Coulomb's Law

Coulomb's Law is the most fundamental law of electrostatics, describing the force between two point charges.

Table of Contents

• Table of Contents
• The Principle of Superposition
• Coulomb's Law
• The Electric Field
• From Discrete to Continuous
• Field from Uniform Sphere
• Summary and Next Steps

The Principle of Superposition

Imagine we have a system of point charges, . Call these charges the "source charges". Then, imagine placing another point charge, , at some point in space. Call this charge the "test charge".

The basis of electrostatics is to ask: what is the force on the test charge due to the source charges ? One extremely powerful principle in electrostatics is known as Principle of Superposition:

Principle of Superposition: The total force on a charge due to multiple charges is the vector sum of the forces on the charge due to each individual charge.

This principle allows us to break down complex systems of charges into simpler systems, and then add up the forces due to each individual charge.

Let be the force on the test charge due to the source charge . Then, the total force on the test charge is:

Coulomb's Law

Given the system of charges described above, the force on the test charge due to the source charge is given by Coulomb's Law:

Coulomb's Law

where:

• is the Coulomb constant, approximately .
• is the charge of the -th source charge.
• is the charge of the test charge.
• is the distance between the -th source charge and the test charge.
• is the unit vector pointing from the -th source charge to the test charge .

The SI unit of charge is the Coulomb (). For historical reasons, is written in terms of the permittivity of free space, , as:

Coulomb's Law is then written as:

The total force from the source charges on the test charge is then:

The Electric Field

Going back to the final expression for the total force on the test charge , let's isolate the test charge :

Define a new quantity, , called the electric field, as:

It acts as a quantity that describes the force per unit charge. Instead of acting on a specific charge at a point, the electric field simply describes the force that would act on a charge of at that point. This makes a function of position in space, and it is a vector field.

Then, the total force on the test charge is simply:

The electric field, as a vector field, can be visualized as a set of vectors at each point in space:

From Discrete to Continuous

As we discussed in the previous page, clasiscal electromagnetism treats charge as a continuous distribution. This means that we can treat the source charges as a continuous charge distribution over space. There are three different types of charge distributions, the linear, surface, and volume charge distributions.

For a linear charge distribution, the charge is distributed along a line. The charge density, , is the charge per unit length, i.e. how much charge is present per unit length of the line. Then, the charge in a small segment of the line, , is:

The total charge in the line is then the integral of the charge density over the length of the line:

And the electric field at a point due to a linear charge distribution is:

A similar definition can be made for surface and volume charge distributions. A table summarizing the three types of charge distributions is given below:

Type of Distribution	Charge Density	Charge	Electric Field
Linear
Surface
Volume

Field from Uniform Sphere

By now, armed with Coulomb's Law and the concept of the electric field, in principle, electrostatics is complete. This means that purely logically, we can derive any result in electrostatics using these two principles.

To demosntrate this, we will now solve some a basic problem involving Coulomb's Law and the electric field. There are two goals for basic problems problems like this:

• To understand how to apply Coulomb's Law and the formula for the electric field to solve problems.
• To realize that this is a f***ing pain in the a** and we need a better way to do this.

The problem is as follows:

A solid sphere of radius has a charge uniformly distributed throughout its surface. Find the electric field at a point that is units above the center of the sphere.

You will need to use the cosine rule to solve this problem. You will end up with two cases, one where the point is inside the sphere and one where the point is outside the sphere.

To solve this problem, we will need to integrate the electric field over the entire surface of the sphere. We shall use spherical coordinates to simplify the calculation.

We shall consider a small surface element on the sphere, then use the surface charge density to calculate the charge in this element. The charge density is the charge per unit area, i.e. how much charge is present per unit area of the surface. Since it is uniformly distributed, the charge density is constant and equal to:

The surface element is a small patch on the sphere shown below:

The area of this surface element is its width times its height, which is . Therefore, the charge in this surface element is:

This charge contributes to the electric field at the point . Call this contribution . The key idea is that the contribution is only along the radial (vertical) direction. This is due to the symmetry of the sphere, which means that there is another point, , on the opposite side of the sphere that contributes an equal and opposite electric field in the horizontal direction. A diagram of this is shown below:

So we only need to consider the vertical component of the electric field; . To find this component, we need:

• The charge in the surface element. We already know this to be .
• The distance from the surface element to the point . We do not know this yet. (I use instead of since looks weird in .)

We can construct a triangle with the surface element, the point , and the center of the sphere. This triangle is shown below:

We can now find the distance using the cosine rule. (Recall that is the angle from the vertical axis to the point .)

With this, we can find the electric field at the point due to the surface element. Remember that only the vertical component of the electric field contributes to the electric field at . We shall use for simplicity.

The total electric field at the point is then the sum of the electric fields due to all the surface elements. This is an integral over the entire surface of the sphere. The bounds are and .

The angle is the angle between the electric field and the vertical axis. From the cosine rule on we have:

To simplify the integral, we can make the subject instead of . By the cosine rule and implicit differentiation, we have:

Now, regarding the bounds of the integral, we can visualize from to . (You could calculate the bounds with brute force, but this is a more elegant way to do it.) At , the charge is at the top of the sphere. The diagram below shows for both and :

Evidently, the bounds are and . (For , the lower bound of is , while for , the lower bound is .) The upper bound is for both cases:

Using both equations, we can rewrite the integral as:

From here, it depends on whether or .

For , we have :

Recall that for a point charge at a distance from the center of the sphere, the electric field is . So the electric field at a point units above the center of the sphere is the same as that of a point charge at the center of the sphere.

For , we have :

So the electric field at a point units below the center of the sphere is zero.

Our conclusions are:

• For , the electric field at a point units above the center of the sphere is the same as that of a point charge at the center of the sphere.
• For , the electric field at a point units below the center of the sphere is zero.

Notice how absolutely painful this was. We had to do a complicated spherical integral, and then we had to use a bunch of trigonometry to simplify the result. Any algebraic mistake would have resulted in a wrong answer. And this is all just for a simple sphere with a uniform charge distribution. Imagine doing this for a more complex charge distribution.

There is a far better way to do this, and that is by using Gauss' Law. In fact, Gauss' Law is so powerful that a person could probably solve this problem completely mentally. We shall discuss Gauss' Law in the next section.

Summary and Next Steps

In this section, we have introduced the concept of the electric field and how it is related to the force on a test charge.

Here are the key points to remember:

• The electric force on a test charge due to a source charge is given by Coulomb's Law:

• The principle of superposition allows us to find the total force on the test charge by summing the forces from all the source charges:

• The electric field is defined as the force per unit charge acting on a test charge at a point:

• For continuous charge distributions, the charge density takes three forms: linear, surface, and volume charge densities.

• The electric field due from a continuous charge distribution can be found by integrating the electric field due to each charge element over the entire distribution.

• Do not use Coulomb's Law to solve problems.

In the next section, we shall introduce Gauss' Law, a powerful tool that allows us to solve electrostatic problems with much, much more ease.