Electric Field Strength

In the previous sections we have studied how we can represent the electric fields around a charge or combination of charges by means of electric field lines. In this representation we see that the electric field strength is represented by how close together the field lines are. In addition to the drawings of the electric field, we would also like to be able to quantify (put a number to) how strong an electric field is and what its direction is at any point in space.

A small test charge $q$ placed near a charge $Q$ will experience a force due to the electric field surrounding $Q$. The magnitude of the force is described by Coulomb's law and depends on the magnitude of the charge $Q$ and the distance of the test charge from $Q$. The closer the test charge $q$ is to the charge $Q$, the greater the force it will experience. Also, at points closer to the charge $Q$, the stronger is its electric field. We define the electric field at a point as the force per unit charge.

Definition: Electric field

The magnitude of the electric field, $E$, at a point can be quantified as the force per unit charge We can write this as:

\[E = \frac{F}{q}\]

where $F$ is the Coulomb force exerted by a charge on a test charge $q$.

The units of the electric field are newtons per coulomb: $\text{N·C$^{-1}$}$.

Since the force $F$ is a vector and $q$ is a scalar, the electric field, $E$, is also a vector; it has a magnitude and a direction at every point.

Given the definition of electric field above and substituting the expression for Coulomb's law for $F$: \begin{align*} E & = \frac{F}{q} \\ & = \frac{kQq}{r^2 q}\\ E & = \frac{kQ}{r^2} \end{align*}we can see that the electric field $E$ only depends on the charge $Q$ and not the magnitude of the test charge.

If the electric field is known, then the electrostatic force on any charge $q$ placed into the field is simply obtained by rearranging the definition equation:\[F=qE.\]

Optional Video by Khan Academy

We can think of the forces between charges as something that comes from a property of space. That property is called the electric field. This video was created by Sal Khan.

See some examples below.

Example: Electric Field 1

Question

Calculate the electric field strength $\text{30}$ $\text{cm}$ from a $\text{5}$ $\text{nC}$ charge.

Step 1: Determine what is required

We need to calculate the electric field a distance from a given charge.

Step 2: Determine what isgiven

We are given the magnitude of the charge and the distance from the charge.

Step 3: Determine how to approach the problem

We will use the equation: $E=k\frac{Q}{{r}^{2}}.$

Step 4: Solve the problem

\begin{align*} E & = \frac{kQ}{r^2} \\ & = \frac{(\text{9.0} \times \text{10}^{\text{9}})(\text{5} \times \text{10}^{-\text{9}})}{\text{0.3}^2} \\ & = \text{4.99} \times \text{10}^{\text{2}}\text{ N·C$^{-1}$} \end{align*}

Example: Electric Field 2

Question

Two charges of ${Q}_{1}=+3\mathrm{nC}$ and ${Q}_{2}=-4\mathrm{nC}$ are separated by a distance of $\text{50}$ $\text{cm}$. What is the electricfield strength at a point that is $\text{20}$ $\text{cm}$ from ${Q}_{1}$ and $\text{50}$ $\text{cm}$ from ${Q}_{2}$? The point lies between ${Q}_{1}$ and ${Q}_{2}$.

Step 1: Determine what is required

We need to calculate the electric field a distance from two given charges.

Step 2: Determine what is given

We are given the magnitude of the charges andthe distances from the charges.

Step 3: Determine how to approach the problem

We will use the equation: $E=k\frac{Q}{{r}^{2}}.$

We need to calculate the electric field for each charge separately and then add them to determine the resultant field.

Step 4: Solve the problem

We first solve for ${Q}_{1}$: \begin{align*} E &= \frac{kQ}{r^2} \\ &= \frac{(\text{9.0} \times \text{10}^{\text{9}})(\text{3} \times \text{10}^{-\text{9}})}{\text{0.2}^2} \\ & = \text{6.74} \times \text{10}^{\text{2}}\text{ N·C$^{-1}$} \end{align*}

Then for ${Q}_{2}$: \begin{align*} E &= \frac{kQ}{r^2} \\ &= \frac{(\text{9.0} \times \text{10}^{\text{9}})(\text{4} \times \text{10}^{-\text{9}})}{\text{0.3}^2} \\ & = \text{3.99} \times \text{10}^{\text{2}}\text{ N·C$^{-1}$} \end{align*}

We need to add the two electric fields because both are in the same direction. The field is away from ${Q}_{1}$ and towards ${Q}_{2}$. Therefore, ${E}_{\text{total}}= \text{6.74} \times \text{10}^{\text{2}} + \text{3.99} \times \text{10}^{\text{2}} = \text{1.08} \times \text{10}^{\text{2}}\text{ N·C$^{-1}$}$

Example: Electric Field 3

Question

Two point charges form a right-angled triangle with the point $A$ at the origin.Their charges are $Q_2 = \text{6} \times \text{10}^{-\text{9}}\text{ C}=\text{6}\text{ nC}$and $Q_3 = -\text{3} \times \text{10}^{-\text{9}}\text{ C}=-\text{3}\text{ nC}$. The distance between$A$ and ${Q}_{2}$is $\text{5} \times \text{10}^{-\text{2}}$ $\text{m}$and the distance between$A$ and ${Q}_{3}$is $\text{3} \times \text{10}^{-\text{2}}$ $\text{m}$. What is the net electric field measured at $A$ from the two charges if they are arranged as shown?

