D. Gravitational, Electric, and Magnetic Fields · Gravitational, Electric and Magnetic FieldsElectric Charge and Coulomb's Law
D2.3 D2.5 D3.2 D2.1About 75 minutes
Learning goals- I can describe electric charge, its unit and its smallest free amount, and the ways objects become charged.
- I can state and use Coulomb's law, F = kq₁q₂/r², including its inverse-square behaviour.
- I can find the net electric force on a charge from several charges in one or two dimensions.
- I can compare the electric and gravitational forces between particles.
Words to know
electric charge (q)
A property of matter that causes electric forces. It comes in two kinds, positive and negative, and is measured in coulombs (C).
elementary charge (e)
The size of the charge on one proton or electron: e = 1.60 × 10⁻¹⁹ C.
law of conservation of charge
Charge cannot be created or destroyed; the total charge of an isolated system stays the same.
Coulomb's law
F = kq₁q₂/r²: the electric force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.
Coulomb's constant (k)
k = 8.99 × 10⁹ N·m²/C².
point charge
A charged object small compared with the distances involved, so it can be treated as if all its charge is at one point. A uniformly charged sphere acts like a point charge at its centre (from outside).
Two forces, one pattern
Rub a balloon on your hair and it sticks to the wall. That pull is the electric force, and it is astonishingly strong. The electric force holds atoms together, makes chemistry happen and stops your hand from passing through a table. Gravity, by comparison, is almost unimaginably weak: the electric force between a proton and an electron is about 10³⁹ times stronger than the gravitational force between them.
In 1785 Charles-Augustin de Coulomb used a torsion balance, the same kind of instrument Cavendish later used for gravity, and found that the electric force follows an inverse-square law, just like gravity.
Remember from Grade 9 (SNC1W, The Study of Electricity): like charges repel and unlike charges attract; objects become charged by friction, contact (conduction) and induction, by gaining or losing electrons; electrons move easily in conductors and hardly at all in insulators.
Electric charge
- Charge comes in two kinds: positive (protons) and negative (electrons). A neutral object has equal amounts.
- Charge is measured in coulombs (C). One coulomb is a very large charge. Static charges are usually microcoulombs (1 μC = 10⁻⁶ C) or nanocoulombs (1 nC = 10⁻⁹ C).
- Charge is conserved: when you rub a balloon on hair, electrons move from the hair to the balloon. The balloon gains exactly as much negative charge as the hair gains positive charge.
- Charge is quantized: every free charge is a whole-number multiple of the elementary charge, e = 1.60 × 10⁻¹⁹ C. So q = Ne, where N is the number of extra (or missing) electrons. You will see how this was discovered in Millikan's Oil Drop Experiment.
Coulomb's law
F = k q₁ q₂ / r² with k = 8.99 × 10⁹ N·m²/C²
- F is the size of the force on each charge (N). The two forces are equal and opposite (Newton's third law).
- q₁, q₂ are the charges (C); r is the distance between their centres (m).
- The force acts along the line joining the charges: repulsive for like charges, attractive for unlike charges.
To avoid sign confusion, calculate the size with the magnitudes of the charges, then decide the direction from "like repel, unlike attract". This works every time, in one or two dimensions.
You may also see k written as 1/(4πε₀), where ε₀ = 8.85 × 10⁻¹² C²/(N·m²) is the permittivity of free space. It is the same number.
Worked exampleExample 1: Two charges
A +3.0 μC charge and a −5.0 μC charge are 0.20 m apart. Find the force on each.
- Use magnitudes: F = kq₁q₂/r² = (8.99 × 10⁹)(3.0 × 10⁻⁶)(5.0 × 10⁻⁶)/(0.20)².
- F = (8.99 × 10⁹)(1.5 × 10⁻¹¹)/0.040 = 0.1349/0.040 = 3.37 N.
- Unlike charges attract.
3.4 N on each charge, each directed toward the other. Small charges a short distance apart already make a noticeable force.
