# Physics with Calculus/Electromagnetism/Electric Fields

### Fields

When dealing with electromagnetic phenomena, it's very helpful to use the concept of a field.

In this context, we will consider a field to be an extended region of space where certain conditions apply to each and every point in that region. Generally, the strength of the conditions will vary depending on position within the field. The fields we will be discussing in this section of the text are electric and magnetic fields.

The Electric Field

We can define the electric field $\mathbf {E}$  to be the electrical force $\mathbf {F} _{e}$  acting on a test charge with a net positive charge of $q_{0}$ , or

$\mathbf {E} \equiv {\frac {\mathbf {F} _{e}}{q_{0}}}$  (3)

Then, an electric field exists at a given point if, at that point, a test charge experiences an electric force. An electric field is measured in units of newtons per coulomb (N/C). Note that a charged particle produces an electric field, and therefore, the test charge must be of very small magnitude so that we can safely ignore its effects on the larger field. The charged particle creating the electric field is called the source particle.

Now that we know the magnitude of the electric field at a given point, we can calculate the force on any particle with a charge of q at that point simply by rearranging equation 3:

$\mathbf {F} _{e}=q\mathbf {E}$  (4)

Consider a particle with charge q placed at a distance of r from our test charge with its charge of $q_{0}$ . Then, by Coulomb's Law, the force exerted on the test particle is

$\mathbf {F} _{12}=k_{e}{\frac {qq_{0}}{r^{2}}}\mathbf {\hat {r}} _{12}$

and using equation (3), we can determine the electric field generated by the particle with charge q at the location of our test charge to be

$\mathbf {E} =k_{e}{\frac {q}{r^{2}}}\mathbf {\hat {r}}$  (3)

where $\mathbf {\hat {r}}$  is a unit vector pointing from q towards our test charge.

### Interpretation

You might suppose that an electric field is just a mathematical device, and you may think it is pretty useless. It certainly seems that way just looking at the field from a single, stationary point charge. Why should we not just see what force it exerts on a particle and be done with it without going through an intermediate field? What if we had 100 charged particles, and we wanted to know the force on another particle? We could add up all the forces, or fields, it is really the same problem, but it is more convenient to factor out the q. This is a relatively weak argument, because it's a just a little bit more convenient.

One real reason is that the equations for electrodynamics are very elegant when written in terms of E. Another is that there is a magnetic force also, which it is useful to keep separate. Finally, and probably most importantly, the electric field is real. For example, there is energy associated with an electric field, which is completely independent of any force it provides. There is also momentum stored in fields, and light is made up of these abstract fields, but light is real. Even though we can't see the field, it is there because physical quantities depend on the field, so it seems; fields are more than just mathematical devices.