Step 1: Determine what is required

We are required to calculate the net electric field at $A$. This field is the sum of the two electric fields - the field from ${Q}_{2}$ at $A$ and from ${Q}_{3}$ at $A$.

Step 2: Determine how to approach the problem

We need to calculate the two fields at $A$, using $E=k\frac{Q}{r^2}$ for the magnitude and determining the direction from the charge signs.
We then need to add up the two fields using our rules for adding vector quantities, because the electric field is a vector quantity.

Step 3: Determine what is given

We are given all the charges and the distances.

Step 4: Calculate the magnitude of the fields.

The magnitude of the field from ${Q}_{2}$ at $A$, which we will call $E_2$, is:

\begin{align*} E_2 & = k\frac{Q_2}{r^2} \\ &= (\text{9.0} \times \text{10}^{\text{9}})\frac{(\text{6} \times \text{10}^{-\text{9}})}{(\text{5} \times \text{10}^{-\text{2}})^2} \\ &= (\text{9.0} \times \text{10}^{\text{9}})\frac{(\text{6} \times \text{10}^{-\text{9}})}{(\text{25} \times \text{10}^{-\text{4}})} \\ &= \text{2.158} \times \text{10}^{\text{4}}\text{ N·C$^{-1}$} \end{align*}

The magnitude of the electric field from ${Q}_{3}$ at $A$, which we will call $E_3$, is:

\begin{align*} E_3 & = k\frac{Q_3}{r^2} \\ &= (\text{9.0} \times \text{10}^{\text{9}})\frac{(\text{3} \times \text{10}^{-\text{9}})}{(\text{3} \times \text{10}^{-\text{2}})^2} \\ &= (\text{9.0} \times \text{10}^{\text{9}})\frac{(\text{3} \times \text{10}^{-\text{9}})}{(\text{9} \times \text{10}^{-\text{4}})} \\ &= \text{2.997} \times \text{10}^{\text{4}}\text{ N·C$^{-1}$} \end{align*}

Step 5: Vector addition of electric fields

We will use precisely the same procedure as before. Determine the vectors on the Cartesian plane, break them into components in the $x$- and $y$-directions, sum components in each direction to get the components of the resultant.

We choose the positive directions to be to the right (the positive $x$-direction) and up (the positive $y$-direction). We know the electric field magnitudes but we need to use the charges to determine the direction. Then we can use the diagram to determine the directions.

The force between a positive test charge and ${Q}_{2}$ is repulsive (like charges). This means that the electric field is to the left, or in the negative $x$-direction.

The force between a positive test charge and ${Q}_{3}$ is attractive(unlike charges) and the electric field will be in the positive $y$-direction.

We can redraw the diagram illustrating the fields to make sure we can visualise the situation:

Step 6: Resultant force

The magnitude of the resultant force acting on $Q_1$ can be calculated from the forces using Pythagoras' theorem because there are only two forces and they act in the $x$- and $y$-directions: \begin{align*} E^{2}_R & = E_{2}^{2} + E_{3}^{2}\ \text{Pythagoras' theorem}\\ E_R &= \sqrt{(\text{2.158} \times \text{10}^{\text{4}})^{2} + (\text{2.997} \times \text{10}^{\text{4}})^{2}}\\ E_R &= \text{3.693} \times \text{10}^{\text{4}}\text{ N·C$^{-1}$} \end{align*} and the angle, $\theta_R$ made with the $x$-axis can be found using trigonometry.

\begin{align*} \tan(\theta_R) &= \frac{\text{y-component}}{\text{x-component}} \\ \tan(\theta_R) &= \frac{\text{2.997} \times \text{10}^{\text{4}}}{\text{2.158} \times \text{10}^{\text{4}}} \\ \theta_R &= \tan^{-1}(\frac{\text{2.997} \times \text{10}^{\text{4}}}{\text{2.158} \times \text{10}^{\text{4}}}) \\ \theta_R &= \text{54.24}\text{°} \end{align*}

The final resultant electric field acting at $A$ is $\text{3.693} \times \text{10}^{\text{4}}$ $\text{N·C$^{-1}$}$ acting at $\text{54.24}$$\text{°}$ to the negative $x$-axis or $\text{125.76}$$\text{°}$to the positive $x$-axis.

Electric Field Strength