An electric field is the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them. It also refers to the physical field for a system of charged particles. Electric fields originate from electric charges and time-varying electric currents. The electric field is defined at each point in space as the force per unit charge that would be experienced by a test charge if held stationary at that point. As the electric field is defined in terms of force, and force is a vector, it follows that an electric field is a vector field.

Consider the charged body A by itself. We say that the charged body A causes an electric field at point P (and at all other points in the neighborhood). This electric field is present at P even if there is no charge at P; it is a consequence of the charge on body A only.

Because the point charge $q_{0}$ would experience a force at any point in the neighborhood of A, the electric field that A produces exists at all points in the region around A. We can likewise say that the point charge $q_{0}$ produces an electric field in the space around it and that this electric field exerts the force $- \overrightarrow{F}_{0}$ on body A. For each force (the force of A on $q_{0}$ and the force of $q_{0}$ on A), one charge sets up an electric field that exerts a force on the second charge. We emphasize that this is an interaction between two charged bodies. A single charge produces an electric field in the surrounding space, but this electric field cannot exert a net force on the charge that created it; a body cannot exert a net force on itself. The electric force on a charged body is exerted by the electric field created by other charged bodies.

$\overrightarrow{E} = \frac{\overrightarrow{F}_{0}}{q_{0}}$

Figure 1.1: The electric force and electric field experienced by a test charge q0 (negative) due to a positively charged body.

Figure 1.1: The electric force and electric field experienced by a test charge q0 (positive) due to a positively charged body.

Electric Field of a Point Charge

$F_{0} = \frac{1}{4\pi \epsilon _{0}}\frac{\left | qq_{0} \right |}{r^{2}}$

$\begin{aligned} E & = \frac{F_{0}}{q_{0}} = \frac{1}{4\pi \epsilon _{0}} \frac{\left | qq_{0} \right |}{r^{2}} \cdot \frac{1}{q_{0}}\\ E & = \frac{1}{4\pi \epsilon _{0}} \frac{\left | q \right |}{r^{2}} \end{aligned}$

$\overrightarrow{E} = \frac{1}{4\pi \epsilon _{0}} \frac{\left | q \right |}{r^{2}} \hat{r}$

Figure 2: The electric field from an isolated (left) positive charge and (right) negative charge. The field strength is proportional to the density of the lines. The field lines are the paths that a charge would follow as it is forced to move within the field. The field lines are simply representations; the field actually permeates all the space between the lines.

In some situations the magnitude and direction of the field (and hence its vector components) have the same values everywhere throughout a certain region; we then say that the field is uniform in this region. An important example of this is the electric field inside a conductor. If there is an electric field within a conductor, the field exerts a force on every charge in the conductor, giving the free charges a net motion. By definition an electrostatic situation is one in which the charges have no net motion. We conclude that in electrostatics the electric field at every point within the material of a conductor must be zero.

Example: Electric-field magnitude for a point charge What is the magnitude of the electric field at a field point 3.0 m from a point charge $q = 5$ nC?

Solution. We are given the magnitude of the charge and the distance from the charge to the field point, so we use $E = \left ( 1 / 4\pi \epsilon _{0} \right ) \left ( \left | q \right | / r^{2}\right )$ to calculate the field magnitude $E$ .

(Coulomb’s constant $1 / 4\pi \epsilon _{0} \approx 9.0 \times 10^{9} N \cdot m^{2}/ C^{2}$ )

$\begin{aligned} E & = \frac{1}{4\pi \epsilon _{0}}\frac{\left | q \right |}{r^{2}} = \left ( 9.0 \times 10^{9} \hspace{0.1cm} N \cdot m^{2} / C^{2} \right ) \left [ \frac{5 \times 10^{-9} \hspace{0.1cm} C}{\left ( 3.0 \hspace{0.1cm} m \right )^{2}} \right ]\\ E & = \[ \boxed{5.0 \hspace{0.1cm} N/C}\] \end{aligned}$

Example: Electric field vector for a point charge

Solution. We must find the electric-field vector $\overrightarrow{E}$ due to a point charge. To do this, first we must find the distance $r$ from the source point $S$ (the position of the charge $q$ , which in this example is at the origin) to the field point $P$ , and then we must obtain an expression for the unit vector $\hat{r} = \overrightarrow{r} / r$ that points from $S$ to $P$ .

