Difficulty: Introductory | Prerequisites: Vector addition, trigonometry (sin, cos, Pythagorean theorem), SI unit prefixes (micro = 10⁻⁶)
This is the starting point for the entire Electricity and Magnetism sequence. Before you can understand electric fields, potential, circuits, or anything else in E&M, you need to know how charged objects push and pull on each other. Coulomb's Law is the force law for electric charges, the same way Newton's law of gravitation is the force law for masses. If you are comfortable with free-body diagrams and vector decomposition from mechanics, you already have the tools; you are simply swapping masses and G for charges and k.
Coulomb's Law gives the force between two point charges. The force is proportional to both charges and inversely proportional to the square of the distance between them. When more than two charges are present, you find the net force on any one charge by computing each pairwise force independently and adding them as vectors.
Coulomb's Law
The electrostatic force between two point charges is F = k |q₁| |q₂| / r², where k ≈ 8.99 × 10⁹ N·m²/C². The force is attractive if the charges have opposite signs, repulsive if same sign.
In simple terms, bigger charges and shorter distances mean stronger forces, and the relationship with distance is an inverse square, so doubling the distance cuts the force to a quarter.
Coulomb's constant (k)
k = 1 / (4πε₀) ≈ 8.99 × 10⁹ N·m²/C². Sometimes written as k_e. This is the proportionality constant in Coulomb's Law, analogous to G in gravitation.
Think of it as the "strength setting" for the electric force. It is large, which is why even microcoulomb charges at short distances produce forces of several newtons.
Superposition principle (forces)
The net force on a charge due to multiple other charges equals the vector sum of each individual pairwise Coulomb force, calculated independently. Each force is unaffected by the presence of the other charges.
In simple terms, you work out each force one at a time, then add them tip-to-tail as vectors. No force "blocks" or modifies any other.
Point charge
An idealised charged object whose size is negligible compared to the distances in the problem. All the charge is treated as concentrated at a single point.
Think of it as the "particle model" from mechanics, but for charge instead of mass.
Microcoulomb (μC)
1 μC = 10⁻⁶ C. A common unit in introductory E&M problems. Convert to coulombs before plugging into any formula.
The magnitude of the force between two point charges is:
F = k |q₁| |q₂| / r²
Direction: along the line joining the two charges. Opposite charges attract (force points toward the other charge); like charges repel (force points away).
Always convert charge to coulombs and distance to metres before calculating. A charge of −5.2 μC becomes −5.2 × 10⁻⁶ C.
Example from the homework: Q₁ = −5.2 μC, Q₂ = 2.6 μC, distance d = 1.6 m.
F = k × (5.2 × 10⁻⁶)(2.6 × 10⁻⁶) / (1.6)²
The charges have opposite sign, so the force is attractive.
When a problem states that two forces are equal in magnitude (F₁₂ = F₃₂), you set up the Coulomb expressions and solve for the unknown.
From the homework: three collinear charges where the force on Q₂ from Q₁ must equal the force on Q₂ from Q₃. Setting k |Q₁||Q₂| / d² = k |Q₃||Q₂| / (3d)² and solving gives Q₃ = 9Q₁ / 16.
The key move: k and |Q₂| cancel from both sides, leaving a clean ratio.
When charges are not collinear, you must resolve each force into x and y components. The procedure is the same as for any force problem in mechanics:
Sketch the geometry and identify the angle θ from the line joining the charges to your chosen axis.
Compute the magnitude using Coulomb's Law.
Resolve: Fₓ = F cos θ, F_y = F sin θ (or the reverse, depending on which angle you use).
Finding the angle and distance from geometry:
If charges are separated by horizontal distance a and vertical distance b, then r = √(a² + b²), sin θ = b / r, cos θ = a / r.
Example: a = 0.09 m, b = 0.12 m gives r = 0.15 m and θ = 53.13°.
Another example: a = 0.12 m, b = 0.05 m gives r = 0.13 m and θ ≈ 22.6°.
Compute each pairwise force vector separately, then sum all x-components and all y-components independently.
Worked pattern from the homework (Problem 2):
Three charges, with test charge q at some position.
Force from Q₂ on q: F₂ₒₙq = k |q| |Q₂| / r², then decompose into Fₓ and F_y using sin θ and cos θ.
Force from Q₁ on q: F₁ₒₙq = k |q| |Q₁| / r₁², purely along one axis (no decomposition needed if collinear on that axis).
Sum: Fₓ,net = Fₓ(from Q₂) + Fₓ(from Q₁), and likewise for y.
Sign discipline matters:
Attractive forces point toward the other charge; repulsive forces point away.
After decomposing, forces pointing left or down pick up a negative sign in the standard x-y convention.
In one homework example, F_y = −9.607 × cos θ = −8.86 N (negative because it points in the −y direction).
Q₁ = −6.6 μC, Q₂ = 5.3 mC, q = 5.1 μC, a = 9 cm, b = 12 cm, θ = 53.13°.
Force of Q₂ on q: magnitude from Coulomb's Law at distance r = 0.15 m, then Fₓ = F cos 53° and F_y = F sin 53°.
Force of Q₁ on q: along the horizontal axis only (distance = 0.09 m), so the full force is in x.
Net: Fₓ,net = Fₓ(Q₂) + Fₓ(Q₁), F_y,net = F_y(Q₂).
