the field around a wire and around a solenoid. Trap: misreading into-page and out-of-page symbols — a cross is a current going away from you (the tail of an arrow), a dot is one coming towards you. Trap: forgetting that the field circles the wire, so the direction at a point above the wire is opposite to that below it. Trap: treating field lines as if they started at the wire. Always add a clear 3D sketch with the wire, the current arrow and at least two field circles labelled with arrows.

53. Exam technique/trapSL & HLData booklet: Yes

    Outline an experiment to investigate how the magnetic flux density near a long straight wire varies with distance, including uncertainties and improvements.

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    Apparatus: a long vertical wire, a low-voltage high-current supply with ammeter and rheostat, a calibrated Hall probe on a clamp, and a metre rule or travelling microscope. Independent variable: perpendicular distance r from the wire axis. Dependent variable: B. Controlled: current I (check the ammeter as the wire warms) and probe orientation (rotate for maximum reading). Method: zero the probe with the current off, then record B for r from about 5 mm to 50 mm. Analysis: B = μ₀I/(2πr) predicts a straight line through the origin for B against 1/r, of gradient μ₀I/(2π). Limitations: the wire is not infinite, r is hard to measure from the axis, and stray fields are comparable with B. Improvements: reverse the current and average, use larger I, and take max and min gradients through the error bars.

54. Exam technique/trapSL & HLData booklet: Yes

    Outline how to determine μ₀ from a solenoid experiment and how to treat the uncertainties.

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    Apparatus: a long solenoid of known length L and turns N, a d.c. supply with ammeter and rheostat, and an axial Hall probe placed at the centre. Independent variable: current I; dependent variable: central flux density B; controlled: N/L, probe position and orientation, and temperature. Method: zero the probe with I = 0, then measure B for I up to about 3 A, reversing the current and averaging to cancel the Earth's field. Analysis: B = μ₀NI/L, so B against I is a straight line through the origin of gradient μ₀N/L, giving μ₀ = gradient × L/N. Uncertainties: N is exact, so combine the fractional uncertainties in the gradient and in L; use uncertainty in gradient = (max − min)/2 from error bars. A non-zero intercept signals a systematic error such as an un-zeroed probe, which repeats cannot remove.

55. DefinitionHL only

    Define electric potential at a point.

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    The electric potential V_e at a point is the work done per unit charge in bringing a small positive test charge from infinity to that point. SI unit volt (V) = J C⁻¹; potential is a scalar. It is defined as zero at infinity. Exam tip: three marking points — "work done per unit charge", "from infinity", and "small positive (test) charge". Around a positive charge V_e is positive and around a negative charge it is negative, so potentials from several charges add algebraically including sign, unlike fields which add as vectors. Contrast gravitational potential, which is always negative because gravity is only attractive.

56. DefinitionHL only

    Define electric potential energy of a system of two point charges.

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    The electric potential energy E_p of a system of two point charges is the work done in bringing the charges from infinite separation to their present separation. SI unit joule (J); a scalar, defined as zero at infinite separation. E_p = kq₁q₂/r, so it is positive for like charges (work must be done against repulsion) and negative for unlike charges (the system is bound). Exam tip: it is a property of the pair, not of one charge — do not say "the energy of the charge". The relation to potential is E_p = qV_e, and the work done moving a charge between two points is W = qΔV_e = ΔE_p.

57. DefinitionHL only

    Define an equipotential surface and state two properties relating it to field lines.

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    An equipotential surface is a surface on which every point has the same electric potential, so no work is done in moving a charge along it (W = qΔV_e = 0 since ΔV_e = 0). Properties: equipotentials are always perpendicular to field lines, and they are closest together where the field is strongest, because E = −ΔV_e/Δr. SI unit of the label on each line: volt. Exam tip: an equipotential can never cross another, and the surface of any conductor in electrostatic equilibrium is an equipotential. For an isolated point charge the equipotentials are concentric spheres (circles in 2-D) whose radii for equal potential steps get further apart as r increases.

58. DefinitionHL only

    Define potential gradient and state its relation to electric field strength.