Worked exampleExample 2: Net force in one dimension
Three charges lie on the x-axis: q₁ = +4.0 μC at x = 0, q₂ = −2.0 μC at x = 0.30 m and q₃ = +1.0 μC at x = 0.50 m. Find the net force on q₃.
- Force from q₁ on q₃: r = 0.50 m. F = (8.99 × 10⁹)(4.0 × 10⁻⁶)(1.0 × 10⁻⁶)/(0.50)² = 0.144 N. Both positive: repulsion, so it pushes q₃ to the right (+x).
- Force from q₂ on q₃: r = 0.20 m. F = (8.99 × 10⁹)(2.0 × 10⁻⁶)(1.0 × 10⁻⁶)/(0.20)² = 0.449 N. Unlike: attraction, so it pulls q₃ to the left (−x).
- Net: +0.144 N − 0.449 N = −0.306 N.
0.31 N to the left (toward q₂). Superposition works exactly as it did for gravity.
Forces on q₃: q₁ pushes it right (0.144 N) and q₂ pulls it left (0.449 N). The net force is 0.31 N to the left.Worked exampleExample 3: Net force in two dimensions
Charge A (+3.0 μC) is at the origin. Charge B (−4.0 μC) is 0.40 m east of A, and charge C (+5.0 μC) is 0.30 m north of A. Find the net electric force on A.
- From B: F_B = (8.99 × 10⁹)(3.0 × 10⁻⁶)(4.0 × 10⁻⁶)/(0.40)² = 0.674 N. Unlike charges attract: A is pulled east, toward B.
- From C: F_C = (8.99 × 10⁹)(3.0 × 10⁻⁶)(5.0 × 10⁻⁶)/(0.30)² = 1.50 N. Like charges repel: A is pushed south, away from C.
- The forces are perpendicular: F_net = √(0.674² + 1.50²) = 1.64 N.
- Direction: θ = tan⁻¹(1.50/0.674) = 66° south of east.
F_net ≈ 1.64 N [E 66° S].
Charge A is attracted east toward B and repelled south away from C. Adding the two forces tip to tail gives 1.64 N at 66° south of east.Electric force versus gravity
Compare the two laws:
- F_g = Gm₁m₂/r² and F_e = kq₁q₂/r². Same form: both inverse-square, both act along the line between centres, both have infinite range, both obey superposition.
- Mass plays the role in gravity that charge plays in electricity.
- Gravity only attracts. Electric forces attract or repel, because charge has two signs.
- k = 8.99 × 10⁹ is about 10²⁰ times larger than G = 6.67 × 10⁻¹¹.
In a hydrogen atom (r = 5.29 × 10⁻¹¹ m), F_e = 8.2 × 10⁻⁸ N while F_g = 3.6 × 10⁻⁴⁷ N, a ratio of about 2.3 × 10³⁹. So why does gravity rule the motion of planets? Because big objects are almost perfectly neutral: their positive and negative charges cancel, but masses never cancel. You will compare all three fields systematically in Comparing Fields and the Fundamental Forces.
Units and signs
Convert prefixes first: μC × 10⁻⁶, nC × 10⁻⁹, pC × 10⁻¹², and cm to m. Then use magnitudes of charge in F = kq₁q₂/r², and get the direction from a diagram.
Just for funThe physicist Richard Feynman liked to point out that if two people standing at arm's length each had just 1% more electrons than protons, the repulsion between them would be roughly comparable to the weight of the entire Earth. Being neutral is a very good idea.
Hands-on activityTest Coulomb's law (simulation and balloons)
Part A: simulation. Open the free PhET simulation "Coulomb's Law" (University of Colorado Boulder, phet.colorado.edu) in a browser.
- Set q₁ = +5 μC and q₂ = +5 μC. Record the force at separations of 2, 3, 4, 5, 6, 8 and 10 cm.
- Make a table of r, F and 1/r². Plot F against r (a curve), then F against 1/r². What shape is the second graph?
