What is Electric Field?

Electric field is defined as force per unit charge.

This means that the magnitude of electric field is a measure of the force that one unit charge exerts on another of like or opposite charge. We often visualize electric field as imaginary lines sourcing from (or sinking to) a point charge like so:

In this image, the dots represent a point charge (+/- q) and the arrows represent electric field. From this image alone it is not clear how a force plays a role in the definition of electric field. This will become more apparent as we begin studying how two (or more) point charges (and the electric field associated with them) interact with each other. The explanation for this becomes more complicated and will introduce some mathematics but is necessary to fully understand electric field.

The units most commonly associated with electric field in engineering applications is volts per meter (V/m). This, however, does not give us a clear picture of the connection between electromagnetism (EM) and Newtonian mechanics. Meaning, defining electric field as force per unit charge can help us relate EM to our every day experiences.

So, then, let us play with this and see where this "force" is really coming from.

First, let us consider only one point charge (say, a positive charge). For our purposes, the picture above does not well describe the radial sourcing of electric field. Let us consider an imaginary sphere surrounding the charge, and imaginary lines of electric field being radiated in all directions:

Now that we have a better view of the charge and electric field lines, let us consider the electric field at a point on the sphere, which is a distance 'r' away from the point charge:

The magnitude of the radius vector I have defined here is equal to the radius of the imaginary sphere. We need to make sure that we are not confusing this vector as an electric field vector - they are not the same. The electric field lines above are imaginary, they just give us an idea of whether the charge is positive or negative (for now). So, to find the electric field strength precisely at the tip of the vector above (at that point on the surface of our sphere) we need to:

1. Find the amount of charge contained in our imaginary sphere
2. Divide the charge enclosed by the total surface area of our sphere
3. Divide by the electric permittivity (this will give us units of electric field)

The third step is what relates the electric field to the materials in space. If we consider the point charge in free space, the electric permittivity is a set, known value (constant). If there are other materials present, the electric permittivity constant is multiplied by another constant - the relative permittivity.

Addressing the steps above, we find the following:

1. The amount of charge enclosed in the imaginary sphere we have defined to be +q (measured in Coulombs [C]).
2. Dividing by the surface area of the sphere gives us
   .
3. The electric permittivity constant is represented by the Greek symbol epsilon (ε₀) and is equal to 8.854x10^-12 [F/m]. We now have the electric field at our point of interest:
   .
   

Doing a units analysis gives the following:

Or, from a different perspective,

So, it is shown above that given information about the charge enclosed, the electric field everywhere can be derived. Now, let us consider two charges (say, of like charge) that are placed a distance 'r' apart from each other. In this case the electric fields are opposing each other, so we imagine them to be similar to the following:

Due to the opposing electric field, a force is observed that acts upon both charges. To reason this phenomenon, consider the electric field due to the first charge (on left) at precisely the point where the second charge (on right) is located. Meaning, we will determine the electric field in the same manner as we did previously, and the 'r' in this case is the distance between the two charges.

So, the electric field due to q1 at the exact position of q2 is:

The subscript on the 'E' is interpreted as "The electric field at the position of charge 2, due to charge 1." To get the force on q2 due to q1, we multiply q2 by the electric field just calculated. So,

A units analysis confirms the result is in units of force:

 Similarly, consider the electrical field due to q2 at the position of q1.

The electric field in this case is

,

and the force on q1 due to q2 is thus

So, it can be concluded that the force felt by each of two charges is

.

This result is commonly known as Coulomb's Law. The fact that the result is positive indicates that the force felt by the two charges is repulsive. If the result were negative (meaning, one of the charges is opposite the other), the force felt by the two charges would be attractive.

This concludes the brief discussion of electric field and forces associated with it.