SL & HL Fields BETA D.2

Electric and Magnetic Fields

This topic explores the mechanics of electrostatic and magnetic forces: how electric charges and magnets establish vector fields in the space around them, how those fields are pictured with field lines, and (at HL) how electric potential, potential energy and equipotential surfaces describe the energy side of the story.

Part of IB Physics (2025-2030 syllabus) — Standard and Higher Level.

Items marked HL are Higher Level only — SL students can skip them.

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Key points

  • Electric charge is a conserved, quantized property; like charges repel and opposite charges attract, with forces calculated using Coulomb's inverse-square law.
  • The electric field strength (EE) at a point is a vector quantity defined as the force per unit positive test charge placed at that location, pointing away from positive charges and toward negative charges.
  • A uniform electric field is created between two parallel plates held at a potential difference, where the field strength is constant in magnitude and direction at all points.
  • Magnetic field lines form continuous closed loops that exit the north pole and enter the south pole of a magnet, representing the direction of force on a theoretical north monopole.
  • HLElectric potential (VeV_e) is a scalar quantity defined as the work done per unit positive test charge to bring it from infinity to a point in space.
  • HLEquipotential surfaces represent regions where the electric potential is constant; these surfaces are always perpendicular to electric field lines, meaning zero work is done when moving a charge along them.
  • HLThe electric field strength is the negative gradient of the potential (E=ΔVeΔrE = -\frac{\Delta V_e}{\Delta r}), so field lines point in the direction of steepest decrease in potential.

Subtopic by subtopic

Electric charge and Coulomb's law

Electric charge is a fundamental property of matter that comes in two types, positive and negative. Charge is quantized, meaning every observed charge is a whole-number multiple of the elementary charge e=1.60×1019 Ce = 1.60 \times 10^{-19}\ \text{C}, and it is conserved: charge can be transferred (for example by friction, when electrons move from one surface to another) but never created or destroyed.

Coulomb's law gives the force between two point charges:

F=q1q24πε0r2F = \frac{q_1 q_2}{4 \pi \varepsilon_0 r^2}

The force is proportional to the product of the charges, falls off with the inverse square of the separation, and acts along the line joining the charges. The constant k=14πε08.99×109 N m2 C2k = \frac{1}{4\pi\varepsilon_0} \approx 8.99 \times 10^9\ \text{N m}^2\ \text{C}^{-2}, where ε0\varepsilon_0 is the permittivity of free space.

  • Like charges repel; opposite charges attract.
  • Doubling the separation cuts the force to one quarter.

For a sense of scale: two 1 μC1\ \mu\text{C} charges held 10 cm10\ \text{cm} apart push on each other with about 0.9 N0.9\ \text{N}, roughly the weight of an apple. You must be able to apply Coulomb's law numerically and add the forces from several charges as vectors (superposition).

Electric field strength and field lines

An electric field is the region around a charged object in which another charge experiences a force. The electric field strength at a point is the force per unit charge on a small positive test charge placed there, with units N C1\text{N C}^{-1} (equivalently V m1\text{V m}^{-1}):

E=FqE = \frac{F}{q}

Because force is a vector, so is EE: it points away from positive charges and toward negative charges. Combining E=F/qE = F/q with Coulomb's law shows that the field of a point charge is radial and weakens with the inverse square of the distance.

Field lines are a visual map of the field:

  • They start on positive charges and end on negative charges.
  • The arrow gives the direction of the force on a positive charge.
  • Line density shows field strength: lines bunch together where the field is strong.
  • Field lines never cross, because the field has a single direction at each point.

An isolated positive point charge, for example, has lines radiating outward symmetrically in all directions. You should be able to sketch the patterns for single charges, pairs of like and unlike charges, and a charged sphere, and calculate net field strengths by vector addition.

Uniform fields between parallel plates

Two parallel conducting plates connected to a power supply create a uniform electric field in the gap: the field has the same magnitude and direction everywhere between the plates (ignoring fringing at the edges). The field points from the positive plate to the negative plate, and its strength is given by:

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

Here VV is the potential difference and dd is the plate separation. Field lines are drawn as equally spaced parallel lines running perpendicular to the plates.

