Source: University Physics: Elec & Mag, UIUC
Tags: Coulomb's law, electric force, superposition, electric field, field lines, charge density, continuous charge distribution, PHY 212
Difficulty: Introductory to Intermediate Prerequisites: Vector addition, basic calculus (integration), Newton's third law.
This is where the course begins: how charged particles exert forces on one another, and how we describe the influence a charge has on the space around it. Coulomb's law gives you the force between two point charges. The electric field extends that idea to continuous distributions of charge, which is where integration comes in. If you are comfortable with vector addition and can set up a basic integral, you have the tools you need. Everything that follows in the course (Gauss's law, potential, capacitance) builds directly on these two topics.
Coulomb's law quantifies the electrostatic force between two point charges; it depends on the product of the charges and falls off as 1/r². The electric field is the force per unit charge at a point in space, and for distributions of charge you calculate it by integrating over infinitesimal charge elements. Superposition says you simply add up (vector sum) the contributions from every charge.
Coulomb's Law
The force between two point charges is F₁,₂ = k·q₁·q₂ / r₁,₂², directed along the unit vector r̂₁,₂ pointing from charge 1 to charge 2. Think of it as the electric version of Newton's law of gravitation, but it can attract or repel depending on the signs of the charges.
Superposition (of forces)
The net force on a charge is the vector sum of the individual forces from every other charge present. In simple terms, each charge acts independently; you add up all the pushes and pulls as vectors.
Electric Field (E)
The force per unit positive test charge at a point in space: E = F / q. Think of it as a map of how strongly, and in which direction, a small positive charge would be pushed if you placed it there.
Electric Field Lines
Visual representations of the electric field. The direction of a field line shows the direction of E; the density of lines (how closely packed they are) shows the magnitude of E. In simple terms, more crowded lines mean a stronger field.
Linear Charge Density (λ)
Charge per unit length, in C/m. Used when charge is spread along a line or wire.
Surface Charge Density (σ)
Charge per unit area, in C/m². Used when charge is spread over a surface.
Volume Charge Density (ρ)
Charge per unit volume, in C/m³. Used when charge fills a three-dimensional region.
Continuous Charge Distribution
Any arrangement where charge is spread out rather than concentrated at discrete points. The summation in superposition becomes an integral: E = ∫ k · dq / r² · r̂.
F₁,₂ = k · q₁ · q₂ / r² · r̂₁,₂, where k = 1 / (4πε₀) ≈ 8.99 × 10⁹ N·m²/C².
r̂₁,₂ is the unit vector pointing from charge 1 to charge 2.
The force acts along the line connecting the two charges.
Newton's third law still applies: the force on charge 1 from charge 2 is equal in magnitude and opposite in direction.
For multiple charges, the net force on charge 1 is: F₁ = F₂₁ + F₃₁ + F₄₁ + ...
Each pairwise force is computed independently using Coulomb's law, then all are added as vectors.
This is a vector sum, so you must resolve into components (x, y, z) before adding.
Defined as E = F / q₁, removing the dependence on the test charge.
For a single point charge Q: E = k · Q / r² · r̂.
For multiple point charges (superposition): E = Σᵢ k · Qᵢ / rᵢ² · r̂ᵢ.
The electric field is a property of space; it exists whether or not a test charge is present.
Direction of a field line = direction of E at that point.
Density of field lines = magnitude of E.
For a point charge: lines are radial, and their density falls off as 1/r².
Opposite charges: field lines run from the positive charge to the negative charge, curving between them.
Same-sign charges: field lines repel outward from both charges and never cross.
When charge is spread continuously, replace the sum with an integral: E = ∫ k · dq / r² · r̂.
The key step is expressing dq in terms of the geometry:
Line: dq = λ · dx (or λ · dl)
Surface: dq = σ · dA
Volume: dq = ρ · dV
r is the distance from the infinitesimal charge element dq to the point where E is being calculated.
A rod of length a with uniform linear charge density λ, evaluated at a point a perpendicular distance h from one end.
