Charge, Coulomb's Law and the Field Idea
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You should be comfortable with vector notation, basic calculus, and the concept of a scalar potential from mechanics. Familiarity with SI units and the permittivity of free space \(\varepsilon_0\) is assumed.
Why this matters
Electromagnetism is the fundamental force governing light, chemistry, and most everyday interactions.
Before we can calculate forces between complex charge distributions, we must establish the basic rules of the game. This lecture introduces the concept of electric charge, the empirical law governing the force between point charges, and the crucial conceptual leap to the electric field. By shifting our focus from direct action-at-a-distance to a field that permeates space, we lay the groundwork for the entire framework of classical electrodynamics. This transition is not merely mathematical; it changes how we visualize physical interactions, preparing us for the more complex phenomena of magnetism and electromagnetic waves that follow in later units.
The Concept of Charge and Force
Electricity begins with a simple observation: certain materials, when rubbed together, attract or repel other objects. In the 18th century Charles du Fay recognised two kinds of electrification, and Benjamin Franklin introduced the labels positive and negative. Like charges repel, and unlike charges attract. This binary nature is fundamental. Charge is a conserved quantity; it cannot be created or destroyed, only transferred. In the SI system, the unit of charge is the coulomb (\(\mathrm{C}\)). The elementary charge, carried by a proton or electron, is approximately \(1.602 \times 10^{-19}\ \mathrm{C}\). While we often speak of "positive" and "negative" charge, these are arbitrary labels. What matters is the interaction: a force exists between any two charged objects. This force is central to the structure of atoms and molecules, binding electrons to nuclei and atoms into molecules, solids and liquids. Understanding this force is the first step in understanding the physical world at a microscopic level.
Coulomb's Law
Charles-Augustin de Coulomb quantified the force between two point charges in the 18th century. He found that the magnitude of the electrostatic force \(F\) between two stationary point charges \(q_1\) and \(q_2\) separated by a distance \(r\) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This is an inverse-square law, similar to Newton's law of gravitation. In vector form, the force \(\vec{F}_{12}\) exerted on charge \(q_1\) by charge \(q_2\) is given by:
Here, \(\varepsilon_0\) is the permittivity of free space, approximately \(8.854 \times 10^{-12}\ \mathrm{F/m}\) (farads per meter). The term \(\frac{1}{4\pi\varepsilon_0}\) is often denoted as \(k\), the Coulomb constant, with a value of approximately \(8.99 \times 10^9\ \mathrm{N\,m^2/C^2}\). The unit vector \(\hat{r}_{12}\) points from \(q_2\) to \(q_1\). If \(q_1\) and \(q_2\) have the same sign, the product \(q_1 q_2\) is positive, and the force is repulsive (pointing away from \(q_2\)). If they have opposite signs, the product is negative, and the force is attractive (pointing towards \(q_2\)). This law is empirical; it is derived from experiment, not from deeper principles, within the scope of classical physics.
From Force to Field
While Coulomb's law allows us to calculate the force between two charges, it becomes cumbersome when dealing with many charges. If we have a source charge \(Q\) and a test charge \(q\), the force on \(q\) is \(\vec{F} = \frac{1}{4\pi\varepsilon_0} \frac{Q q}{r^2} \hat{r}\). Notice that the force is proportional to the test charge \(q\). This suggests that the source charge \(Q\) creates something in the space around it that exerts a force on any charge placed there. This "something" is the electric field \(\vec{E}\). We define the electric field as the force per unit positive test charge, in the limit that the test charge is vanishingly small so that it does not disturb the source distribution:
For a point source charge \(Q\), substituting Coulomb's law gives:
The electric field is a vector field; it has a magnitude and a direction at every point in space. The direction of \(\vec{E}\) is the direction of the force that would be exerted on a positive test charge. For a positive source charge, \(\vec{E}\) points radially outward. For a negative source charge, \(\vec{E}\) points radially inward. The field concept allows us to separate the source (which creates the field) from the test object (which responds to the field). This separation is crucial for understanding how fields propagate and interact.
Superposition and Continuous Distributions
When multiple source charges are present, the total electric field at a point is the vector sum of the fields produced by each individual charge. This is the principle of superposition. For a set of discrete point charges \(q_i\) located at positions \(\vec{r}_i\), the total field at \(\vec{r}\) is:
In the case of a continuous charge distribution, we replace the sum with an integral. If the charge is distributed with a volume charge density \(\rho(\vec{r}')\), the contribution to the field from a small volume element \(d^3r'\) containing charge \(dq = \rho(\vec{r}') d^3r'\) is:
Integrating over the entire source volume gives the total field:
This integral form is the general result for electrostatics. It allows us to calculate the field for any arbitrary charge distribution, provided we know the density \(\rho(\vec{r}')\). The vector \(\vec{r}-\vec{r}'\) points from the source element to the field point, and its magnitude cubed in the denominator accounts for both the inverse-square law and the direction.