- Keep r = 4 cm. Double q₁, then double both charges. Record what happens to F.
- Use the slope of your F vs 1/r² graph (slope = kq₁q₂) to calculate k. Compare with 8.99 × 10⁹ N·m²/C².
Part B: at home. You need two round balloons, thread, tape, a ruler and a wool sweater or your hair (on a dry day).
- Blow up the balloons (a grown-up should supervise; keep balloon pieces away from young children, who can choke on them). Hang each from a 50 cm thread taped to the top of a door frame, side by side and touching.
- Rub both balloons equally on the sweater. Let them hang. Measure how far apart their centres settle.
- Rub them more. Does the separation grow?
What to look for: Part A: F against 1/r² is a straight line through the origin, which confirms the inverse-square law. Doubling one charge doubles F; doubling both makes it 4 times as large. Your k should be within a few percent of 8.99 × 10⁹. Part B: the balloons repel and hang apart; more charge gives a larger separation, because a larger electric force pushes the balloons farther apart until the sideways pull of each thread balances it.
Your turn
1Find the force between two 1.0 μC charges 1.0 m apart.
F = (8.99 × 10⁹)(1.0 × 10⁻⁶)²/(1.0)² = 8.99 × 10⁻³ N, repulsive.
2How many extra electrons are on an object with a charge of −1.0 μC?
N = q/e = 1.0 × 10⁻⁶/1.60 × 10⁻¹⁹ = 6.25 × 10¹² electrons.
3Two charges repel with 0.80 N. What is the force if the distance between them is doubled and one charge is tripled?
F ∝ q₁q₂/r²: (3)/(2)² = 3/4. New force = 0.75 × 0.80 = 0.60 N, still repulsive.
4At what distance would two 1.0 μC charges repel with a force of 1.0 N?
r = √(kq₁q₂/F) = √((8.99 × 10⁹)(1.0 × 10⁻¹²)/1.0) = 0.095 m (9.5 cm).
5Two identical small metal spheres carry +6.0 μC and −2.0 μC. They are touched together and then separated to 0.10 m. Find the force between them.
Total charge = +4.0 μC, shared equally: +2.0 μC each. F = (8.99 × 10⁹)(2.0 × 10⁻⁶)²/(0.10)² = 3.6 N, repulsive.
6Calculate the ratio of the electric force to the gravitational force between two protons. Does it depend on the distance?
F_e/F_g = ke²/(Gm_p²) = (8.99 × 10⁹)(1.60 × 10⁻¹⁹)²/((6.67 × 10⁻¹¹)(1.67 × 10⁻²⁷)²) = 1.24 × 10³⁶. It does not depend on distance, since both forces are inverse-square and r² cancels.
More practice: warm-up
1What is the value of the elementary charge?
2What happens to the electric force if the distance between two charges is tripled?
3Convert 25 nC to coulombs.
4Name one way the electric force is different from the gravitational force.
It can be attractive or repulsive (gravity only attracts); it is also far stronger between particles.
5What is the value of k, with units?
More practice: on your own
1A +2.0 μC charge is at x = 0 and a +8.0 μC charge is at x = 0.60 m. Where between them would a third charge feel no net force?
Set kq₁/x² = kq₂/(0.60 − x)²: 2/x² = 8/(0.60 − x)², so (0.60 − x)/x = 2, giving x = 0.20 m from the +2.0 μC charge. (The third charge's size and sign do not matter.)
2Three +2.0 μC charges sit at the corners of an equilateral triangle with sides 0.10 m. Find the net force on one charge.
Each pair: F = (8.99 × 10⁹)(2.0 × 10⁻⁶)²/(0.10)² = 3.6 N, repulsive. The two forces are 60° apart; the net is along the bisector, away from the triangle: 2(3.6)cos 30° = 6.2 N, directed away from the centre.
3An alpha particle (charge +2e) is 1.0 × 10⁻¹⁴ m from the centre of a gold nucleus (charge +79e). Find the repulsive force.