A charge qq in the gap feels a constant force F=qEF = qE, so it undergoes constant acceleration, just like a projectile in a gravitational field. A charged particle entering the gap sideways therefore follows a parabolic path; this is how older oscilloscopes and inkjet printers steer charged particles.

A classic experimental setup balances the electric force against weight: a charged droplet hangs stationary between horizontal plates when qE=mgqE = mg.

You must be able to:

  • Calculate EE from the supply voltage and separation.
  • Find the force and acceleration on a charge in the gap.
  • Recognise that the field strength does not depend on where the charge sits between the plates.

Magnetic fields and field lines

Magnetic fields are produced by magnets and by moving charges: every electric current has a magnetic field looping around it. A magnetic field is a region in which a magnetic material, a current-carrying conductor, or a moving charge experiences a force.

Field lines visualize the field. Outside a bar magnet they emerge from the north pole and enter the south pole; inside the magnet they run from south back to north, so every line is a continuous closed loop. This is a key contrast with electric field lines: there are no magnetic monopoles, so magnetic field lines never start or stop on anything.

The arrow shows the direction a small compass needle (or a free north pole) would point, and the line density indicates the field strength.

You should be able to sketch the standard patterns:

  • A single bar magnet.
  • Two magnets attracting or repelling.
  • The concentric circles around a straight current-carrying wire (direction given by the right-hand grip rule).
  • The bar-magnet-like field of a solenoid.

Iron filings scattered around a magnet line up along the field and make the pattern visible, a simple demonstration worth remembering when describing field shapes.

Electric potential and potential energyHL

Electric potential VeV_e at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point. It is a scalar measured in volts (1 V=1 J C11\ \text{V} = 1\ \text{J C}^{-1}), which makes combining contributions from several charges easy: simply add the values, with no components needed. For a point charge:

Ve=q4πε0rV_e = \frac{q}{4 \pi \varepsilon_0 r}

Note the 1/r1/r dependence rather than 1/r21/r^2, and that the sign of VeV_e follows the sign of the charge.

The electric potential energy of two point charges is:

Ep=q1q24πε0rE_p = \frac{q_1 q_2}{4 \pi \varepsilon_0 r}

Do not confuse the two quantities: potential is energy per unit charge, while EpE_p is an energy measured in joules. When a charge qq moves between two points, the work done by the field is W=qΔVeW = q\Delta V_e; this is exactly how accelerating a particle through a potential difference gives it kinetic energy, as in an electron gun.

Field and potential are linked by E=ΔVeΔrE = -\frac{\Delta V_e}{\Delta r}, so a rapidly changing potential means a strong field. You must be able to calculate VeV_e and EpE_p for point-charge systems and use W=qΔVeW = q\Delta V_e in energy problems.

Equipotential surfacesHL

An equipotential surface joins all points that share the same electric potential. Because the potential does not change along the surface, zero work is done when a charge is moved anywhere on it, regardless of the path taken.

Around a point charge the equipotentials are concentric spheres; between parallel plates they are flat planes parallel to the plates. They behave like contour lines on a map: closely spaced equipotentials mean the potential is changing quickly, which signals a strong field.

The relationship between fields and potentials makes this precise. The electric field strength vector is the negative spatial gradient of the electric potential:

E=ΔVeΔrE = -\frac{\Delta V_e}{\Delta r}

This implies that electric field lines must always point in the direction of the steepest decrease in electric potential. Consequently, equipotential surfaces must cross electric field lines at exactly 9090^{\circ}: if there were a non-zero component of the electric field parallel to an equipotential surface, a force would act along the surface, and work would be required to move a charge along it, violating the definition of an equipotential.

You should be able to sketch equipotential patterns for point charges, charge pairs and parallel plates, read potential differences from them, and use the gradient relation to estimate field strength from equipotential spacing.