The derivation
Electric Field from Coulomb's Law
The electrostatic field \(\vec{E}(\vec{r})\) is defined as the electric force per unit stationary test charge in the limit that the test charge is vanishingly small, \(\vec{E}(\vec{r}) = \lim_{q\to 0}\vec{F}(\vec{r})/q\). Combining this definition with Coulomb's law and the superposition of forces yields, for a source charge density \(\rho(\vec{r}')\) at rest, \(\displaystyle \vec{E}(\vec{r}) = \frac{1}{4\pi\varepsilon_0}\int \rho(\vec{r}')\,\frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^{3}}\,d^{3}r'\).
A common misconception is that the electric field is a physical substance that "flows" from the charge. In reality, the field is a mathematical construct that describes the force per unit charge. However, in classical electrodynamics, the field carries energy and momentum, making it physically real in a dynamic sense. Even for static fields, the energy of a charge configuration can be regarded as stored in the field, with density \(\tfrac12\varepsilon_0 E^2\). This distinction becomes critical when considering electromagnetic waves, where the field propagates through empty space without any material medium. C
Worked example
Problem. Two point charges, \(q_1 = +5.0\ \mu\mathrm{C}\) and \(q_2 = -3.0\ \mu\mathrm{C}\), are located on the x-axis at \(x_1 = 0\ \mathrm{m}\) and \(x_2 = 4.0\ \mathrm{m}\), respectively. Calculate the electric field at the point \(P\) located at \(x = 1.0\ \mathrm{m}\).
- Identify the position vectors. The field point is \(\vec{r} = 1.0\ \hat{i}\ \mathrm{m}\). The source positions are \(\vec{r}_1 = 0\ \hat{i}\ \mathrm{m}\) and \(\vec{r}_2 = 4.0\ \hat{i}\ \mathrm{m}\).
- Calculate the vector from \(q_1\) to \(P\): \(\vec{r} - \vec{r}_1 = 1.0\ \hat{i}\ \mathrm{m}\). The distance is \(r_1 = 1.0\ \mathrm{m}\). The unit vector is \(\hat{r}_1 = \hat{i}\).
- Calculate the vector from \(q_2\) to \(P\): \(\vec{r} - \vec{r}_2 = 1.0 - 4.0 = -3.0\ \hat{i}\ \mathrm{m}\). The distance is \(r_2 = 3.0\ \mathrm{m}\). The unit vector is \(\hat{r}_2 = -\hat{i}\).
- Compute the field due to \(q_1\): \(\vec{E}_1 = \frac{1}{4\pi\varepsilon_0} \frac{q_1}{r_1^2} \hat{r}_1 = (8.99 \times 10^9) \frac{5.0 \times 10^{-6}}{1.0^2} \hat{i} = 44950\ \hat{i}\ \mathrm{N/C}\).
- Compute the field due to \(q_2\): \(\vec{E}_2 = \frac{1}{4\pi\varepsilon_0} \frac{q_2}{r_2^2} \hat{r}_2 = (8.99 \times 10^9) \frac{-3.0 \times 10^{-6}}{3.0^2} (-\hat{i}) = (8.99 \times 10^9) \frac{-3.0 \times 10^{-6}}{9.0} (-\hat{i}) = 2996.7\ \hat{i}\ \mathrm{N/C}\).
- Sum the fields: \(\vec{E}_{\text{total}} = \vec{E}_1 + \vec{E}_2 = (44950 + 2996.7) \hat{i} = 47946.7\ \hat{i}\ \mathrm{N/C}\).
Units check. The Coulomb constant has units \(\mathrm{N\,m^2/C^2}\). Charge is in \(\mathrm{C}\), distance squared is in \(\mathrm{m^2}\). The result is \(\frac{\mathrm{N\,m^2}}{\mathrm{C^2}} \cdot \frac{\mathrm{C}}{\mathrm{m^2}} = \mathrm{N/C}\). This is consistent with the definition of electric field as force per unit charge.
Three quick tests
- What is the direction of the electric field produced by a negative point charge?
Answer
Radially inward, towards the charge. - If the distance between two point charges is doubled, by what factor does the electrostatic force change?
Answer
It decreases by a factor of 4 (inverse square law). - Does the electric field depend on the magnitude of the test charge used to measure it?
Answer
No, the field is defined as force per unit charge, so it is independent of the test charge magnitude (provided it is small enough not to disturb the source).
Problem set for PU-102 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.