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    The potential gradient at a point is the rate of change of electric potential with distance in a given direction, ΔV_e/Δr, with unit V m⁻¹. Electric field strength is the negative of the potential gradient: E = −ΔV_e/Δr. SI unit V m⁻¹ (= N C⁻¹); E is a vector, V_e a scalar. Exam tip: the minus sign is a marking point — the field points from high potential towards low potential, that is, in the direction of steepest decrease of V_e. On a V_e–r graph, E is minus the gradient of the tangent; a flat region (zero gradient) means zero field, which is why the field inside a hollow conductor is zero while its potential is constant and non-zero.

59. DefinitionHL only

    Compare gravitational and electric fields: state three similarities and three differences.

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    Similarities: both are inverse-square fields (g = GM/r², E = kQ/r²) with 1/r potentials; both are conservative, so work done is path-independent and potential is defined as zero at infinity; both use the same graphical tools — field lines, equipotentials and E = −ΔV/Δr. Differences: gravitational force is always attractive so gravitational potential is always negative, while electric force can be attractive or repulsive and V_e may be either sign; the source of a gravitational field is mass (only positive) whereas the source of an electric field is charge (two signs, so shielding and neutral points are possible); the electric interaction is about 10³⁶ times stronger for a pair of protons, so gravity dominates only for large neutral masses.

60. EquationHL onlyData booklet: Yes

    State the equation for the electric potential due to a point charge and define every symbol.

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    V_e = kq/r = q/(4πε₀r). V_e is the electric potential in volts (J C⁻¹); q is the source charge in coulombs, carrying its sign; r is the distance from the charge in metres; k = 8.99 × 10⁹ N m² C⁻². Valid for a point charge, and outside a uniformly charged conducting sphere with r measured from the centre; inside such a sphere V_e is constant and equal to its surface value kq/R. Data booklet: Yes. Note V_e ∝ 1/r while E ∝ 1/r². Common misuse: squaring r, or ignoring the sign of q so that potentials near a negative charge come out positive. Sanity check: 5.0 nC at 0.10 m gives V_e = 8.99 × 10⁹ × 5.0 × 10⁻⁹/0.10 = 4.5 × 10² V.

61. EquationHL onlyData booklet: Yes

    State the equation for the electric potential energy of two point charges and its relation to potential.

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    E_p = kq₁q₂/r = q₁q₂/(4πε₀r). E_p is the potential energy of the pair in joules; q₁ and q₂ are the charges in coulombs with their signs; r is their separation in metres. Zero is taken at infinite separation, so E_p > 0 for like charges and E_p < 0 for unlike (bound) charges. Data booklet: Yes. Related forms: E_p = qV_e and W = ΔE_p = qΔV_e. Note E_p ∝ 1/r while the force F ∝ 1/r², and F = −ΔE_p/Δr. Common misuse: dropping a negative sign and reporting a bound system as having positive energy. Sanity check: a proton and an electron 0.53 × 10⁻¹⁰ m apart give E_p = 8.99 × 10⁹ × (−(1.60 × 10⁻¹⁹)²)/5.3 × 10⁻¹¹ ≈ −4.3 × 10⁻¹⁸ J ≈ −27 eV.

62. EquationHL onlyData booklet: Yes

    State the equation relating electric field strength to potential gradient and give its uniform-field special case.

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    E = −ΔV_e/Δr. E is the field strength in V m⁻¹ (= N C⁻¹); ΔV_e is the change in potential in volts over the small displacement Δr in metres, taken along the direction of E. Valid generally; for a uniform field between parallel plates the gradient is constant, giving the familiar E = V/d. Data booklet: Yes. Rearranged: ΔV_e = −EΔr, so the potential drops by Ed on moving a distance d along the field. Common misuse: omitting the minus sign, or dividing the potential at a point by the distance from the charge — for a point charge that would give kq/r², which happens to be correct in magnitude only by coincidence of the 1/r form. Sanity check: potential falling 120 V over 4.0 mm gives E = 3.0 × 10⁴ V m⁻¹ directed from high to low potential.

63. EquationHL onlyData booklet: Yes

    State the equation for the work done in moving a charge in an electric field and state one key property.

