# 3. The Electric Field

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# Applying Field Model to Electric Force

Another way to determine the electric force is via the field model. From here on in, we will use the field model more frequently than the direct model. In the field model, we analyze the interaction between two charges in two steps: \[\text{Charge 1} \xrightarrow{\text{creates field}} \mathbf{E} \xrightarrow{\text{exerts force on}} \text{Charge 2}\]That is, instead of directly calculating the force Charge 1 exerts on Charge 2, we think of the Charge 1 creating an electric field (\mathbf{E}\) which exerts a force on Charge 2. Once we have determined the electric field produced by the source charge, we can ignore it; we can determine the force on the test charge entirely from the field \(\mathbf{E}\).

As with the gravitational field \(\mathbf{g}\), the electric field \(\mathbf{E}\) exists in all points of space, and may or may not change over time. To define \(\mathbf{E}\) for all space, you must know both the magnitude and direction of \(\mathbf{E}\) at all points. The property of having both a magnitude and direction at every point means \(\mathbf{E}\) is a vector field. We can represent these properties with either field vectors or field lines (more on that below).

Determining the field created by the source charge does not involve the test charge in any capacity. Likewise, determining the force created by the field does not involve the source charge. At this stage, we no longer consider the source charge but instead consider the field the source charge produced. In the direct model, we consider the interaction between the two charges directly. In the field model, we consider each charge in a separate step.

# Electric Field and Force

Now that we have defined the ideas of electric fields and forces, let’s explore the relationship between the two more exactly. We previously determined that the electric force can be calculated from: \[|\mathbf{F}_{\text{electric } Q \text{ on } q}| = \dfrac{kQq}{r^2}\]In defining our electric field, we must ensure that the field does not depend on the test charge. Consider \(Q\) to be the source charge and \(q\) to be the test charge. The simplest construction we could make without the test charge is \(kQ/r^2\), which would be the electric force on a test charge of \(q=1\text{ C}\). To determine the force on any other test charge, we could take the value \(kQ/r^2\) and multiply by the magnitude of the test charge. This idea should sound very familiar, as it is exactly what we did in defining a gravitational field!

Thus, we find that the electric field is given by: \[\mathbf{E}_{\text{of source charge } Q} = \begin{cases} \text{Magnitude} & = |kQ/r^2| \\ \text{Direction} & \text{towards } -Q \text{; away from } +Q \end{cases}\] The units of the electric field are* Newtons per Coulomb* (N/C). There are other equivalent units such as *Volts per meter *(V/m) that we will see later.

The magnitude of the \(\mathbf{E}\) field is an absolute value–it's just a length of a vector. The electric field’s direction at any point shows the direction a *positive* charge would feel an electric force. A negative charge feels a force in the direction *opposite* to the electric field. As a consequence of this convention, positive charges repelling other positive charges tells us that the electric field *starts* on a positive charge and points away. A negative charge would attract a positive charge, and therefore negative charges create electric fields that point inward; the electric field *ends* on negative charges. The total electric field is taken by combining the electric fields of all the source particles and superposing them.

To determine the strength of the electric force, multiply the field \(\mathbf{E}\) by the magnitude of the test charge: \[\mathbf{F}_{\text{field on } q} =q \mathbf{E}\]The units work out nicely to give Newtons, the expected units of force. If we combine the two steps of the field model of electric forces, we find: \[|\mathbf{F}_{\text{field from } Q \text{ on } q}| = |q \mathbf{E}_{\text{from } Q}| = \left| q \left( \dfrac{kQ}{r^2} \right) \right| = \left| \dfrac{kQq}{r^2} \right|\] This is merely the *direct model *of electric forces. It is a nice result that we mathematically obtain the same answer for the magnitude of the force on \(q\) using either the direct or field method.

Example # 1

In the direct model of force, we found that the force is attractive when the charges are different signs and repulsive if the charges are alike. How can we use the electric field to determine the direction of the force?

Solution

We know that positive charges create electric fields that point outwards in all directions of space. If we place a second positive charge in the field, we will have a repulsive force that also points away from the source charge:

If instead we place a negative charge in the field, we find an attractive force that points toward the source charge, in the opposite direction to the field:

At least in the case of two charges in space, we find that that the force is in the same direction as the field for positive source charges, and opposite the field for negative charges.

