Electrostatic vs. Gravitational Forces Revision Notes | A Level AQA Physics

1. The Surprising Scale of Forces Inside an Atom

Picture a hydrogen atom: one proton and one electron, separated by roughly $5.3 \times 10^{-11}$ m. Two forces act between them simultaneously — gravity pulling them together by mass, and the electrostatic force pulling them together by charge. Both follow an inverse-square law. Both depend on separation in exactly the same way.

So which one actually matters? The answer is striking. The electrostatic force between a proton and an electron is roughly $10^{39}$ times larger than the gravitational force between them. Gravity, the force that holds galaxies together, is essentially irrelevant inside an atom.

This comparison is a core AQA A Level requirement, and it rewards careful calculation. The key insight — that the ratio of the two forces is independent of separation — is the kind of analytical point that earns marks in extended-response questions.

2. The Two Force Laws Side by Side

Both forces share the same mathematical skeleton. Recognising this structure is the first step to comparing them efficiently.

Notice the structural parallel: both laws have a constant in the numerator multiplied by two properties of the objects, divided by $r^2$. This means when you form the ratio $F_E / F_G$, the $r^2$ terms cancel completely. The ratio is the same at any separation — a powerful result worth stating explicitly in exam answers.

3. Calculating the Force Ratio for a Proton and Electron

The standard AQA comparison uses a proton and an electron — the most common pair of subatomic particles in any atom. You need to know their key data, or be able to use values given in an exam question.

To find the ratio, form $F_E / F_G$ by dividing Coulomb's law by Newton's law. The $r^2$ cancels, leaving a clean expression involving only constants, charges, and masses. The step-by-step below walks through this calculation in full.

Worked example: Calculating the Ratio of Electrostatic to Gravitational Force

Calculate the ratio $F_E / F_G$ for a proton and an electron. Use: $k = 8.99 \times 10^9$ N m² C⁻², $G = 6.67 \times 10^{-11}$ N m² kg⁻², $e = 1.60 \times 10^{-19}$ C, $m_p = 1.67 \times 10^{-27}$ kg, $m_e = 9.11 \times 10^{-31}$ kg.

1. $$\frac{F_E}{F_G} = \frac{\frac{kQ_1Q_2}{r^2}}{\frac{Gm_1m_2}{r^2}} = \frac{kQ_1Q_2}{Gm_1m_2}$$ — The $r^2$ terms cancel because both laws are inverse-square. This confirms the ratio is independent of separation — a key point to state in exam answers. [M1]
2. Numerator: $k \times e^2 = 8.99 \times 10^9 \times (1.60 \times 10^{-19})^2$ $(1.60 \times 10^{-19})^2 = 1.60 \times 10^{-19} \times 1.60 \times 10^{-19} = 2.56 \times 10^{-38}$ Numerator $= 8.99 \times 10^9 \times 2.56 \times 10^{-38} = 2.301 \times 10^{-28}$ N m² — Both charges have magnitude $e = 1.60 \times 10^{-19}$ C, so $Q_1 Q_2 = e^2$. Squaring: $1.60 \times 1.60 = 2.56$ and $10^{-19} \times 10^{-19} = 10^{-38}$. [M1]
3. Denominator: $G \times m_p \times m_e = 6.67 \times 10^{-11} \times 1.67 \times 10^{-27} \times 9.11 \times 10^{-31}$ First: $6.67 \times 1.67 = 11.14$, so $6.67 \times 10^{-11} \times 1.67 \times 10^{-27} = 11.14 \times 10^{-38} = 1.114 \times 10^{-37}$ Then: $1.114 \times 10^{-37} \times 9.11 \times 10^{-31} = 10.15 \times 10^{-68} = 1.015 \times 10^{-67}$ N m² — Multiply the three values carefully, keeping track of powers of ten. $10^{-11} \times 10^{-27} \times 10^{-31} = 10^{-69}$; combined with the coefficient gives $\approx 1.015 \times 10^{-67}$. [M1]
4. $$\frac{F_E}{F_G} = \frac{2.301 \times 10^{-28}}{1.015 \times 10^{-67}} = \frac{2.301}{1.015} \times 10^{-28-(-67)} = 2.27 \times 10^{39}$$ — Dividing the coefficients: $2.301 \div 1.015 \approx 2.27$. Subtracting exponents: $-28 - (-67) = +39$. So the ratio is approximately $2.3 \times 10^{39}$. [A1]
5. $$\frac{F_E}{F_G} \approx 2.3 \times 10^{39}$$ — The electrostatic force between a proton and an electron is approximately $2.3 \times 10^{39}$ times greater than the gravitational force between them. Gravity is completely negligible at this scale. This result holds at any separation because $r^2$ cancelled in step 1.

Answer: $F_E / F_G \approx 2.3 \times 10^{39}$ — the electrostatic force dominates by a factor of around $10^{39}$.

4. Why the Ratio Is So Large — A Physical Interpretation

The ratio $\approx 10^{39}$ is not a coincidence of units — it reflects a genuine physical asymmetry. The charge on a proton ($1.60 \times 10^{-19}$ C) is tiny in SI terms, yet the Coulomb constant $k \approx 9 \times 10^9$ N m² C⁻² is enormous. Meanwhile, the proton mass ($1.67 \times 10^{-27}$ kg) is vanishingly small, and $G = 6.67 \times 10^{-11}$ N m² kg⁻² is also tiny.

The net effect is that $k e^2$ (the electrostatic numerator) is vastly larger than $G m_p m_e$ (the gravitational numerator). Charge interacts through a much stronger fundamental interaction than mass does at these scales.

This has a direct consequence for atomic structure. Electrons are bound to nuclei by the electrostatic attraction, not by gravity. If gravity were the only force, atoms as we know them could not exist — the gravitational pull between an electron and a proton is far too weak to maintain a bound orbit at atomic scales.

5. Comparing the Two Forces — Key Similarities and Differences

AQA exam questions often ask you to compare these two forces directly. A strong answer addresses both similarities and differences with precise language. The table below gives you the framework.

One similarity that examiners frequently reward is the shared inverse-square form. Both forces decrease with the square of separation, which is why their ratio is constant. Stating this explicitly — and linking it to the cancellation of $r^2$ — demonstrates the analytical depth expected at A Level.

6. Exam Technique and Mark-Winning Strategies

Questions on this topic tend to fall into three categories: calculation of the ratio, explanation of why one force dominates, and comparison of the two force laws. Each requires a slightly different approach.

For calculation questions, the most efficient method is to write the ratio expression first, cancel $r^2$ immediately, then substitute. Avoid calculating $F_E$ and $F_G$ separately and then dividing — you risk arithmetic errors and lose the opportunity to demonstrate the cancellation, which is itself a mark-winning step.

For explanation questions using the command word 'explain', structure your answer as: observation + 'because' + scientific principle. For example: 'The electrostatic force dominates because the ratio $k e^2 / G m_p m_e$ is approximately $10^{39}$, meaning the gravitational force is negligible by comparison.' Avoid vague statements like 'charge is stronger than mass' — examiners want quantitative justification or reference to the constants.

The central lesson here is that the enormous difference in magnitude between these two forces — roughly $10^{39}$ — is not a quirk of the particular separation chosen. It is a fixed property of the charges and masses of the proton and electron, encoded in the ratio $k e^2 / G m_p m_e$. This is the quantitative foundation for understanding why electrostatic forces govern atomic structure while gravity operates at the scale of planets and galaxies.