The Electric field

Prerequisites

When we studied the gravitational force in the case of flat-earth gravity, we learned that the gravitational force from the earth onto an object was proportional to the object's mass. That made it convenient to write the gravitational force — the object's weight — as a product of something belonging only to the object (its mass) times something depending only on the earth (we called it "$g$").  This was convenient because in the case of free-fall the only significant force on the object was its weight. Put that force into Newton's second law for $F_{net}$, and the factor of the object's mass cancelled out. The "$g$", which we thought of as a force per unit mass (N/kg), in that special case turned into an acceleration. This gave us the neat result that all objects (at least those for whom resistive forces could be ignored) accelerated in the same way independent of their mass. We called "$g$", the gravitational field.

It turns out that for electric forces, this sort of trick is even more important. We don't get that nice cancellation that we had in gravity, because the parameter that couples a charge to electric forces is charge, not mass. But there's another fact to take into account. Remember two facts about electricity:

  1. Electric forces are WAY stronger than gravitational forces.
  2. All matter is made up of electric charges.

Of course it seems like gravity is stronger because the only object whose gravity we feel is the earth's — and it's huge compared to anything else we interact with. Since all matter is made up of electric charges, there are lots of possibilities for objects exerting forces on us.

We can do the same kind of thing with electricity as we did with gravity — factor out a charge instead of a mass. That lets us compare the effect of a set of charged objects acting on different objects, just like g let us compare the effect of the earth's gravity acting on different objects ($W_{E→m} = mg$), but more than that, it lets us take into account the effect of large numbers of sources of electric forces, some of which we might not even know about!  Before we explain this complicated sounding statement, let's see how we can take out the effect of a particular charge from an electric force.

Test charges and source charges

Suppose we have a large number of charges.  We'll label them 0, 1, ...N and assume that they are at locations $\overrightarrow{r}_0, \overrightarrow{r}_1, \overrightarrow{r}_2, ....\overrightarrow{r}_N$,  Now suppose also that we are particularly interested in what happens to charge 0.  We might at some time be interested in the other charges, but for right now, that's the one we want to look at.

We call a charged object a test charge when we are interested in finding out about the electric forces it feels and when we can ignore the effects of our charge on all the other charged objects. This might happen if those other objects are held down by forces that are much stronger than the force from our charge, or because our charge is pretty small and doesn't exert a noticeable electric force on the other charges. 

We call all the other charged objects that exert electric forces on our test charge source charges.

So our picture is that we are interested in the forces exerted by charges 1,...N on charge 0.  Note that there is no real difference between a source charge and a test charge — just the spotlight of our particular attention and interest!

Force on a test charge

So suppose we take charge 0 to be our test charge and charges 1,..., N to be our source charges.  We'll write the charge on object i to be qi, i= 0,...N.  The electric force on the charge 0 is then given by (see Coulomb's law -- vector character)

$$\overrightarrow{F}^E_{q_0} =  \frac{k_Cq_0q_1}{r^2_{01}} \hat{r}_{1 \rightarrow 0} + \frac{k_Cq_0q_2}{r^2_{02}} \hat{r}_{2 \rightarrow 0} + \frac{k_Cq_0q_3}{r^2_{03}} \hat{r}_{3 \rightarrow 0} ... + \frac{k_Cq_0q_N}{r^2_{0N}} \hat{r}_{N \rightarrow 0}$$

This is the net electric force acting on $q_0$.  It's just the Coulomb's law forces on $q_0$ from each of the charges $q_1, q_2, q_3, ...q_N$. The unit vectors are dimensionless arrows of unit length pointing from the position of the indicated charge to the position of charge $q_0$. This gives each force the proper direction and lets the forces be added correctly as vectors.

We notice that this is a real mess, but we also can see that it depends on the charge of $q_0$ very simply: Every term has one factor of $q_0$.  We can factor this out to get

$$\overrightarrow{F}^E_{q_0} =  q_0 \bigg( \frac{k_Cq_1}{r^2_{01}} \hat{r}_{1 \rightarrow 0} + \frac{k_Cq_2}{r^2_{02}} \hat{r}_{2 \rightarrow 0} + \frac{k_Cq_3}{r^2_{03}} \hat{r}_{3 \rightarrow 0} ... + \frac{k_Cq_N}{r^2_{0N}} \hat{r}_{N \rightarrow 0} \bigg)$$

Although this expression is still a mess, it lets us separate out the dependence of the force q0feels on its charge.  The stuff in the brackets does NOT depend on the value of q0, only on its position.

