Fe/Fg = k q² / (G m²) = (q/m)² / (4π ε0 G), and the distance cancels. Two electrons: . Two protons: . The forces are equal when q/m = sqrt(4π ε0 G) = C/kg. The book gives 4 × 10^42, 1 × 10^36 and 0.86 × 10^-10 C/kg. For your sliders Fe/Fg = .
Problem statement
Calculate the ratio of the electrostatic force to the gravitational force between two electrons, and between two protons. At what value of the specific charge q/m of a particle would these two forces become equal in size, for two identical particles?
We assume the particles are points in empty space. This means no other charges or masses matter, and the size of the particle does not enter the formulas.
We assume classical physics. At very small distances quantum effects take over, and the live warnings tell you where.
We assume the two particles are identical, with the same charge q and the same mass m. This means one pair of symbols is enough.
We assume we compare the sizes (absolute values) of the forces. The electric force repels like charges and gravity attracts, so their directions are opposite, but the ratio uses only the sizes.
Symbols: q is the charge, m the mass, r the distance between the particles, k = 1/(4π ε0) = 8.988 × 10^9 N m²/C², and G = 6.674 × 10^-11 N m²/kg² (the book calls it γ). The charge of one electron or proton has size e = 1.602 × 10^-19 C. Also m(electron) = 9.109 × 10^-31 kg and m(proton) = 1.6726 × 10^-27 kg.
The big idea in plain words
Both forces get weaker as the square of the distance. So when we divide one by the other, the distance disappears. What is left depends only on what the particle is made of: its charge and its mass. Even better, it depends only on the single number q/m, the charge per kilogram. Square it and multiply by a fixed constant, and you have the ratio.
The solution from one viewpoint
Reference frames do not apply. The particles are treated as at rest, and any frame at rest with them gives the same two forces.
First we write the two forces, because the ratio needs both. Coulomb's law gives the electric force and Newton's law of gravitation gives the gravity force.
Next we divide. Both have r² in the bottom, so r cancels, and that is why the ratio does not depend on the distance.
Now the two particles of the problem. For two electrons, q = 1.602 × 10^-19 C and m = 9.109 × 10^-31 kg. For two protons, the charge is the same size but the mass is 1836 times larger, so the ratio is 1836² = 3.37 million times smaller.
Finally we set Fe = Fg for identical particles. Then k q² = G m², so (q/m)² = G/k.
Compare this with an electron, whose q/m is 1.76 × 10^11 C/kg. That is 2 × 10^21 times the critical value, and squaring it gives the 4 × 10^42. For the particles you set above, q/m is C/kg.
Book answers: 4 × 10^42 for two electrons, 1 × 10^36 for two protons, and q/m = 0.86 × 10^-10 C/kg for equal forces. This page gives , and C/kg. The values differ only by rounding: the book keeps one figure and uses 1/(4π ε0) = 9 × 10^9.
Five methods
1. Force method
We want the ratio of the two forces on one particle, and the q/m where they are equal. Free-body diagram: the other particle pushes this one away with the electric force Fe, and pulls it in with gravity Fg. They act along the same line, in opposite directions. Newton's second law for one particle then reads m a = Fe − Fg, with outward as positive. If the forces are equal, a = 0, and each particle feels no net push.
Result: electrons , protons , equal forces at C/kg.
2. Energy method
Not suitable as the main method, because nothing moves and no energy is exchanged. It still works as a side check. The electric potential energy is k q²/r and the gravity potential energy is −G m²/r. Both go as 1/r, so the size of their ratio is the same number, k q² / (G m²) = . The side panel shows both energies.
3. Graph method
Plot both forces against distance on log-log axes. Each is a straight line of slope −2, so the lines are parallel. The vertical gap between them is the logarithm of the ratio, and it is the same at every distance. Read it straight from the first graph. The second graph plots the ratio against q/m: a straight line of slope 2, and the point where it crosses 1 is the answer to the last part.
4. Calculus method
Not suitable. There is nothing to integrate or differentiate in the answer. If you differentiate the two potential energies to get the forces, you only recover the formulas we started with.
5. Check method
Units: k/G has units kg²/C², and (q/m)² has units C²/kg², so the product has no units, as a ratio should. Scaling: the proton is 1836 times heavier, so its ratio must be 1836² = 3.37 × 10^6 times smaller than the electron's. Dividing by 3.37 × 10^6 gives , which matches. Limits: if q goes to 0 the ratio goes to 0, and if m goes to 0 it goes to infinity, as it should.
Fastest: the force method with the single variable q/m. Most reliable: the check method, because the scaling test catches a wrong mass or a missing square.
What each slider does
| Slider | Changes the answer? | What it changes |
|---|---|---|
| Charge q (in e) | Yes, as q squared | Fe, ratio, q/m, energies, arrow lengths |
| Mass m | Yes, as 1 over m squared | Fg, ratio, q/m, particle size, arrow lengths |
| Distance r | No, it cancels | Fe and Fg both change by the same factor, energies, spacing in the picture, warnings |
| Target Fe/Fg (given-answer mode) | Sets the answer | Solves the mass m for the chosen charge |
| Sign of charge | No, q is squared | Colour and sign in the picture |
| Zoom | No | Pixels per decade of distance in the picture only |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Coulomb constant | k = 1/(4π ε0) | 8.988 × 10^9 N m²/C² |
| Gravitational constant | G (book: γ) | 6.674 × 10^-11 N m²/kg² |
| Elementary charge | e | 1.602 × 10^-19 C |
| Electron mass | me | 9.109 × 10^-31 kg |
| Proton mass | mp | 1.6726 × 10^-27 kg |
| Your charge and mass now | q, m | C and kg |
Exam traps
- Putting a distance in the answer. It cancels, so the ratio has no r.
- Using the electron mass for the proton pair, or the other way round.
- Forgetting to square q or m. Both are squared in the ratio.
- Mixing 4π ε0 and k. Use one or the other: k = 1/(4π ε0).
- Calling the critical q/m the charge of a real particle. A real electron or proton has a q/m at least 10^21 times too big (electron) or 10^18 times too big (proton).
- Reporting a ratio of 4 × 10^42 without saying it is a pure number with no units.