Formulae

F=q1q24πε0r2F = \frac{q_1 q_2}{4 \pi \varepsilon_0 r^2}

To calculate the electrostatic force of attraction or repulsion between two point charges separated by distance rr.

E=FqE = \frac{F}{q}

To determine the electric field strength when the force acting on a known charge is given.

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

To calculate the uniform electric field strength between two parallel conductive plates separated by distance dd.

Ve=kQrV_e = \frac{kQ}{r}HL

To calculate the absolute electric potential at a distance rr from a point charge QQ, where k=14πε0=8.99×109 N m2 C2k = \frac{1}{4\pi\varepsilon_0} = 8.99 \times 10^9\ \text{N m}^2\ \text{C}^{-2}.

Ep=q1q24πε0rE_p = \frac{q_1 q_2}{4 \pi \varepsilon_0 r}HL

To calculate the electrostatic potential energy of a two-charge system.

W=qΔVeW = q \Delta V_eHL

To calculate the work done on a charge, or the kinetic energy it gains, when it moves through a potential difference ΔVe\Delta V_e.

E=ΔVeΔrE = -\frac{\Delta V_e}{\Delta r}HL

To relate electric field strength to how rapidly the potential changes with position; the minus sign shows the field points towards decreasing potential.

Definitions

Electric Field Strength (EE)
The electrostatic force experienced per unit positive charge by a small test charge placed at that point in space (E=FqE = \frac{F}{q}).
Coulomb's Law
The electrostatic force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of their separation distance.
Magnetic Field
A region of space in which a magnetic material, a current-carrying conductor, or a moving charge experiences a force.
Electric Potential (VeV_e)HL
The work done per unit positive charge in bringing a small test charge from infinity to a specific point in an electric field.
Equipotential SurfaceHL
A continuous surface composed of points that all share the same electric potential, ensuring no work is done when a charge is moved along it.

Worked examples

1

Two point charges, q1=+3.0×106 Cq_1 = +3.0 \times 10^{-6}\ \text{C} and q2=6.0×106 Cq_2 = -6.0 \times 10^{-6}\ \text{C}, are placed in a vacuum at a separation distance of 0.15 m0.15\ \text{m}. Calculate the electrostatic force exerted on q1q_1 by q2q_2, and state its direction relative to q1q_1. (Use ε0=8.85×1012 F m1\varepsilon_0 = 8.85 \times 10^{-12}\ \text{F m}^{-1}).

  1. 1
    State Coulomb's Law: F=q1q24πε0r2F = \frac{|q_1 q_2|}{4 \pi \varepsilon_0 r^2}.
  2. 2
    Substitute values: F=(3.0×106 C)×(6.0×106 C)4π×(8.85×1012 F m1)×(0.15 m)2F = \frac{(3.0 \times 10^{-6}\ \text{C}) \times (6.0 \times 10^{-6}\ \text{C})}{4 \pi \times (8.85 \times 10^{-12}\ \text{F m}^{-1}) \times (0.15\ \text{m})^2}.
  3. 3
    Calculate the denominator: 4π×8.85×1012×0.02252.502×1012 F m1 m24 \pi \times 8.85 \times 10^{-12} \times 0.0225 \approx 2.502 \times 10^{-12}\ \text{F m}^{-1}\ \text{m}^2.
  4. 4
    Divide the numerator product by the denominator: F=1.8×10112.502×10127.2 NF = \frac{1.8 \times 10^{-11}}{2.502 \times 10^{-12}} \approx 7.2\ \text{N}.
  5. 5
    Determine direction: Since the charges are of opposite signs, the force is attractive, pulling q1q_1 directly toward q2q_2.

Answer: 7.2 N7.2\ \text{N} directed towards q2q_2

2

Two horizontal parallel plates are separated by 12 mm12\ \text{mm} and connected to a 600 V600\ \text{V} supply. A small oil droplet between the plates carries a charge of q=+3.2×1019 Cq = +3.2 \times 10^{-19}\ \text{C}. Calculate (a) the electric field strength between the plates and (b) the magnitude of the electric force on the droplet.