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    W = qΔV_e, where W is the work done in joules, q is the charge moved in coulombs (with sign) and ΔV_e = V_final − V_initial is the change in electric potential in volts. Data booklet: Yes. Key property: the electrostatic field is conservative, so W depends only on the end points and not on the path — and W = 0 for any movement along an equipotential, or around any closed loop. Positive W means external work must be done; negative W means the field does the work and the kinetic energy increases. Common misuse: using the magnitude of the charge for an electron and getting the sign of the energy change wrong. Sanity check: moving +2.0 μC from 300 V to 800 V requires W = 2.0 × 10⁻⁶ × 500 = 1.0 × 10⁻³ J.

64. EquationHL onlyData booklet: Yes

    Compare the equation set for electric fields with that for gravitational fields.

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    Electric: F = kq₁q₂/r², E = kq/r², V_e = kq/r, E_p = kq₁q₂/r, E = −ΔV_e/Δr, W = qΔV_e. Gravitational: F = Gm₁m₂/r², g = GM/r², V_g = −GM/r, E_p = −Gm₁m₂/r, g = −ΔV_g/Δr, W = mΔV_g. Units: E in N C⁻¹ and g in N kg⁻¹; V_e in J C⁻¹ and V_g in J kg⁻¹. All are data booklet equations. The mapping is k ↔ G, q ↔ m, with an extra minus sign in the gravitational potential expressions because gravity is purely attractive. Common misuse: writing V_e = −kq/r by false analogy — the sign is already carried by q. Sanity check: for two protons F_electric/F_gravity = ke²/(Gm_p²) ≈ 1.2 × 10³⁶.

65. Graph/diagramHL only

    Sketch and interpret graphs of electric potential V_e against distance r for a positive and for a negative point charge.

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    Axes: V_e/V (y) against r/m (x). For +q: a positive curve V_e ∝ 1/r falling towards zero as r → ∞, asymptotic to both axes. For −q: the mirror image below the axis, rising towards zero from below. The gradient of the tangent gives −E, so the field is strongest where the curve is steepest; for +q the gradient is negative, so E is positive (radially outwards). The area under an E–r graph between two radii equals the potential difference. For a charged conducting sphere of radius R, V_e is constant at kq/R for r ≤ R (flat line, hence zero field inside) and then follows the 1/r curve. Doubling q doubles every ordinate without changing the shape.

66. Graph/diagramHL only

    Describe equipotential diagrams for a point charge, a dipole and parallel plates, and how field lines relate to them.

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    Point charge: concentric circles (spheres) labelled in volts; for equal potential steps the spacing increases with r, showing the field weakening. Dipole (+q and −q): distorted closed loops around each charge with the straight perpendicular bisector as the V_e = 0 equipotential; potential is positive near + and negative near −. Parallel plates: straight lines parallel to the plates, equally spaced, running from V at one plate to 0 at the other. In every case equipotentials cross field lines at 90°, and closer equipotentials mean a larger potential gradient and so a stronger field. Exam tip: label values on the lines, keep the spacing consistent with the field pattern, and never let two equipotentials intersect.

67. Graph/diagramHL only

    Sketch and interpret graphs of electric potential energy E_p against separation r for a pair of like charges and a pair of unlike charges.

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    Axes: E_p/J (y) against r/m (x). Like charges: a positive curve E_p = kq₁q₂/r decreasing towards zero as r → ∞ — no minimum, so the system is unbound and always repels. Unlike charges: a negative curve rising towards zero from below; the system is bound and energy equal to |E_p| must be supplied to separate the charges to infinity. In both cases the force is F = −ΔE_p/Δr, i.e. minus the gradient of the tangent, so a steep curve means a large force; the area under an F–r graph equals the change in E_p. Increasing the magnitude of either charge scales the whole curve vertically. This is the shape used for alpha-particle closest approach: set ½mv² = kQq/d.

68. Graph/diagramHL only

    Explain how a graph is used to determine the charge on a sphere from potential measurements, and compare the potential graphs for gravitational and electric fields.