It turns out that the relationship between force and field direction explored in the previous example generalizes. No matter how complicated the charge configuration, if we know the direction of the electric field, we can easily determine the direction of the electric force. \[\mathbf{F}_{\text{of field on charge } q} = \begin{cases} \text{Magnitude} & = q|\mathbf{E}| \\ \text{Direction} & = \text{ along } \mathbf{E} \text{ field vector for } +q \text{; opposite } \mathbf{E} \text{ field vector for } -q \end{cases}\]

Before proceeding, we should pause for a moment. We have provided an equation to calculate the electric field created by a single charge. If we would like to find the electric field created by two charges at a certain location in space, we know we can use the principle of superposition to add the various fields at that specific location. Note that we must add the fields as vectors; simply adding the fields’ magnitudes is not correct (depending on orientation, adding two fields with equal magnitude can result in a net field of zero, of double the individual fields, or anything in between). We could extend this process to calculate the field from three charges, 10 charges, or 10,000 charges. Every unique configuration of charge will have an equation for the total electric field, but \(mathbf{E}= kQ/r^2\) is a result that applies individually to point or spherical charges..

# Representing the Electric Field

So far, we've explored three ways to represent fields. In this section, we will expand on two of these representations for the case of electric fields: *vector map* and* field line map*.

## Vector Map

As we learned in What are Fields?, to create a map of the electric field surrounding a source charge \(Q\), we must evaluate the magnitude and direction of the electric field vector at various evenly spaced points on a grid surrounding the source charge, but even this can be a daunting task (two examples are below)! You should begin to be comfortable with calling this “map” of electric field vectors the “\(mathbf{E}\) field.” Some very long \(\text{E}\) vectors (i.e. vectors at points very close to the charge \(Q\)) have been omitted for clarity. One advantage of the vector map representation is that it highlights field vectors at well-defined points in space. Compare the below maps to those for gravitational fields from earlier.

## Field Lines

A second way we can represent the electric field is through field lines, which may offer a shortcut for drawing. As we've established , create field lines by connecting field vectors together. Electric field lines always start from a positive charge and end on a negative charge (or start/end at infinity, like for gravitational fields). The direction of the field vector can be determined from the field line; the field vector is tangent to the field line (see diagram). (diagram showing a complicated field line map, showing several sample field vectors).

Example # 2

Explain how you can determine the strength of the electric field using **a)** the vector map representation and **b)** the field lines representation. You may find the relationships between vector representations useful.

Solution

**a) **In the vector map representation, a field vector is provided at many sample locations throughout a region. Locate the field vectors closest to the place of interest. Interpolate these vectors to get a rough vector for your position.

The length of the vectors indicates magnitude of the electric field at their respective positions. Note that the actual scale of the vectors is somewhat arbitrary and chosen for clarity (i.e. is 1 cm on the map equal to 10 N/C? Or 1,000 N/C?) The actual magnitude of the \(\mathbf{E}\) field could be determined from a scale, but we can determine relative strength of the field from the relative lengths of the vectors on the map.

**b)** In the field lines representation, the density of the field lines determines the strength of the electric field. Locate the point of interest, and see how many field lines are nearby. Consider placing a quarter on the paper at that point. How many lines does your quarter cover? How does this compare to how many lines your quarter would cover elsewhere on the page? Like the vector map, the *relative *strength of the field can be determined (by how dense the lines are) but it not possible to determine an *actual* magnitude from the field lines representation alone.

Example # 3

In two previous examples (here and here), you developed a vector field map and a field line map for gravitational field created by two spherical balls of equal mass. Now suppose that each ball carries a net charge, and that the electric field they generate can be represented by the exact same maps. Determine the signs of the charges.

Solution

Looking at the field line map (reproduced below), it is clear that the field lines end on the charges. Field lines end on negative charges, so each charged ball must carry a *negative charge.*

Hands-On

The simulation below allows you to fix positive and negative charges in the space provided. Place a few charges and examine how the \(\mathbf{E}\) field responds to different arrangements of charge. Try concentrating the charges or spreading them out. Look at the \(\mathbf{E}\) field close to the charges or far away. Based on what you learned in this section, try to predict the different ways you can get the \(\mathbf{E}\) field to change. Come back to this simulation after reading Electric Potential.