This is sort of like what happens with weight — the mass of the object we are looking at factors out — but because in the electric case we have many sources, not just one (the earth, for gravity) the stuff in brackets is NOT a constant. Our field depends on the position of the charge $q_0$ through all the distances and the direction factors. Of course it also depends on all the source charges, but we are considering them fixed — for now.

The Electric field

So what turns out to be a convenient representation of the force acting on a test charge is the following:

$$\overrightarrow{F}^E_{q_0} =  q_0 \overrightarrow{E}(\overrightarrow{r}_0)$$

$$\overrightarrow{E}(\overrightarrow{r}_0) = \frac{k_Cq_1}{r^2_{01}} \hat{r}_{1 \rightarrow 0} + \frac{k_Cq_2}{r^2_{02}} \hat{r}_{2 \rightarrow 0} + \frac{k_Cq_3}{r^2_{03}} \hat{r}_{3 \rightarrow 0} ... + \frac{k_Cq_N}{r^2_{0N}} \hat{r}_{N \rightarrow 0} $$

After we have factored out the charge of the test charge from the force, we call what is left the Electric Field or E-Field  for short.  It does NOT depend on the charge of the test charge but it DOES depend on the position of the test charge.

This turns out to be very convenient. We might use one charge to measure the E-field and then use the value we have measured to figure out what would happen to a different charge when we put it at that place. This is particularly useful if we don't actually know where the source charges are. (We typically do this by using the energy analog of the E-field, the electrostatic potential, but the idea is the same.)

Note that we have labeled our test charge 0 to stress that it really isn't any different from any of the source charges -- just that we are paying attention to it (and have somehow nailed the source charges down).  If we don't need to keep track of the specific source charges, we typically tend to drop the 0 giving the equation

$$\overrightarrow{F}^E_{q} =  q \overrightarrow{E}(\overrightarrow{r})$$

Read this as:

The electric force acting on a charge $q$ placed at a point $\overrightarrow{r}$ is equal to the value of the charge (including sign) times the electric field at that point.

Thinking about the electric field

Note that we consider the electric field to be a function of space.  Imagine that you want to map the electric field in a region of space.  You take a test charge q, go to each point in space, measure the electric force q feels at that point, and drop a little arrow there in the direction of that force with a magnitude of the force divided by q

$$\overrightarrow{E}(\overrightarrow{r}) = \frac{\overrightarrow{F}^E_{q} }{q}$$

Be careful!  Often, the fact that the electric force on $q$ is proportional to $q$ is not made explicit.  You might simply see the equation for the electric field as given by: $E = F/q$.  This makes it look as if the electric field depends on the charge of the test charge.  But the whole reason for dividing by $q$ is because $F$ includes a factor of $q$ that we are cancelling out.  As long as you remember what the terms mean physically (and don't treat this like a math equation) you should be OK

This is the "Hansel and Gretel model of the field".  It's like dropping crumbs along your path (with little arrows on the crumbs). Lots of physical variable are treated this way — described as a function that varies over space. Any such function is called a field. Examples include the temperature field (the temperature can be different at different points in space) — this is a scalar (i.e., it's a number without a direction), and the wind velocity field, telling how fast the wind is blowing at each point in space.  For more examples and an activity using this, see the page, The concept of field .

A good place to get some sense of what an electric field means is with the PhET simulation, Charges and Fields. A sample image (analogous to the one shown in the reading, The gravitational field) built using that sim is shown below for the example of a dipole (an important configuration for biology and chemistry). The workout will give you more practice with the concept. 

The little arrows show the  E-field at the position of the little white dot at their centers. The strength of the field is represented by the intensity of the arrows, darker meaning stronger.

Workout: Electric field

 

Joe Redish 10/13/11

 

Article 390
Last Modified: May 14, 2019