Coulomb's Law — Electrostatic Force Problems Solved
Coulomb's law is the first equation in electrostatics, and every later idea — electric field, potential, capacitance, the structure of the atom — is built on it. The formula is short. What costs students marks is the unit conversion, the squared distance, and above all the fact that force is a vector: with three charges you cannot simply add the two forces as numbers. This article works through each of those, with a full vector-addition problem solved step by step.
The law
In words: the force between two point charges at rest is proportional to the product of the charges and inversely proportional to the square of the distance between them. It acts along the line joining them — repulsive for like charges, attractive for unlike.
| Symbol | Meaning | SI unit |
|---|---|---|
| F | Magnitude of the force on each charge (they are equal and opposite) | newton, N |
| q₁, q₂ | The two point charges | coulomb, C |
| r | Distance between the charges | metre, m |
| k | Coulomb constant, 8.988 × 10⁹ (usually taken as 9 × 10⁹ in exams) | N·m²·C⁻² |
| ε₀ | Permittivity of free space, 8.854 × 10⁻¹² | C²·N⁻¹·m⁻² |
Useful conversions you will need in almost every problem: 1 μC = 10⁻⁶ C, 1 nC = 10⁻⁹ C, 1 cm = 10⁻² m. The charge on one electron or proton has magnitude e = 1.602 × 10⁻¹⁹ C.
In a medium other than vacuum, the force is reduced. If the medium has relative permittivity εr (also written K, the dielectric constant), then
For air εr is very close to 1, which is why school problems in air use the vacuum formula unchanged. For water near room temperature the static value is about 80.
Worked example 1 — the basic calculation
Question: Two point charges q₁ = +2 μC and q₂ = +3 μC are 30 cm apart in air. Find the force between them.
Step 1 — convert to SI. q₁ = 2 × 10⁻⁶ C, q₂ = 3 × 10⁻⁶ C, r = 30 cm = 0.30 m
Step 2 — square the distance. r² = (0.30)² = 0.09 m²
Step 3 — multiply the charges. q₁q₂ = (2 × 10⁻⁶)(3 × 10⁻⁶) = 6 × 10⁻¹² C²
Step 4 — put it together (using k = 9 × 10⁹):
F = (9 × 10⁹ × 6 × 10⁻¹²) / 0.09 = 0.054 / 0.09 = 0.60 N
Step 5 — state the direction. Both charges are positive, so the force is repulsive, along the line joining them.
Cross-check with the precise constant: k = 8.988 × 10⁹ gives 0.05393/0.09 = 0.599 N — the same to two significant figures ✔
Worked example 2 — distance and medium
(a) The same two charges are moved to 60 cm apart. Since F depends on 1/r², doubling r divides the force by 2² = 4: F = 0.60 / 4 = 0.15 N
Direct check: r² = (0.60)² = 0.36; F = 0.054/0.36 = 0.15 N ✔
(b) The original pair (30 cm apart) is immersed in water, for which εr ≈ 80:
Fwater = 0.60 / 80 = 0.0075 N = 7.5 mN
Water molecules polarise and partly screen the charges, so the force falls by a factor of about eighty. That is why ionic solids such as NaCl dissolve in water but not in petrol.
The superposition principle — why three charges need vectors
When more than two charges are present, the rule is simple to state and easy to get wrong:
Each pair acts as if the others were not there — a third charge never changes the force between the first two — but the results must be combined as vectors. The reliable routine is:
- Find the magnitude of each individual force using F = k|q₁q₂|/r².
- Decide each direction separately from the signs — like charges push apart, unlike charges pull together.
- Resolve each force into x and y components.
- Add all the x components; add all the y components.
- Combine: F = √(ΣFx² + ΣFy²), and direction tan θ = |ΣFy| / |ΣFx|.
Worked example 3 — three charges, full vector addition
Question: A charge q₃ = +1 μC sits at the origin O(0, 0). A charge q₁ = +3 μC is at A(0.30 m, 0) on the x-axis, and a charge q₂ = +4 μC is at B(0, 0.30 m) on the y-axis. Find the net force on q₃. (Take k = 9 × 10⁹, air.)
