Fields (HL) — IB Physics HL HL Study Guide

For: IB Physics HL candidates sitting IB Physics HL.

Covers: Gravitational field strength and potential, escape velocity, electric fields and potential, equipotentials and field lines, cross-comparison of gravitational and electric fields, and HL-only orbital mechanics with Kepler's laws.

You should already know: IGCSE Physics, basic algebra and calculus.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Physics HL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Is Fields (HL)?

Fields are regions of space where a test object (mass for gravitational fields, charge for electric fields) experiences a non-contact force, without direct physical contact with the source of the field. This HL topic expands on SL field content with calculus-derived potential definitions, quantitative cross-field comparisons, and orbital mechanics applications, making up ~11% of total HL exam marks across Papers 1, 2, and 3 per IBO syllabus guidelines. Examiners frequently test derivation skills (e.g., escape velocity, Kepler's third law) and graphical interpretation of field lines and equipotentials in this topic.

2. Gravitational fields — strength, potential, escape velocity

Gravitational fields are produced by all objects with mass, and always exert attractive forces on other masses.

Key Definitions

1. Gravitational field strength ($g$): The force per unit mass acting on a small test mass placed at a point in the field, defined as: $$g=\frac{F_g}{m}$$ For a point source of mass $M$, field strength at distance $r$ from the center of the source follows the inverse square law: $$g=\frac{GM}{r^2}$$ Units are ${\text{N kg}}^{-1}$ (equivalent to ${\text{m s}}^{-2}$), and $g$ is a vector pointing towards the center of the source mass. $G=6.67\times 10^{-11}{\text{ N m}}^2{\text{ kg}}^{-2}$ is the universal gravitational constant.
2. Gravitational potential ($V_g$): The work done per unit mass to move a test mass from infinity (where potential is defined as zero) to a given point in the field. It is a scalar quantity: $$V_g=-\frac{GM}{r}$$ Units are ${\text{J kg}}^{-1}$. The negative sign indicates that gravity does work on the test mass as it moves from infinity to the point, so no external work input is required.
3. Escape velocity ($v_{esc}$): The minimum speed required for an object to escape the gravitational pull of a source mass without further propulsion, derived by equating the kinetic energy of the object to the magnitude of its gravitational potential energy at the surface: $$\frac{1}{2}m{v}_{esc}^2=\frac{GMm}{r}$$ Cancel the mass of the escaping object $m$ to get: $$v_{esc}=\sqrt{\frac{2GM}{r}}$$

Worked Example

Calculate the escape velocity from Mars, where $M=6.42\times 10^{23}\text{ kg}$ and radius $r=3.39\times 10^6\text{ m}$.

1. Substitute values into the escape velocity formula: $$v_{esc}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 6.42\times 10^{23}}{3.39\times 10^6}}$$
2. Simplify to get $v_{esc}\approx 5.0{\text{ km s}}^{-1}$.

3. Electric fields and potential

Electric fields are produced by charged objects, and exert forces on other charged objects, either attractive or repulsive depending on charge sign.

Key Definitions

1. Electric field strength ($E$): The force per unit positive test charge placed at a point in the field, defined as: $$E=\frac{F_e}{q}$$ For a point source of charge $Q$, field strength at distance $r$ from the center of the source follows the inverse square law: $$E=\frac{kQ}{r^2}$$ Units are ${\text{N C}}^{-1}$ (equivalent to ${\text{V m}}^{-1}$), and $E$ is a vector pointing away from positive charges and towards negative charges. $k=\frac{1}{4\pi {ϵ}_0}\approx 9.0\times 10^9{\text{ N m}}^2{\text{ C}}^{-2}$ is Coulomb's constant.
2. Electric potential ($V_e$): The work done per unit positive test charge to move a test charge from infinity to a given point in the field. It is a scalar quantity: $$V_e=\frac{kQ}{r}$$ Units are volts ($\text{V}={\text{J C}}^{-1}$). Potential is positive around positive charges and negative around negative charges, as external work is required to move a positive test charge towards a positive source.

Worked Example

Find the electric potential 0.2 m from a $-4\mu \text{C}$ point charge.

1. Substitute values into the potential formula, noting the negative charge: $$V_e=\frac{9.0\times 10^9\times (-4\times 10^{-6})}{0.2}=-1.8\times 10^5\text{ V}$$

4. Equipotentials and field lines

Field lines and equipotentials are graphical tools used to visualize field properties, and examiners frequently ask you to draw or interpret these representations.