$r = \sqrt{x^{2} + y^{2}} = \sqrt{\left ( 1.2 \hspace{0.1cm} m \right )^{2} + \left ( - 1.6 \hspace{0.1cm} m \right )^{2}} = 2.0 \hspace{0.1cm} m$

The unit vector $\hat{r}$ is then

$\begin{aligned} \hat{r} & = \frac{\overrightarrow{r}}{r} = \frac{x\hat{i} + y\hat{j}}{r} \\ & = \frac{\left ( 1.2 \hspace{0.1cm} m \right )\hat{i} + \left ( - 1.6 \hspace{0.1cm} m \right )\hat{j}}{2.0 \hspace{0.1cm} m} = 0.60 \hat{i} - 0.80\hat{j} \end{aligned}$

The electric field vector is

$\begin{aligned} \overrightarrow{E} & = \frac{1}{4\pi \epsilon _{0}}\frac{q}{r^{2}}\hat{r} \\ & = \left ( 9.0 \times 10^{9} N \cdot m^{2}/C^{2} \right )\frac{-8.0 \times 10^{-9} C}{\left ( 2.0 \hspace{0.1cm} m \right )^{2}}\left ( 0.60\hat{i} - 0.80\hat{j} \right ) \\ & = \left ( - 11 \hspace{0.1cn} N/C \right )\hat{i} + \left ( 14 \hspace{0.1cm} N/C \right )\hat{j} \end{aligned}$

$\begin{aligned} E & = \sqrt{\left ( - 11 \hspace{0.1cm} N/C \right )^{2} + \left ( 14 \hspace{0.1cm} N/C \right )^{2}} \\ E & = \[ \boxed{17.8 \hspace{0.1cm} N/C}\] \end{aligned}$

and the direction in degrees is

$\begin{aligned} \theta & = arctan \left ( \frac{14 \hspace{0.1cm} N/C}{-11 \hspace{0.1cm} N/C} \right ) \\ & = \[ \boxed{-51.8^{\circ}}\] \end{aligned}$

Charge Distribution and Superposition of Electric Fields

$\overrightarrow{F}_{0} = \overrightarrow{F}_{1} + \overrightarrow{F}_{2} + \overrightarrow{F}_{3} + ... = q_{0}\overrightarrow{E}_{1} + q_{0}\overrightarrow{E}_{2} + q_{0}\overrightarrow{E}_{3} + ...$

$\overrightarrow{E} = \overrightarrow{F}_{0} / q_{0} = \overrightarrow{E}_{1} + \overrightarrow{E}_{2} + \overrightarrow{E}_{3} + ...$

The total electric field at P is the vector sum of the fields at P due to each point charge in the charge distribution. This statement is the principle of superposition of electric fields.

Example: Uniformly charged rod

(a) What is the linear charge density of the rod?
(b) Find the electric field at point $P$ a distance $a$ from the end of the rod.
(c) If $P$ were very far from the rod compared to $L$ , the rod would look like a point charge. Show that the answer in (b) reduces to the electric field of a point charge for $a \gg L$ .

Solution.
(a) The linear charge density $\lambda$ is the charge per unit length, given by the total charge divided by the total length.
$\lambda = - q / L$

Since the rod is uniformly charged, $\lambda$ is a constant.

(b) We use $E = \left ( 1 / 4\pi \epsilon _{0} \right )\left ( \left | q \right | \right / r^{2})$ .

First, we must obtain an expression for the infinitesimal amount of charge $dq$ on the infinitesimal length $dx$ of the rod:
$dq = \lambda dx$

Assuming P is at (0,0), we obtain an expression for the electric field by integrating with respect to $x$ , through the whole length of the rod, from $x = - \left ( L + a \right )$ to $x = -a$ .
$\begin{aligned} \overrightarrow{E} & = \frac{1}{4\pi \epsilon _{0}}\frac{\left | q \right |}{r^{2}} \hat{r} = \frac{1}{4\pi \epsilon _{0}} \int \frac{dq}{r^{2}} \hat{r}\\ \overrightarrow{E} & = \frac{1}{4\pi \epsilon _{0}} \int_{-\left ( L + a \right )}^{-a} \frac{\lambda }{x^{2}} dx \hspace{0.1cm} \hat{i} \hspace{1.5cm} \left ( \hat{r} \rightarrow \hat{i}, r \rightarrow x, \hspace{0.1cm} and \hspace{0.1cm} substituting \hspace{0.1cm} dq = \lambda dx \right )\\ \overrightarrow{E} & = \frac{\lambda }{4\pi \epsilon _{0}} \int_{-\left ( L + a \right )}^{-a} \frac{1}{x^{2}} dx = - \frac{\lambda }{4\pi \epsilon _{0}}\frac{1}{x} \bigg|_{-\left ( L + a \right )}^{-a} \hat{i} = - \frac{\lambda }{4\pi \epsilon _{0}} \left [ \frac{1}{L + a} - \frac{1}{a}\right ]\hat{i}\\ \overrightarrow{E} & = \frac{q}{4\pi \epsilon _{0} L}\left [ \frac{-L}{a\left ( L + a \right )} \right ]\hat{i} \hspace{1.5cm} \left ( substituting \hspace{0.1cm} \lambda = -q/L \right ) \\ \overrightarrow{E} & = \[ \boxed{\frac{q}{4\pi \epsilon _{0}}\frac{1}{a\left ( L + a \right )}\left ( -\hat{i} \right )}\] \end{aligned}$

The $-\hat{i}$ in the result confirms that the direction of the field at $P$ points towards the rod since it has a negative charge.

