8.3 - Electric Fields

8.3.1 - The Electric Field

Gravitational Fields and Electrical Fields

It might sound strange that we are discussing gravitational fields. After all, shouldn't we be talking about electric fields? Well, as you've seen in previous topics, the concepts from mechanics are not all that different from E&M.

When studying gravitational fields, we knew that the gravitational force $\mathbf{F}_{g}$ acting on a particle of mass $m$ could be defined as $\mathbf{g} = \frac{\mathbf{F}_{g}}{m}$. Similarly, electric fields can act on a particle of charge. And just like how a gravitational field exists around an object, an electrical field exists in the region of space around a charged object which is called the source charge. When a secondary object, called the test charge enters the field, an electric force acts upon it. That may sound strange, but consider the following: think about the Earth and the Moon for a bit. The Earth has a gravitational field all around it. If we removed the Moon and all other objects, then Earth's gravitational field wouldn't exert a force on anything. Sure, the field is still present, but there's nothing to exert a force on. If we add the Moon back, you'll see that Earth's gravitational field acts on it. Similarly, a charged particle has an electric field and when another charged particle enters it, an electric force will act on the particle that entered.

Mathematically Defining the Electric Field

Key Equation

An electric field is defined as $\mathbf{E} = \frac{\mathbf{F}_{e}}{q_{0}}$, where: $\mathbf{E}$ is the electric field vector, or the symbol representing the field $\mathbf{F}_{e}$ is the electrical force, as seen in Coulomb’s law $q_{0}$ is the positive test charge which we placed in the field (basically the electrical equivalent of the Moon being placed in Earth’s gravitational field)

This equation says that the electric field from the source charge at the location of the test charge is equivalent to the electric force on the test charge per unit of charge.

warning

The equation above only works for particles, or an object of zero size. For objects that have a defined size, things can get significantly more complicated and will often involve calculus or more complex math.

It’s important to note that $\mathbf{E}$ is the electric field of the source charge, not the test charge. Furthermore, the existence of the electric field is always part of the source charge and is not dependent on the test charge existing. Remember: Earth still has a gravitational field, even if we removed the Moon. The test charge is only there to help us detect the electric field, just as the Moon helps us detect Earth’s gravitational field.

Key Equation

We can also write the electric field equation as $\mathbf{F}_{e} = q\mathbf{E}$

The equation above tells us the force on a charged particle placed in an electric field. Just like before, directionality still applies. So, if $q$ is positive, then the force is in the same direction of the field. If $q$ is negative, then the force and field are in opposite directions.

$\mathbf{E}$ is a vector and is in newtons per coulomb, or N/C. The direction of $\mathbf{E}$ is the direction of the force on the test charge. For example, if $\mathbf{E}$ is in the positive direction, then the force on our test charge is also in the positive direction.

We can state that an electric field exists as a specific point/location if a test charge at that point/location experiences an electric force. If you know the magnitude and direction of the field at a point, you can calculate the force exerted on any particle at that same point using the aforementioned key equation.

It’s important to note that these equations only work if we assume that $q_{0}$ (the test charge) is extremely small and doesn’t affect the charge distribution of the electric field. If you were to have a particle with a significant amount of charge ($q^{'}_{0} \gg q_{0}$), then the electric fields end up changing. In that case, we can’t use the equations above. This is because the ratio of the force to the test charge is different. In other words,

$\frac{F^{\prime}_{e}}{q^{\prime}_{0}} \not\equiv \frac{F_{e}}{q_{0}}$

Now, the equations will not work for the new test case. So remember: $q_{0}$ has to be extremely small.

Determining the Direction of an Electric Field

Consider the diagram above. Here, we have a source charge $q$ and a test charge $q_{0}$. We have placed $q_{0}$ at a distance of $r$ away from $q$, and we will use this test charge to find out the direction of $q$’s electric force and thus the direction of its field.

Notice how the force vector $\mathbf{F}$ is pointing away from $q$. Because $q$ is positive, the force of the test charge will be directed away from $q$. Remember, the test charge is considered as positive, since this is the standard convention. Both charges are positive and are repelling, so the vector is directed away.

We know that the force exerted by $q$ on the test charge is the following (Coulomb’s law):

$\mathbf{F_{e}} = k_{e}\frac{qq_{0}}{r^{2}}\mathbf{\hat{r}}$

$\mathbf{\hat{r}}$ is a unit vector and it’s directed from $q$ to $q_{0}$.

Furthermore, we know that an electric field is defined as $\mathbf{E} = \frac{\mathbf{F}_{e}}{q_{0}}$. When we substitute $\mathbf{F_{e}}$ in, we get the following:

$\mathbf{E} = \frac{k_{e}\frac{qq_{0}}{r^{2}}\mathbf{\hat{r}}}{q_{0}} = \frac{k_{e}\frac{q\cancel{q_{0}}}{r^{2}}\mathbf{\hat{r}}}{\cancel{q_{0}}} = k_{e}\frac{q}{r^{2}}\mathbf{\hat{r}}$

Key Equation

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

Remember how we said that the electric field of a charged object is not dependent on the test charge’s presence, similar to how Earth’s gravitational field is not dependent on the Moon’s presence? Well, you’ll see that in the equation above, $q_{0}$ (the test charge) isn’t even there!

Now, consider the following diagram:

The diagram above represents our equation for the electric field. You’ll see that the electric field at point $P$, where $q_{0}$ used to be, is in the same direction as the electric force. Below, you’ll see what happens to the electric field if $q$ is negatively charged. In fact, if you plug in a negative value for $q$ into the electric field equation, you’ll get a negative vector answer.

