Find the gravitational field strength 2.0 × 10^7\ m from the centre of a planet of mass 5.0 × 10^24\ kg (G = 6.67 × 10^-11). [2 marks]

WJEC Eduqas Eduqas 2021-style practice: Two point charges of +3.0\ nC and -3.0\ nC are separated by 0.040\ m in a vacuum. Calculate the magnitude of the electrostatic force between them. Take 14π_0 = 8.99 × 10^9\ N m^2\ C^-2.

Coulomb's law: F = 14π_0Q_1 Q_2r^2. F = (8.99 × 10^9)(3.0 × 10^-9)(3.0 × 10^-9)(0.040)^2 = (8.99 × 10^9)9.0 × 10^-181.6 × 10^-3. F = (8.99 × 10^9)(5.625 × 10^-15) = 5.1 × 10^-5\ N (attractive, since the charges are opposite). Markers reward Coulomb's law, correct powers of ten, and the force about 5.1 × 10^-5\ N (attractive).

EnglandPhysicsSyllabus dot point

How do the inverse-square laws for gravity and electrostatics compare, and how do field strength and potential describe them?

Electrostatic and gravitational fields: Newton's law of gravitation and Coulomb's law, gravitational and electric field strength, the inverse-square law, and gravitational and electric potential.

A focused answer to the Eduqas A-Level Physics Component 2 fields content, covering Newton's law of gravitation and Coulomb's law, gravitational and electric field strength, the inverse-square law, and gravitational and electric potential, treated side by side as Eduqas presents them.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Newton's law of gravitation gives the attractive force between masses, $F = \frac{GMm}{r^2}$ ; Coulomb's law gives the force between charges, $F = \frac{1}{4\pi\varepsilon_0}\frac{Q_1 Q_2}{r^2}$ . Both are inverse-square laws. Gravitational field strength is the force per unit mass, $g = \frac{F}{m} = \frac{GM}{r^2}$ ; electric field strength is the force per unit positive charge, $E = \frac{F}{Q}$ , equal to $\frac{V}{d}$ in a uniform field. Gravitational potential is the work done per unit mass to bring a mass from infinity, $V = -\frac{GM}{r}$ (always negative); electric potential is the work per unit positive charge, $V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$ . The crucial difference: gravity is always attractive, but electrostatic forces can attract or repel.

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What this dot point is asking

Eduqas wants you to state Newton's law of gravitation and Coulomb's law, recognise both as inverse-square laws, define gravitational and electric field strength, and define gravitational and electric potential, drawing out the parallels and the key difference between the two fields.

The answer

Newton's law and Coulomb's law

Field strength

The inverse-square law

Potential

Comparing gravitational and electric force on a scale

Compare the gravitational and electrostatic forces between a proton and an electron $5.3 \times 10^{-11}\ \text{m}$ apart (the hydrogen atom). Take $m_p = 1.67 \times 10^{-27}\ \text{kg}$ , $m_e = 9.11 \times 10^{-31}\ \text{kg}$ , $e = 1.6 \times 10^{-19}\ \text{C}$ , $G = 6.67 \times 10^{-11}$ and $\frac{1}{4\pi\varepsilon_0} = 8.99 \times 10^{9}$ .

step 1: Electrostatic force (Coulomb's law)

$F_E = (8.99 \times 10^{9})\dfrac{(1.6 \times 10^{-19})^2}{(5.3 \times 10^{-11})^2} = (8.99 \times 10^{9})\dfrac{2.56 \times 10^{-38}}{2.81 \times 10^{-21}} = 8.2 \times 10^{-8}\ \text{N}$ .

step 2: Gravitational force (Newton's law)

$F_G = (6.67 \times 10^{-11})\dfrac{(1.67 \times 10^{-27})(9.11 \times 10^{-31})}{(5.3 \times 10^{-11})^2} = 3.6 \times 10^{-47}\ \text{N}$ .

step 3: Compare

The ratio $\dfrac{F_E}{F_G} \approx 2 \times 10^{39}$ : the electrostatic force is about $10^{39}$ times stronger, which is why gravity is negligible inside atoms but dominates on astronomical scales (where charges cancel).

Examples in context

The inverse-square gravitational field governs the orbits of planets, moons and satellites and the escape speed from a planet. Coulomb's law underpins the structure of atoms, the forces in chemical bonds, and the operation of capacitors, particle accelerators and electrostatic precipitators that clean smoke. The deep parallel between the two fields is one of the unifying ideas of physics, while their key difference (gravity only attracts) explains why large-scale structure is shaped by gravity.

Try this

Q1. State Newton's law of gravitation. [1 mark]

Cue. The gravitational force between two point masses is $F = \frac{GMm}{r^2}$ , attractive and along the line joining them.

Q2. Find the gravitational field strength $2.0 \times 10^{7}\ \text{m}$ from the centre of a planet of mass $5.0 \times 10^{24}\ \text{kg}$ ( $G = 6.67 \times 10^{-11}$ ). [2 marks]

Cue. $g = \frac{GM}{r^2} = \frac{(6.67 \times 10^{-11})(5.0 \times 10^{24})}{(2.0 \times 10^{7})^2} = 0.83\ \text{N kg}^{-1}$ .

Q3. State the key difference between gravitational and electrostatic forces. [1 mark]

Cue. Gravity is always attractive; electrostatic forces can be either attractive or repulsive.

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20194 marksCalculate the gravitational field strength at the surface of a planet of mass

6.4 \times 10^{23}\ \text{kg}

and radius

3.4 \times 10^{6}\ \text{m}

. Take

G = 6.67 \times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}

Show worked answer →

Gravitational field strength at the surface: $g = \dfrac{GM}{r^2}$ .

$g = \dfrac{(6.67 \times 10^{-11})(6.4 \times 10^{23})}{(3.4 \times 10^{6})^2} = \dfrac{4.27 \times 10^{13}}{1.156 \times 10^{13}} = 3.7\ \text{N kg}^{-1}$ .

Markers reward $g = \frac{GM}{r^2}$ , correct substitution and squaring of the radius, and the field strength about $3.7\ \text{N kg}^{-1}$ (close to the value for Mars).

Eduqas 20214 marksTwo point charges of

+3.0\ \text{nC}

and

-3.0\ \text{nC}

are separated by

0.040\ \text{m}

in a vacuum. Calculate the magnitude of the electrostatic force between them. Take

\dfrac{1}{4\pi\varepsilon_0} = 8.99 \times 10^{9}\ \text{N m}^2\ \text{C}^{-2}

Show worked answer →

Coulomb's law: $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q_1 Q_2}{r^2}$ .

$F = (8.99 \times 10^{9})\dfrac{(3.0 \times 10^{-9})(3.0 \times 10^{-9})}{(0.040)^2} = (8.99 \times 10^{9})\dfrac{9.0 \times 10^{-18}}{1.6 \times 10^{-3}}$ .

$F = (8.99 \times 10^{9})(5.625 \times 10^{-15}) = 5.1 \times 10^{-5}\ \text{N}$ (attractive, since the charges are opposite).

Markers reward Coulomb's law, correct powers of ten, and the force about $5.1 \times 10^{-5}\ \text{N}$ (attractive).

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Sources & how we know this

Eduqas GCE AS/A Level Physics specification (A720QS) — WJEC Eduqas (2015)