Coulomb's Law
F = k |q₁||q₂| / r²
where k = 8.99 × 10⁹ N·m²/C²
Vector components of a force at angle θ to the x-axis
Fₓ = F cos θ
F_y = F sin θ
Distance between two points separated by a and b
r = √(a² + b²)
Net force (superposition)
F_net,x = ΣFₓ,i
F_net,y = ΣF_y,i
|F_net| = √(F_net,x² + F_net,y²)
Coulomb's Law is the reason a photocopier works: charged toner particles are attracted to oppositely charged regions on the drum. It is also the basis for electrostatic precipitators in power-plant smokestacks, which use electric forces to pull particulate pollution out of exhaust gases.
Forgetting to convert μC to C. A charge of 4.8 μC is 4.8 × 10⁻⁶ C, not 4.8. Missing this factor of 10⁻⁶ throws the answer off by twelve orders of magnitude.
Treating Coulomb's Law as a scalar equation when charges are off-axis. The magnitude formula gives you the size of the force, but you still need to decompose into components. Skipping the vector step is the single most common error on multi-charge problems.
Confusing which angle to use. If θ is measured from the x-axis, then Fₓ = F cos θ and F_y = F sin θ. If you measure from the y-axis instead, the trig functions swap. Draw the triangle and label it before writing anything.
Dropping the sign from an attractive force. If Q₁ is negative and q is positive, the force on q points toward Q₁. That direction determines the sign of each component. The magnitude from Coulomb's Law is always positive; the signs come from the geometry.
⚠️ Nearly every exam in the first third of E&M has at least one Coulomb's Law problem requiring vector addition. The algebra is straightforward; the marks are lost on sign errors and unit conversions.
⚠️ Professors frequently test whether you can find an unknown charge given a force-balance condition (as in Problem 1). The trick is recognising that k and one of the charges cancel.
⚠️ Problems with three or more charges at the vertices of a triangle are extremely common. They test your ability to find r from the Pythagorean theorem, pick the correct angle, and decompose forces.
T/F: The Coulomb force between two charges is always repulsive.
False. It is repulsive for like charges and attractive for opposite charges.
T/F: If you double both charges, the force quadruples.
True. F is proportional to |q₁||q₂|, so doubling both multiplies the force by 2 × 2 = 4.
Fill in the blank: If the distance between two charges is tripled, the force becomes ______ of its original value.
One-ninth (1/9). Inverse-square law: (1/3)² = 1/9.
T/F: The superposition principle means the force between Q₁ and Q₂ changes when a third charge Q₃ is introduced.
False. Each pairwise force is independent. Q₃ adds its own force to the net total, but does not alter the Q₁–Q₂ interaction.
Q: Two charges, q₁ = +3.0 μC and q₂ = −6.0 μC, are 0.20 m apart. What is the magnitude of the force between them, and is it attractive or repulsive?
A: F = (8.99 × 10⁹)(3.0 × 10⁻⁶)(6.0 × 10⁻⁶) / (0.20)² = 4.05 N. The charges have opposite sign, so the force is attractive.
Q: Three charges lie on a line: Q₁ at x = 0, Q₂ at x = d, Q₃ at x = 3d. If Q₂ experiences equal magnitude forces from Q₁ and Q₃, express Q₃ in terms of Q₁ and d.
A: Set k|Q₁||Q₂|/d² = k|Q₃||Q₂|/(2d)². Cancel k and |Q₂|: |Q₃| = 4|Q₁|. (Note: in the homework variant where Q₃ is at x = 5d from Q₁, the ratio changes accordingly because the distance from Q₂ to Q₃ changes.)
Q: Charge q sits at the origin. Charge Q₁ is at (a, 0) and charge Q₂ is at (0, b). Write expressions for the x and y components of the net Coulomb force on q.
A: F from Q₁ is purely in the x-direction: Fₓ₁ = kqQ₁/a² (positive means repulsive to the left if charges are like-sign). F from Q₂ is purely in the y-direction: F_y₂ = kqQ₂/b². Net: Fₓ = Fₓ₁, F_y = F_y₂. The signs depend on the charge signs and which direction is positive.
Q: Two charges of +2.0 μC and −5.0 μC are placed at the corners of a right triangle with legs 3 cm and 4 cm. A third charge of +1.0 μC sits at the right-angle corner. What is the distance from the third charge to each of the other two, and what angle does the hypotenuse subtend at the right-angle corner?
A: The distances are 3 cm and 4 cm (the legs of the triangle). The hypotenuse is 5 cm. The angle at the right-angle corner between the two legs is 90°, so each force is along a leg and the components separate cleanly into x and y.
This material connects directly to electric fields (next topic): Coulomb's Law gives the force, and dividing by the test charge gives the electric field, E = F/q. The vector superposition method you practise here is identical to the one used for adding electric field contributions from multiple charges.
It also connects to gravitational force from mechanics. Both are inverse-square laws; the mathematics is nearly identical, but electric forces can be attractive or repulsive (gravity is only attractive) and are vastly stronger at the atomic scale.
Coulomb's Law, electric force, electrostatic force, point charge, superposition principle, vector addition of forces, inverse-square law, microcoulomb, Coulomb's constant, k = 8.99e9, force between charges, 2D force decomposition, net electric force, charge interaction, PHYS 212, University Physics E&M