F = k q² / r², where q = f × (M/A) × NA × Z × e is the net charge of one sphere. For your sliders F = N. For the book case (1 g of copper, f = 1 percent, r = 1 m) F = 1.74 × 10^15 N.
Problem statement
What would be the force between two copper spheres, each of mass 1 g, a distance of 1 m apart, if the total electron charge in each sphere differed from the total nuclear charge by one percent?
We assume each neutral copper atom has 29 protons in its nucleus and 29 electrons around it. This means the nuclear charge of a sphere is the number of atoms times 29 times e, and we use it to find the charge scale.
We assume only the electrons change. The nuclei stay as they are, so the sphere's net charge is one percent of its nuclear charge.
We assume each sphere acts like a point charge at its centre. This is true when the distance is much larger than the sphere's radius, and the live warnings tell you when it fails.
We assume the spheres sit in a vacuum, so no air can carry the charge away. We use this in Coulomb's law with k = 8.988 × 10^9 N m²/C².
Symbols: M is the mass of one sphere, A = 63.54 g/mol is the mass of one mole of copper, NA = 6.022 × 10^23 per mole is the number of atoms in a mole, Z = 29 is the number of electrons in a copper atom, e = 1.602 × 10^-19 C is the size of the charge of one electron, f is the imbalance as a fraction, r is the distance between the centres, and k is the Coulomb constant.
The big idea in plain words
Matter is almost perfectly neutral, because the electrons cancel the nuclei. So we first find how much charge is hiding in the nuclei of one gram of copper. It is huge. Then one percent of that huge number is still a huge charge. Putting that charge into Coulomb's law gives an enormous force. The answer tells us how well matter hides its electric charge.
The solution from one viewpoint
Reference frames do not apply. Both spheres are at rest, and an observer at rest with them sees the same Coulomb force.
First we count the atoms, because each atom brings 29 positive charges. The mass of one mole of copper is A grams and a mole holds NA atoms. So M grams hold M/A moles.
Next we find the total nuclear charge. Every atom has Z protons, and each proton carries the charge e. So the nuclei carry N Z e in total, and the electrons carry the same amount with the opposite sign when the sphere is neutral.
Now we find the net charge. The electrons differ from the nuclei by a fraction f of Q, so the sphere is left with a net charge q equal to f times Q.
Finally we use Coulomb's law, which says two point charges push or pull with a force equal to k times the product of the charges divided by the distance squared. Both spheres have the same charge, so the product is q squared, and like charges push apart.
This force is about kg-weight, which is why a one percent imbalance never happens in real matter. The page's warnings show this live.
Book answer: F = 1.74 × 10^15 N for 1 g of copper, f = 1 percent and r = 1 m. This page gives N for those values, which differs only by rounding. The book prints the charge as 4.298 × 10^2 C in one line and 4.398 × 10^2 C in the next. The second one gives its final 1.74 × 10^15 N, and this page gets 440.3 C with e = 1.602 × 10^-19 C.
Five methods
1. Force method
We want the force on one sphere from the other. Free-body diagram: the other sphere pushes this one away along the line joining them, with the Coulomb force F. Gravity between them is far smaller, G M²/r² = N. The working is the one above: count atoms, find Q, find q, then use F = k q² / r².
2. Energy method
Not suitable. No work is asked for, and the force at one distance is the thing we want. As a side result, the electric potential energy of the pair is k q² / r = J, and the force is the energy divided by r.
3. Graph method
Plot the force against the imbalance f on log-log axes. It is a straight line of slope 2, because F goes as f². Plot it against the distance and it is a line of slope −2. Read the answer from the point where your f and your r meet the lines. The graphs on the canvas show this.
4. Calculus method
Not suitable. There is nothing to integrate or differentiate in the answer. A sphere of uniform charge acts as a point charge, which comes from integrating over the sphere, and that is a standard result we use as given.
5. Check method
Units: k q² / r² has units (N m²/C²) × C² / m², which is N. Limits: if f goes to zero, q and F go to zero. If f doubles, F becomes four times larger. If r doubles, F falls to one quarter. Size: the charge C is enormous, because one coulomb is already a huge charge in everyday life, so the huge force is expected.
Fastest: the force method with the numbers above. Most reliable: the check method, because the f² and 1/r² tests catch a missing square.
What each slider does
| Control | Changes the answer? | What it changes |
|---|---|---|
| Mass M | Yes, as M squared | Atoms, Q, q, force, sphere radius, electron dots, warnings |
| Imbalance f | Yes, as f squared | q, force, missing electrons, surface field, dots |
| Distance r | Yes, as 1 over r squared | Force, spacing in the picture, point-charge warnings. Does not change q. |
| Target force (given-answer mode) | Sets the answer | Solves the imbalance f |
| Metal | Yes | Atoms per gram, Z, density, radius, force |
| Sign of charge | No, q is squared | Colour and signs |
| Zoom | No | Pixels per metre in the picture only |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Coulomb constant | k | 8.988 × 10^9 N m²/C² |
| Elementary charge | e | 1.602 × 10^-19 C |
| Avogadro number | NA | 6.022 × 10^23 per mol |
| Copper: molar mass, Z, density | A, Z, ρ | 63.54 g/mol, 29, 8960 kg/m³ |
| Air breakdown field | Eair | 3 × 10^6 V/m |
| Your metal now | A, Z | g/mol, |
Exam traps
- Using the charge of the whole sphere as if it were the net charge. The net charge is only f times the nuclear charge.
- Forgetting the factor Z. Each copper atom brings 29 electrons, not 1.
- Using the mass in kilograms with a molar mass in grams. Keep both in grams.
- Squaring only part of the charge in k q² / r², or forgetting to square the percentage as a fraction (1 percent is 0.01).
- Giving the force as a size without noting that like charges repel.