Electric Fields & Coulomb's Law

What is an Electric Field?

An electric field is a region of space around a charged particle where a force is exerted on other charged particles. It is how charges "communicate" their pushing or pulling power across an empty vacuum without physically touching.

Electric fields are vector fields — at every point in space, they possess both a magnitude (how strong the push is) and a direction (which way the push goes).

Learning Goals: By the end of this guide, you should be able to:

Calculate electrostatic force using Coulomb's Law.

Determine electric field strength at a distance from a point charge.

Draw accurate electric field line diagrams.

Solve problems involving the uniform electric field between parallel plates.

Coulomb's Law: Force Between Charges

Coulomb's Law states that the magnitude of the electrostatic force $F$ between two point charges $q_1$ and $q_2$ is directly proportional to the product of their charges, and inversely proportional to the square of the distance $r$ between them.

F = \frac{k \cdot |q_1| \cdot |q_2|}{r^2} \quad \text{or} \quad F = \frac{1}{4\pi\varepsilon_0} \frac{|q_1| |q_2|}{r^2}

The Constants

$k \approx 8.99 \times 10^9\ \text{N}\cdot\text{m}^2/\text{C}^2$ (Coulomb's constant).
$\varepsilon_0 \approx 8.85 \times 10^{-12}\ \text{F/m}$ (permittivity of free space).

Direction Rule: Like charges repel (the force pushes them apart). Opposite charges attract (the force pulls them together).

Electric Field Strength ( $E$ )

Electric field strength (or intensity) is defined as the force per unit positive charge acting on a stationary test charge at that point in the field.

E = \frac{F}{q} \qquad \text{(Units: N/C or V/m)}

Field around a Point Charge

Substituting Coulomb's law into the definition above tells us the field generated by a single source charge $Q$ at a distance $r$ :

E = \frac{k |Q|}{r^2}

The field gets rapidly weaker the further you move away, following the inverse-square law.

Representing The Field: Field Lines

We visualise these invisible fields using electric field lines. They follow strict geometrical rules:

Lines always originate on positive charges and terminate on negative charges (or at infinity).
The lines must always meet the surface of a conductor at exactly $90^\circ$ (perpendicular).
The lines never cross. If they did, a test charge at the intersection would have two different directions of force simultaneously, which is impossible.
The density of the lines (how close together they are drawn) represents the magnitude of the field strength $E$ .

(In the Vectora simulation platform, you can place multiple test charges to instantly visualise complex, curving resultant field maps based on these exact rules).

Uniform Electric Fields (Parallel Plates)

When two flat, parallel metal plates are connected to a voltage supply $V$ and separated by a distance $d$ , an incredibly useful thing happens: they create a uniform electric field.

In the central region between the plates:

The field lines are perfectly straight, parallel, and equally spaced.
The field strength $E$ is constant everywhere.

E = \frac{V}{d}

Because the field is uniform, any charge $q$ placed anywhere between the plates experiences a constant force $F = qE$ , leading to constant acceleration kinematic problems!

Worked Examples

Example 1: Net force on a charge in a 1D line

Question: A $+2\mu\text{C}$ charge is placed $3\ \text{m}$ to the right of a $-5\mu\text{C}$ charge. Find the electrostatic force experienced by the $+2\mu\text{C}$ charge. Step 1: Convert microcoulombs ( $\mu\text{C}$ ) to standard Coulombs: $1\mu\text{C} = 10^{-6}\text{C}$ . Step 2: Apply Coulomb's Law magnitude:

F = \frac{(8.99\times10^9)(5\times10^{-6})(2\times10^{-6})}{3^2} = \frac{0.0899}{9} \approx 0.010\ \text{N}

Step 3: Determine direction. They are opposite charges, so they attract. The force on the rightmost charge is directed to the left.

Example 2: The Electron Gun (Parallel Plates)

Question: An electron ( $q = 1.6 \times 10^{-19}\ \text{C}$ , $m = 9.11 \times 10^{-31}\ \text{kg}$ ) is placed between two parallel plates $0.05\ \text{m}$ apart with a $500\ \text{V}$ potential difference. Find its acceleration. Step 1: Calculate uniform field strength:

E = \frac{V}{d} = \frac{500}{0.05} = 10,000\ \text{V/m}\ (\text{or N/C})

Step 2: Calculate force on the electron:

F = qE = (1.6 \times 10^{-19})(10,000) = 1.6 \times 10^{-15}\ \text{N}

Step 3: Use Newton's Second Law ( $F = ma$ ) to find acceleration:

a = \frac{F}{m} = \frac{1.6 \times 10^{-15}}{9.11 \times 10^{-31}} = 1.76 \times 10^{15}\ \text{m/s}^2

Common Mistakes

Forgetting to convert prefix units — The standard unit for distance is the metre ( $m$ ) and charge is the Coulomb ( $C$ ). You must convert $\mu\text{C}$ , $\text{nC}$ , and $cm$ before doing the math, or you will be off by millions.
Plugging $r$ in without squaring it — In $F = \frac{kq_1q_2}{r^2}$ , the distance must be squared. The force drops off dramatically at a distance. If you double the distance, the force divides by 4.
Plugging negative signs into the magnitude formula — Use $F = \frac{k|q_1||q_2|}{r^2}$ with absolute absolute values to find the magnitude (size) of the force. Only after finding the magnitude should you determine the direction using the "opposites attract, likes repel" logic. Do not mix minus signs into your magnitude calculation.

Exam Tips (A-Level / AP / IB)

Superposition principle: For 2D field problems (like finding the force from three charges arranged in a triangle), you must use vector addition to sum the $x$ and $y$ force components separately.
Zero-field point: An exam classic is asking where the net electric field is zero between two charges.
- If the charges are the same sign, it is between them (closer to the weaker charge).
- If they are opposite signs, it is outside them (closer to the weaker charge).
Remember that the force of gravity is $10^{39}$ times weaker than electrostatic force. In problems concerning electrons and protons on an atomic scale, always ignore gravity unless the question explicitly asks about it (like an oil-drop experiment).

Frequently Asked Questions

Are electric field lines real?

No. Electric field lines are a visual, conceptual model created by Michael Faraday in the 19th century to help us intuit how invisible space influences charge. The field itself is a real, physical entity carrying momentum and energy, but the "lines" are just a drawing framework.

Why is the field inside a hollow metal sphere zero?

In a conductor, charges are completely free to move. If there were an electric field inside the metal, the free electrons would instantly be pushed by that field until they reached the surface, rearranging themselves to cancel out the specific field that moved them. This state is called electrostatic equilibrium, creating a perfect Faraday Cage.

Vector Decomposition — Crucial for solving 2D electrostatic force problems via the superposition principle.
Photoelectric Effect — How atomic electrons interact with the electromagnetic fields of incoming light particles.
Motion Graphs — Describing the resulting constant acceleration patterns of particles injected between uniform plates.