Set up: dq / r² = λ dx / [(a − x)² + h²], then resolve into x- and y-components.
The x-component integrates as: Eₓ(P) = λk ∫₀ᵃ (a − x) dx / [(a − x)² + h²]^(3/2).
Result: Eₓ = λk/h · [1 − h / √(h² + a²)].
The approach is always the same: choose coordinates, express dq and r, split into components, integrate.
Quantity | Formula |
|---|---|
Coulomb's law | F = k · q₁ · q₂ / r² |
Coulomb constant | k = 1/(4πε₀) ≈ 8.99 × 10⁹ N·m²/C² |
Permittivity of free space | ε₀ ≈ 8.85 × 10⁻¹² C²/(N·m²) |
Electric field (point charge) | E = k · Q / r² |
Superposition (E) | E = Σ k · Qᵢ / rᵢ² · r̂ᵢ |
Continuous distribution | E = ∫ k · dq / r² · r̂ |
Linear charge density | λ = Q / L |
Surface charge density | σ = Q / A |
Volume charge density | ρ = Q / V |
Coulomb's law is the basis for understanding electrostatic precipitators (used to remove soot and dust from industrial exhaust), and it governs how toner particles are attracted to paper in laser printers. Electric field analysis underpins the design of cathode-ray tubes, ink-jet printers, and the deflection plates in oscilloscopes.
Students often assume the electric field exists only where charges are present. The field permeates all of space around a charge, whether or not another charge is there to feel it.
Forgetting that Coulomb's law and superposition are vector equations. You cannot simply add magnitudes; you must resolve into components first.
Confusing the r in Coulomb's law (distance between two specific charges) with the r in a continuous-distribution integral (distance from dq to the field point). They refer to different distances and must be set up carefully for each problem.
Thinking that electric field lines can cross. They cannot; at any point in space the field has a single direction.
⚠️ Setting up the integral for continuous distributions is a very common exam problem. Practice identifying dq, r, and the correct limits of integration.
⚠️ Vector addition errors (forgetting to decompose into components, or adding magnitudes instead of components) are the most frequent source of lost marks.
⚠️ Know how field-line diagrams differ for like charges vs. opposite charges; sketching questions appear regularly.
⚠️ Remember that k = 1/(4πε₀). Problems may give you ε₀ instead of k.
Q: True or false – The electric force between two charges doubles if you double the distance between them.
A: False. The force goes as 1/r², so doubling the distance reduces the force by a factor of four.
Q: Fill in the blank – The electric field at a point in space is defined as E = ___ / ___.
A: F / q (force on a test charge divided by the test charge).
Q: True or false – For a continuous charge distribution, you replace the summation in superposition with an integral.
A: True.
Q: Two point charges, +3 μC and −5 μC, are separated by 0.2 m. What is the magnitude of the force between them?
A: F = k · |q₁·q₂| / r² = (8.99 × 10⁹)(3 × 10⁻⁶)(5 × 10⁻⁶) / (0.2)² = 3.37 N. The force is attractive (opposite signs).
Q: A point charge Q creates an electric field of 500 N/C at a distance of 0.1 m. What is Q?
A: E = kQ/r², so Q = E·r²/k = 500 · (0.01) / (8.99 × 10⁹) ≈ 5.56 × 10⁻¹⁰ C ≈ 0.556 nC.
Q: Why must you break the electric field into components when applying superposition?
A: Because the electric field is a vector. Fields from different charges point in different directions, and only by resolving into x- and y-components (or similar) can you correctly add them.
Q: For a uniformly charged rod, what does dq equal in terms of linear charge density?
A: dq = λ · dx (or λ · dl), where λ = Q/L is the charge per unit length.
This material connects directly to Gauss's law, which provides a shortcut for calculating E when the charge distribution has high symmetry (spherical, cylindrical, planar). It also underpins electric potential, since V is defined through the integral of E. Every formula for capacitance later in the course starts by finding E first.
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