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    Measure V_e with a probe at several distances r from the centre of a charged sphere. Plotting V_e against r gives a curve; instead plot V_e against 1/r (x-axis m⁻¹) to get a straight line through the origin of gradient kq, so q = gradient/k, with the uncertainty in q from the max/min gradient lines. Comparison: the electric V_e–r graph lies above the axis for a positive charge and below for a negative one, whereas the gravitational V_g–r graph is always negative, rising asymptotically to zero — reflecting that gravity is purely attractive. Both have the same 1/r magnitude dependence and both take zero at infinity, so the two curves are identical in shape apart from sign.

69. Concept/explainHL onlyData booklet: Yes

    Explain what is meant by the electric potential at a point and why it is a scalar.

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    • The electric potential at a point is the work done per unit charge in bringing a small positive test charge from infinity to that point
    • For a point charge V_e = kq/r, and the zero of potential is taken at infinity, where the field is zero
    • Potential is a scalar, so potentials due to several charges add algebraically, each keeping the sign of its own charge
    • Near a positive charge V_e is positive and falls to zero at infinity; near a negative charge V_e is negative and rises to zero
    • The unit is the volt, J C⁻¹
    • V_e can be zero at a point where E is not zero, for example midway between equal and opposite charges
    • Exam tip: never resolve potentials into components.
70. Concept/explainHL onlyData booklet: Yes

    Explain the sign of the electric potential energy of a pair of charges and what it means physically.

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    • For two point charges E_p = kq₁q₂/r, with the zero taken when the charges are infinitely far apart
    • Like charges give a positive E_p: work had to be done against repulsion to assemble them, and releasing them converts that energy to kinetic energy as they fly apart
    • Unlike charges give a negative E_p: the field did the work in bringing them together, so the pair is bound and energy must be supplied to separate them to infinity
    • The magnitude of E_p falls as 1/r, more slowly than the force, which falls as 1/r²
    • E_p is a scalar measured in joules, and for a group of charges you sum over every pair
    • Exam tip: the negative sign is not an error to be dropped; it signals a bound system.
71. Worked problemHL onlyData booklet: Yes

    Charges of +6.0 nC and −2.0 nC are fixed in place. Point P is 0.30 m from the first and 0.20 m from the second. Determine the potential at P and the work needed to bring a +5.0 nC charge from infinity to P.

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    Potential is a scalar, so add the contributions algebraically with signs, using V_e = kq/r. From the positive charge: V₁ = (8.99 × 10⁹ × 6.0 × 10⁻⁹)/0.30 = 53.94/0.30 = 180 V. From the negative charge: V₂ = (8.99 × 10⁹ × (−2.0 × 10⁻⁹))/0.20 = −17.98/0.20 = −89.9 V. Total: V_e = 180 − 89.9 = 90 V (2 s.f.). Work: W = qΔV_e = 5.0 × 10⁻⁹ × 89.9 = 4.5 × 10⁻⁷ J, and it is positive, so external work must be done pushing the positive charge in. Check/Trap: do not resolve into components — that is for fields, not potentials. The distances used are the actual distances from each charge to P, not the separation of the charges. Since infinity is the zero of potential, ΔV_e equals V_e at P.

72. Worked problemHL onlyData booklet: Yes

    Two charges of +2.0 μC and +3.0 μC are held 0.15 m apart. Determine their electric potential energy, and the kinetic energy released if they are freed and move to a separation of 0.30 m.

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    Use E_p = kq₁q₂/r with the zero at infinite separation. Initial: E_p1 = (8.99 × 10⁹ × 2.0 × 10⁻⁶ × 3.0 × 10⁻⁶)/0.15 = (5.394 × 10⁻²)/0.15 = 0.36 J. At double the separation E_p ∝ 1/r, so E_p2 = 0.36/2 = 0.18 J. By conservation of energy the loss in potential energy appears as kinetic energy: ΔE_k = 0.36 − 0.18 = 0.18 J shared between the two charges according to momentum conservation. Check/Trap: E_p is positive because both charges are positive, and it decreases as they separate — a positive E_p that grew on separation would be a sign error. Do not use the field formula with r²; potential energy falls as 1/r. To separate them to infinity you would have to supply the full 0.36 J.