F = k(2e)(79e)/r² = (8.99 × 10⁹)(2)(79)(1.60 × 10⁻¹⁹)²/(1.0 × 10⁻¹⁴)² = 364 N. Huge for one tiny particle: its mass is only 6.6 × 10⁻²⁷ kg.
4Two point charges attract with a force of 5.0 × 10⁻³ N when 0.30 m apart. One is +4.0 × 10⁻⁸ C. Find the other.
q₂ = Fr²/(kq₁) = (5.0 × 10⁻³)(0.090)/((8.99 × 10⁹)(4.0 × 10⁻⁸)) = 1.25 × 10⁻⁶ C. Attraction means unlike charges: q₂ = −1.3 × 10⁻⁶ C (−1.3 μC).
5Explain why a charged balloon attracts a neutral wall.
The balloon's negative charge repels electrons in the wall's surface slightly away (the wall's molecules become polarized). The wall's nearer surface is slightly positive and the far side slightly negative. Since the force depends on 1/r², the attraction to the nearer positive side is stronger than the repulsion from the farther negative side, so there is a net attraction.
Challenge
1What equal charge would Earth and the Moon need for their electric repulsion to cancel their gravitational attraction?
Set kq²/r² = GM_EM_M/r²: q = √(GM_EM_M/k) = √((6.67 × 10⁻¹¹)(5.97 × 10²⁴)(7.35 × 10²²)/(8.99 × 10⁹)) = 5.7 × 10¹³ C. (r cancels.)
2Two small spheres, each of mass 2.0 g, hang from 0.50 m threads from the same point. They carry equal charges and hang with each thread at 10° to the vertical. Find the charge on each.
Separation r = 2(0.50 sin 10°) = 0.1736 m. For each sphere, T cos 10° = mg and T sin 10° = F_e, so F_e = mg tan 10° = (0.0020)(9.8)(0.1763) = 3.456 × 10⁻³ N. q = r√(F_e/k) = 0.1736 × √(3.456 × 10⁻³/8.99 × 10⁹) = 0.1736 × 6.200 × 10⁻⁷ = 1.1 × 10⁻⁷ C (0.11 μC) each, same sign.
3Four charges of +1.0 μC sit at the corners of a square with sides 0.20 m. Find the net force on one of them.
Two neighbours at 0.20 m: F = (8.99 × 10⁹)(10⁻¹²)/(0.04) = 0.225 N each, at 90°, sum along the diagonal = 0.225√2 = 0.318 N. Opposite corner at 0.20√2 m: F = (8.99 × 10⁻³)/(0.08) = 0.112 N along the diagonal. Net = 0.43 N, directed along the diagonal away from the centre.
Quiz
1. Two charges are 3.0 cm apart. If they are moved to 6.0 cm apart, the force between them becomes
2. A neutral object has 5.0 × 10¹² electrons removed. Its charge is
3. Which statement about the electric force between two point charges is correct?
4. Coulomb's law and Newton's law of gravitation are similar because both
5. Why do planets move under gravity rather than electric forces, even though the electric force is much stronger?
6. Two identical metal spheres with charges +8 μC and +2 μC touch and are separated. Each now has
7. The constant k in Coulomb's law has the value
Big idea (D2.3, D3.2): Coulomb's law has the same inverse-square form as gravity, but charge can be positive or negative, so electric forces attract or repel and can cancel.
How to help: Ask your teen to draw a quick sketch for every problem and mark the force directions before calculating. Encourage the habit of calculating sizes with positive numbers and getting direction from "like repel, unlike attract".
Common difficulties:
- Forgetting to convert μC to C and cm to m.
- Plugging in negative charges and then misreading the sign of the answer as a direction.
- Adding forces from different charges as plain numbers when they point in different directions.
- Squaring the distance incorrectly (0.20² = 0.040, not 0.40).
Safety: supervise balloon use with younger children nearby (choking hazard from broken balloons). The PhET simulation is free and needs only a browser.