Key Rules

1. Field lines: The tangent to a field line at any point gives the direction of the field at that point, and the density of field lines is proportional to the magnitude of field strength. Field lines never cross, as the field can only have one direction at a single point.
2. Equipotentials: Surfaces where the potential is constant. No work is done to move a test mass or charge along an equipotential surface, as there is no component of field force parallel to the surface.
3. Relation between field lines and equipotentials: Field lines are always perpendicular to equipotential surfaces at every point. For uniform fields (e.g., between two parallel charged plates), equipotentials are equally spaced parallel lines. For radial fields (e.g., around a point mass or charge), equipotentials are concentric spheres, spaced closer together near the source where field strength is higher.

Worked Example

Sketch the equipotentials around a positive point charge: draw 4 concentric circles centered on the charge, with spacing between circles increasing as distance from the charge increases, to reflect the inverse square falloff of potential. Label the potential values as decreasing as you move away from the charge, approaching zero at infinity.

5. Comparison of $g,E$ fields

Gravitational and electric fields share core structural similarities, but have key differences that drive their behavior at different scales.

Similarities

• Both follow the inverse square law for point sources, with field strength proportional to $1/r^2$.
• Potential is defined relative to infinity, with work done to move a test object equal to the product of the test object's property (mass/charge) and the change in potential.
• Field lines are always perpendicular to equipotential surfaces for both field types.

Differences

Worked Comparison

Calculate the ratio of gravitational force to electric force between a proton and electron in a hydrogen atom, where $m_p=1.67\times 10^{-27}\text{ kg}$, $m_e=9.11\times 10^{-31}\text{ kg}$, $e=1.6\times 10^{-19}\text{ C}$: $$\frac{F_g}{F_e}=\frac{G{m}_p{m}_e}{k{e}^2}\approx \frac{6.67\times 10^{-11}\times 1.67\times 10^{-27}\times 9.11\times 10^{-31}}{9\times 10^9\times (1.6\times 10^{-19}{)}^2}\approx 10^{-40}$$ Gravitational force is negligible at atomic scales, as expected.

6. Orbits and Kepler's laws (HL)

Orbital motion describes the movement of a smaller mass around a larger central mass (e.g., planets around the Sun, satellites around Earth) under gravitational attraction, and is an HL-only extension of gravitational field content.

Kepler's Three Laws

1. Law of Orbits: All planets move in elliptical orbits with the Sun at one of the two foci. Circular orbits are a special case of elliptical orbits, and most IB exam questions use circular orbits for simplicity.
2. Law of Areas: A line connecting the orbiting body to the central mass sweeps out equal areas in equal time intervals. This implies orbiting bodies move faster when closer to the central mass, a consequence of conservation of angular momentum.
3. Law of Periods: For circular orbits, the square of the orbital period $T$ is proportional to the cube of the orbital radius $r$ (distance from the center of the central mass). The derivation comes from equating gravitational force to centripetal force: $$\frac{GMm}{r^2}=m\frac{4{\pi }^{2}r}{T^2}$$ Cancel the orbiting mass $m$ and rearrange to get: $$T^2=\frac{4{\pi }^2}{GM}{r}^3$$ The constant of proportionality depends only on the mass of the central body $M$, so it is the same for all objects orbiting the same central mass.

Worked Example

Calculate the orbital period of a geostationary satellite orbiting Earth, where $M=5.97\times 10^{24}\text{ kg}$ and orbital radius $r=4.2\times 10^7\text{ m}$:

1. Substitute into Kepler's third law: $$T=\sqrt{\frac{4{\pi }^2{r}^3}{GM}}=\sqrt{\frac{4{\pi }^2\times (4.2\times 10^7{)}^3}{6.67\times 10^{-11}\times 5.97\times 10^{24}}}$$
2. Simplify to get $T\approx 86400\text{ s}=24\text{ hours}$, matching the rotation period of Earth, as required for geostationary orbits.

7. Common Pitfalls (and how to avoid them)

• Pitfall 1: Using height above the surface instead of distance from the center of the source in inverse square law formulas. Why it happens: Questions often give height above a planet's surface instead of total orbital radius. Correct move: Always add the radius of the central object to the given height to get the correct $r$ value.
• Pitfall 2: Treating potential as a vector and adding direction. Why it happens: Students confuse potential (scalar) with field strength (vector). Correct move: Potential has sign but no direction; simply add algebraic values for multiple sources, no need for vector addition.
• Pitfall 3: Mixing up escape velocity and orbital velocity formulas. Why it happens: Both depend on $GM/r$, but have different energy conditions. Correct move: Remember $v_{orbit}=\sqrt{\frac{GM}{r}}$ (KE = ½ |PE| for stable orbit) and $v_{esc}=\sqrt{2}{v}_{orbit}$ (KE = |PE| for escape).
• Pitfall 4: Applying the same Kepler's third law constant to orbits around different central bodies. Why it happens: Students forget the constant depends only on the central mass. Correct move: Recalculate the proportionality constant if the central body changes (e.g., switching from orbiting Earth to orbiting Mars).
• Pitfall 5: Drawing field lines parallel to equipotential surfaces. Why it happens: Misremembering the work rule for equipotentials. Correct move: Field lines are always perpendicular to equipotentials, as no work is done moving along an equipotential, so no field component can exist parallel to the surface.