  1. 1
    Convert the plate separation to SI units: d=12 mm=0.012 md = 12\ \text{mm} = 0.012\ \text{m}.
  2. 2
    Calculate the field strength: E=Vd=600 V0.012 m=5.0×104 V m1E = \frac{V}{d} = \frac{600\ \text{V}}{0.012\ \text{m}} = 5.0 \times 10^4\ \text{V m}^{-1}.
  3. 3
    Note that this field is uniform, so the droplet experiences the same field strength at every position between the plates.
  4. 4
    Calculate the force: F=qE=(3.2×1019 C)×(5.0×104 V m1)=1.6×1014 NF = qE = (3.2 \times 10^{-19}\ \text{C}) \times (5.0 \times 10^4\ \text{V m}^{-1}) = 1.6 \times 10^{-14}\ \text{N}.
  5. 5
    State the direction: the droplet is positively charged, so the force acts along the field, from the positive plate towards the negative plate.

Answer: E=5.0×104 V m1E = 5.0 \times 10^4\ \text{V m}^{-1}; F=1.6×1014 NF = 1.6 \times 10^{-14}\ \text{N} towards the negative plate

3

An alpha particle (q=+3.2×1019 Cq = +3.2 \times 10^{-19}\ \text{C}, m=6.6×1027 kgm = 6.6 \times 10^{-27}\ \text{kg}) is accelerated from rest through a potential difference of 5.0×103 V5.0 \times 10^3\ \text{V} created between two parallel plates. Calculate its final velocity.HL

  1. 1
    Use the work-energy relation: the change in kinetic energy equals the work done on the particle by the field: ΔEk=W=qΔVe\Delta E_k = W = q \Delta V_e.
  2. 2
    Calculate kinetic energy gained: Ek=(3.2×1019 C)×(5.0×103 V)=1.6×1015 JE_k = (3.2 \times 10^{-19}\ \text{C}) \times (5.0 \times 10^3\ \text{V}) = 1.6 \times 10^{-15}\ \text{J}.
  3. 3
    Equate kinetic energy to mechanical velocity: Ek=12mv2E_k = \frac{1}{2} m v^2.
  4. 4
    Rearrange to solve for velocity: v=2Ekmv = \sqrt{\frac{2 E_k}{m}}.
  5. 5
    Substitute values: v=2×1.6×1015 J6.6×1027 kg6.96×105 m s1v = \sqrt{\frac{2 \times 1.6 \times 10^{-15}\ \text{J}}{6.6 \times 10^{-27}\ \text{kg}}} \approx 6.96 \times 10^5\ \text{m s}^{-1}.

Answer: 7.0×105 m s17.0 \times 10^5\ \text{m s}^{-1}

Common mistakes

  • ×Treating electric field strength as a scalar when finding the net field from multiple charges. Remember to resolve fields into vectors and add them components-wise, rather than just adding their scalar magnitudes.
  • ×Drawing magnetic field lines starting or ending on surfaces like electric field lines do. Magnetic field lines must form continuous closed loops and never intersect.
  • ×Confusing electric potential (VeV_e) with electric potential energy (EpE_p). Potential is energy *per unit charge* (1 V=1 J C11\ \text{V} = 1\ \text{J C}^{-1}), whereas potential energy is measured in joules (J\text{J}).
  • ×Assuming the electric field strength (EE) varies between parallel plates. The field is completely uniform (constant magnitude and direction) at all positions between the plates, excluding edge effects.

Exam tips

  • Under the command term **describe**, when explaining field lines, make sure to explicitly state that the density of the lines represents the field magnitude and that the arrows indicate force direction.
  • Under the command term **sketch**, always ensure your electric field lines are drawn perpendicular to conductor surfaces and that equipotential surfaces (HL) are drawn orthogonal to those field lines.
  • When asked to **determine** work done on a charge, remember that no work is done when a charge moves along an equipotential line, regardless of path length or shape.
  • If a question asks you to **calculate** the force on a charge within parallel plates, verify that you use F=qEF = qE directly, where E=VdE = \frac{V}{d} is independent of position.

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