73. Worked problemHL onlyData booklet: Yes

    The potential near an isolated charge is 180 V at r = 0.10 m and 90 V at r = 0.20 m. Estimate the field strength at r = 0.15 m from the gradient, and compare with the exact value.

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    Gradient estimate: E = −ΔV_e/Δr = −(90 − 180)/(0.20 − 0.10) = 90/0.10 = 9.0 × 10² V m⁻¹, directed radially outward since V_e falls with r. Exact value: from V_e = kq/r, kq = 180 × 0.10 = 18 V m, so q = 18/(8.99 × 10⁹) = 2.0 nC. Then E = kq/r² = 18/(0.15)² = 18/0.0225 = 8.0 × 10² V m⁻¹. Check/Trap: the chord gradient overestimates by about 12 % because V_e against r is a curve, not a straight line; the tangent at r = 0.15 m gives the exact answer. The minus sign only tells you the field points down the potential gradient — quote a positive magnitude with a stated direction. Note V_e ∝ 1/r halves when r doubles, whereas E ∝ 1/r² quarters.

74. Worked problemHL onlyData booklet: Yes

    An alpha particle of kinetic energy 5.0 MeV is fired head-on at a gold nucleus (Z = 79). Determine the distance of closest approach.

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    At closest approach the alpha momentarily stops, so all its kinetic energy has become electric potential energy: E_k = kq₁q₂/d. Charges: alpha q₁ = 2e = 3.20 × 10⁻¹⁹ C; gold q₂ = 79e = 79 × 1.60 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷ C. Energy: E_k = 5.0 × 10⁶ × 1.60 × 10⁻¹⁹ = 8.0 × 10⁻¹³ J. Rearrange: d = kq₁q₂/E_k = (8.99 × 10⁹ × 3.20 × 10⁻¹⁹ × 1.264 × 10⁻¹⁷)/(8.0 × 10⁻¹³) = (3.64 × 10⁻²⁶)/(8.0 × 10⁻¹³) = 4.5 × 10⁻¹⁴ m. Check/Trap: convert MeV to joules before use. The answer is larger than a nuclear radius of about 7 × 10⁻¹⁵ m, so the alpha never reaches the nucleus and the electrostatic model is valid. Assume the heavy nucleus stays at rest; use E_p = kq₁q₂/r, not the 1/r² force law.

75. Exam technique/trapHL onlyData booklet: Yes

    Explain the traps in electric potential questions and the correct sign conventions.

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    Trap: resolving potentials into components. V_e is a scalar, so add contributions algebraically, each carrying the sign of its own charge. Trap: confusing V_e = 0 with E = 0. Midway between equal and opposite charges the potential is zero but the field is a maximum; between equal like charges the field is zero but the potential is not. Trap: dropping the minus sign in E = −ΔV_e/Δr, or reporting a negative field magnitude — the sign only records that the field points from high to low potential. Trap: using r² in V_e = kq/r or in E_p = kq₁q₂/r; only forces and field strengths carry the square. Trap: sign of work. W = qΔV_e is the work done by an external agent; if the answer is negative, the field did the work and the charge lost potential energy.

76. Exam technique/trapHL only

    State the command-term expectations for field questions and the phrases that lose marks.

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    State: give the answer alone, for example the field inside the conductor is zero. Outline: a brief account of the main points, no derivation. Describe: say what happens without a cause. Explain: give a causal chain, for example free electrons redistribute until their field cancels the applied field, hence E = 0. Determine: a numerical answer with working, unit and significant figures. Deduce: reach a conclusion showing the reasoning, for example the null point must lie nearer the smaller charge. Sketch: a labelled shape with correct features and axes, no scale needed. Suggest: apply the physics to an unfamiliar context. Compare and contrast: paired statements about both, in the same sentence. Phrases that lose marks: the field gets weaker because it spreads out (say flux crosses an area ∝ r²), charges cancel out, metal blocks the field, and potential is the energy at a point.

Practise this topic with exam-style questions: D.2 Electric and magnetic fields questions (SL) · D.2 Electric and magnetic fields questions (HL) · all flashcards

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