Calculating the Electric Field due to a Group of Point Charges

What do we do if we want to find out the electric field at a point $P$ due to multiple point charges? Well, we know that $\mathbf{E}$ is a vector. Therefore, we take the vector sum of all the electric field vectors at $P$ to find the net electric field at that point. This is just like how we add up force vectors to find a net force on an object!

Key Equation

The electric field at point $P$ due to a group of source charges is the following vector sum:

$\mathbf{E} = k_{e}\sum_{i}^{}\frac{q_{i}}{r^{2}_{i}}\hat{\mathbf{r}}_{i}$.

• $r_{i}$ is the distance between the $i$th source charge $q_{i}$ and point $P$
• $\hat{\mathbf{r}}_{i}$ is a unit vector directed from $q_{i}$ toward $P$

8.3.2 - Electric Fields of Continuous Charge Distributions

Let’s say that we have a group of charges and we want to calculate the electric field at a certain location or point. We already know that we can take the vector sum of all the electric fields, as demonstrated previously. However, it is often that these groups of electric charges are so close together that they can be modeled as continuous. In order words, these charges that are closely spaced and packed together are the same as a total charge which is continuously distributed along a line, surface, or throughout a volume. You can think of a continuous charge distribution like a bundle of charges all wrapped up together.

When we want to find out the electric field created from this “bundle,” we will have to divide the bundle into small parts that have a small charge $\Delta q$. Next, we use the equation for calculating the electric field for one source charge (see 8.3.1) to calculate the electric field at a point $P$:

$\Delta \mathbf{E} = k_{e}\frac{\Delta q}{r^{2}}\mathbf{\hat{r}}$

Then, we find the total electric field at the point $P$ by summing everything up:

$\mathbf{E} \approx k_{e}\sum_{i}^{}\frac{\Delta q_{i}}{r^{2}_{i}}\mathbf{\hat{r}}_{i}$

Note that the field is an approximation.

However, we know that the charge distribution is continuous, so we the following:

$\mathbf{E} = k_{e}\space\lim_{\Delta q_{i} \to 0} \space\sum_{i}^{}\frac{\Delta q_{i}}{r^{2}_{i}}\space\mathbf{\hat{r}}_{i} = k_{e}\int_{}^{} \frac{dq}{r^{2}}\space\mathbf{\hat{r}}$

You can think of this like finding the center of mass for a uniformly distributed element. For example, finding the center of mass of a plank of wood which has a uniform mass distribution. In fact, just like how you used the density equations for center of mass, you can also use it for charge. This concept is called charge density.

• A charge of $Q$ that is uniformly distributed throughout a volume $V$ is called a volume charge density ($\rho$).
  • $\rho = \frac{Q}{V}$
  • $\rho$ is coulombs per cubic meter ($C/m^{3}$)
• A charge of $Q$ that is uniformly distributed on a surface of area $A$ is called a surface charge density($\sigma$).
  • $\sigma = \frac{Q}{A}$
  • $\sigma$ is coulombs per square meter ($C/m^{2}$)
• A charge of $Q$ that is uniformly distributed along a line of length $l$ is called a linear charge density ($\lambda$).
  • $\lambda = \frac{Q}{l}$
  • $\lambda$ is coulombs per meter ($C/m$)

And just like how you can have something uniformly distributed, you can also have the charge be nonuniformlydistributed. This nonuniform distribution can be over a volume, surface, or line. The amount of charge $dq$ in a small volume/surface/line are the following:

• Volume: $dq = \rho dV$
• Surface: $dq = \sigma dV$
• Line: $dq = \lambda dV$

Suggested Content

A video covering integrals in the context of electric fields. It also covers something called electric potential, just ignore that for now.

Useful Resources

8.3.1

• Professor Dave Explains video on fields
  • The first part of the video is about charge. Skip it and go to the section about electric fields.
• Khan Academy video explaining fields
• Crash Course video about electric fields
• Khan Academy article about electricl fields
• Explanation of the electric field equation

8.3.2

• Video about integrals and electric fields (same as the Suggested Content)
• Lecture about continuous charge distribution field calculations
• Article about continuous distributions

Equations and Constants

Equations

• $\mathbf{E} = \frac{\mathbf{F}_{e}}{q_{0}}$
  • $\mathbf{E}$ = the electric field vector
  • $\mathbf{F}_{e}$ = the electrical force
  • $q_{1}$ and $q_{2}$ = the two charges
  • $q_{0}$ = the positive test charge
• $\mathbf{F}_{e} = q\mathbf{E}$
  • Alternate form of the equation above
• $\mathbf{E} = k_{e}\frac{q}{r^{2}}\mathbf{\hat{r}}$
  • $k_{e}$ = Coulomb's constant
  • $q$ = the magnitude of force between the point charge ($q$) and the test charge
  • $r^{2}$ = the distance between the point charge ($q$) and the point of interest
• $\mathbf{E} = k_{e}\sum_{i}^{}\frac{q_{i}}{r^{2}_{i}}\hat{\mathbf{r}}_{i}$.
  • $r_{i}$ = the distance between the $i$th source charge $q_{i}$ and point $P$
  • $\hat{\mathbf{r}}_{i}$ = a unit vector directed from $q_{i}$ toward $P$
  • All other variables are the same as the equation above
• Charge density formulas:
  • Volume: $dq = \rho dV$
  • Surface: $dq = \sigma dV$
  • Line: $dq = \lambda dV$

Constants

• $k_{e} = 8.9875 \times 10^{9} \text{ N} \cdot \text{m}^{2}/\text{C}^{2} = \frac{1}{4\pi\epsilon_{0}}$