8. Practice Questions (IB Physics HL Style)

Question 1

A planet has twice the mass of Earth and 1.5 times the radius of Earth. Earth's surface gravitational field strength is $9.81{\text{ m s}}^{-2}$ and Earth's escape velocity is $11.2{\text{ km s}}^{-1}$. (a) Calculate the gravitational field strength at the surface of the planet. (b) Calculate the escape velocity from the surface of the planet.

Solution

(a) Use the ratio method for $g=GM/r^2$: $$\frac{g_p}{g_e}=\frac{M_p{r}_e^2}{M_e{r}_p^2}=\frac{2M_e{r}_e^2}{M_e(1.5r_e{)}^2}=\frac{2}{2.25}\approx 0.889$$ $$g_p=0.889\times 9.81\approx 8.72{\text{ m s}}^{-2}$$

(b) Use the ratio method for $v_{esc}=\sqrt{2GM/r}$: $$\frac{v_p}{v_e}=\sqrt{\frac{M_p{r}_e}{M_e{r}_p}}=\sqrt{\frac{2M_e{r}_e}{M_{e}1.5r_e}}=\sqrt{\frac{2}{1.5}}\approx 1.155$$ $$v_p=1.155\times 11.2\approx 12.9{\text{ km s}}^{-1}$$

Question 2

Two point charges, $+3\mu \text{C}$ and $-5\mu \text{C}$, are placed 0.8 m apart in a vacuum. (a) Find the distance from the $+3\mu \text{C}$ charge to the point between the two charges where electric potential is zero. (b) Is the electric field strength zero at this point? Justify your answer.

Solution

(a) Let $x$ be the distance from the $+3\mu \text{C}$ charge, so $0.8-x$ is the distance from the $-5\mu \text{C}$ charge. Set total potential to zero: $$\frac{k\times 3\times 10^{-6}}{x}+\frac{k\times (-5)\times 10^{-6}}{0.8-x}=0$$ Cancel $k$ and $10^{-6}$, rearrange: $$\frac{3}{x}=\frac{5}{0.8-x}⟹2.4-3x=5x⟹x=0.3\text{ m}$$

(b) No, electric field is not zero. The field from the $+3\mu \text{C}$ charge points away from the positive charge, towards the $-5\mu \text{C}$ charge (to the right). The field from the $-5\mu \text{C}$ charge points towards the negative charge (also to the right). The two vectors add, so they cannot cancel out.

Question 3

A moon orbits a gas giant with a period of 18 Earth days, at an orbital radius of $3.2\times 10^6\text{ km}$. A second moon orbits the same gas giant at an orbital radius of $9.6\times 10^6\text{ km}$. Calculate the orbital period of the second moon in Earth days, using Kepler's third law.

Solution

Use the proportionality $T^2\propto r^3$, so $\frac{T_1^2}{r_1^3}=\frac{T_2^2}{r_2^3}$: $$T_2=T_1\times {\left(\frac{r_2}{r_1}\right)}^{3/2}=18\times {\left(\frac{9.6\times 10^6}{3.2\times 10^6}\right)}^{1.5}=18\times 3^{1.5}$$ $$T_2=18\times 5.196\approx 93.5\text{ Earth days}$$

9. Quick Reference Cheatsheet

10. What's Next

Mastery of Fields HL content is a prerequisite for multiple later sections of the IB Physics HL syllabus, including electromagnetism (where electric fields combine with magnetic fields to produce Lorentz forces on moving charges), nuclear physics (calculating electrostatic potential barriers for fusion reactions), and astrophysics (applying orbital mechanics to exoplanet detection and dark matter mass calculations). The inverse square law and potential definitions you learned here also appear in Paper 3 Option D (Astrophysics) and Option C (Imaging) questions that use electric field configurations for particle detectors and lens systems.

If you struggle with any derivation steps, practice question solutions, or mark scheme interpretation for this topic, you can ask Ollie, our AI tutor, for personalized explanations and additional targeted practice problems anytime. You can also access more IB Physics HL study guides, full past paper walkthroughs, and timed self-assessment quizzes on the homepage to test your understanding ahead of your exams.