Step 1 — force on q₃ due to q₁. r = 0.30 m, r² = 0.09 m²;
q₃q₁ = (1 × 10⁻⁶)(3 × 10⁻⁶) = 3 × 10⁻¹² C²
F₁ = (9 × 10⁹ × 3 × 10⁻¹²)/0.09 = 0.027/0.09 = 0.30 N
Both charges are positive, so q₁ pushes q₃ away. As q₁ lies in the +x direction from the origin, this force points along −x.
Step 2 — force on q₃ due to q₂. r = 0.30 m again;
q₃q₂ = (1 × 10⁻⁶)(4 × 10⁻⁶) = 4 × 10⁻¹² C²
F₂ = (9 × 10⁹ × 4 × 10⁻¹²)/0.09 = 0.036/0.09 = 0.40 N, and since q₂ lies in
the +y direction, this force points along −y.
Step 3 — components. The two forces are already perpendicular, so no resolving is needed: ΣFx = −0.30 N, ΣFy = −0.40 N
Step 4 — magnitude. F = √( (−0.30)² + (−0.40)² ) = √(0.09 + 0.16) = √0.25 = 0.50 N
Step 5 — direction. tan θ = 0.40 / 0.30 = 1.3333 → θ = 53.13°. Both components are negative, so the resultant points into the third quadrant: 0.50 N at 53.13° from the −x axis, turned towards −y. Physically q₃ is pushed away from both neighbours, so it moves off diagonally.
Sanity check: 0.30 and 0.40 form a 3–4–5 right triangle, so 0.50 N is exactly what a scale drawing would give ✔. The plain sum 0.30 + 0.40 = 0.70 N is wrong — that would need the two forces to point the same way.
Now change one sign. If q₁ were −3 μC, its force becomes an attraction pulling q₃ towards A, that is along +x. The components become (+0.30, −0.40): the magnitude is still 0.50 N, but the direction is now 53.13° below the +x axis. Same size, completely different answer — which is why the direction step can never be skipped.
How strong is it? Electric force versus gravity
In a hydrogen atom the electron and proton sit about 5.29 × 10⁻¹¹ m apart, so r² = 2.798 × 10⁻²¹ m². The electrostatic attraction is FE = 8.988 × 10⁹ × (1.602 × 10⁻¹⁹)² / 2.798 × 10⁻²¹ = 8.24 × 10⁻⁸ N. The gravitational pull between the same pair (G = 6.674 × 10⁻¹¹, me = 9.109 × 10⁻³¹ kg, mp = 1.673 × 10⁻²⁷ kg) is only 3.63 × 10⁻⁴⁷ N — a ratio of about 2.3 × 10³⁹, and since r² cancels, that factor holds at any separation. This is why gravity is ignored in atomic and chemical problems, and why chemistry is, at bottom, electrostatics.
Common mistakes that cost marks
- Leaving charges in μC or nC. Forgetting the 10⁻⁶ on two charges makes the answer 10¹² times too large. Convert before substituting, every single time.
- Leaving the distance in centimetres. 30 in place of 0.30 changes r² by a factor of 10 000.
- Forgetting to square r. The most common single-mark loss in this chapter.
- Adding forces as numbers. In example 3 the plain sum 0.70 N is wrong; the vector answer is 0.50 N. Only add magnitudes when the forces are along the same line.
- Putting the minus sign of a negative charge into the formula. Use |q₁q₂| for the size, then decide attraction or repulsion from the signs separately — otherwise you get a "negative force" you cannot interpret.
- Applying the law to charged spheres that are close together. It is a point-charge law; spheres behave like point charges at their centres only while far apart compared with their size.
- Confusing k with ε₀. k = 1/(4πε₀) ≈ 9 × 10⁹, while ε₀ ≈ 8.854 × 10⁻¹².
Where this appears in exams
| Exam / subject | Typical use |
|---|---|
| CBSE/ICSE Class 12 Physics | Electric Charges and Fields — force between charges, superposition, equilibrium positions |
| JEE Main & Advanced | Multi-charge geometries, charge in equilibrium, force with a dielectric present |
| NEET Physics | Direct numerical substitution and inverse-square reasoning questions |
| Class 11 Chemistry | Why ionic bonds form, lattice energy trends, effect of a solvent on ion pairing |
Check your electrostatics answer instantly. Enter the two charges and their separation and the Coulomb's Law tool returns the force — so you can confirm each magnitude before building it into a longer superposition problem, the stage where one wrong number quietly ruins every later step.
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