(c) When $a \gg L$ , $L$ . becomes very small compared to $a + L$ . Hence we can write $L + a \approx a$ and the expression for the electric field becomes

$\overrightarrow{E} = \frac{1}{4\pi \epsilon _{0}}\frac{q}{a^{2}}\left ( -\hat{i} \right )$

which is the electric field of a point charge.

Solution. This is a problem in the superposition of electric fields. Each bit of charge around the ring produces an electric field at an arbitrary point on the x-axis; our target variable is the total field at this point due to all such bits of charge.

We divide the ring into infinitesimal segments

The ring is uniformly charged, hence $\lambda$ is a constant.

The charge in a segment of length $ds$ is

$dQ = \lambda ds$

$dE = \frac{1}{4\pi \epsilon _{0}}\frac{dQ}{x^{2} + a^{2}}$

$cos \hspace{0.1cm} a = x/r = x / \left ( x^{2} + a^{2} \right )^{1/2}$

$dE_{x} = dE \hspace{0.1cm} cos \hspace{0.1cm} a = \frac{1}{4\pi \epsilon _{0}}\frac{dQ}{x^{2} + a^{2}}\frac{x}{\sqrt{x^{2} + a^{2}}}$

Substituting $dQ = \lambda ds$ ,

$dE_{x} = \frac{1}{4\pi \epsilon _{0}}\frac{\lambda x}{\left ( x^{2} + a^{2} \right )^{3/2}}ds$

$\begin{aligned} E_{x} & = \int dE_{x} = \frac{1}{4\pi \epsilon _{0}} \frac{\lambda x}{\left ( x^{2} + a^{2} \right )^{3/2}} \int_{0}^{2\pi a} ds \\ & = \frac{1}{4\pi \epsilon _{0}}\frac{\lambda x}{\left ( x^{2} + a^{2} \right )^{3/2}}\left ( 2\pi a \right ) \end{aligned}$

Substituting $\lambda = Q / 2\pi a$ ,

$\begin{aligned} E_{x} & = \frac{1}{4\pi \epsilon _{0}} \frac{x}{\left ( x^{2} + a^{2}\right )^{3/2}} \frac{Q}{2\pi a} \left ( 2\pi a \right )\\ \overrightarrow{E} & = E_{x}\hat{i} = \[ \boxed{\frac{1}{4\pi \epsilon _{0}}\frac{Qx}{\left ( x^{2} + a^{2}\right )^{3/2}}\hat{i}}\] \end{aligned}$

$\overrightarrow{E} = \frac{1}{4\pi \epsilon _{0}}\frac{Q}{x^{2}}\hat{i}$

When the ring is so far away that its radius is negligible in comparison to the distance x, its field is the same as that of a point charge.

Example: Uniformly charged disk

We represent the charge distribution as a collection of concentric rings of charge $dQ$ . In the previous example, we obtained the field on the axis of a single uniformly charged ring. All we need to do to obtain the field at a point along the axis of the disk is to integrate the contributions of concentric rings on the same plane.

$\begin{aligned} A & = \pi r^{2}\\ dA & = 2\pi r \hspace{0.1cm} dr \end{aligned}$

$\sigma = dQ / dA$

so the charge of the ring is

$dQ = \sigma \hspace{0.1cm} dA = \sigma 2\pi r \hspace{0.1cm} dr$

Let $u = x^{2} + r^{2} \rightarrow du = 2r \hspace{0.1cm} dr$ .

So the integral becomes

$\begin{aligned} E_{x} & = \frac{\sigma x}{4\epsilon _{0}}\int_{x^{2}}^{x^{2} + R^{2}} \frac{du}{u^{3/2}} = \frac{\sigma x}{4\epsilon _{0}} \left ( -2 \right )u^{-1/2} \bigg|_{x^{2}}^{x^{2} + R^{2}}\\ E_{x} & = \frac{\sigma x }{2\epsilon _{0}} \left [ - \frac{1}{\sqrt{x^{2} + R^{2}}} + \frac{1}{x} \right ]\\ \overrightarrow{E} & = E_{x}\hat{i} = \[ \boxed{\frac{\sigma }{2\epsilon _{0}} \left [ 1 - \frac{1}{\sqrt{\left ( R^{2} / x^{2} \right )+ 1}} \right ] \hat{i}}\] \end{aligned}$

In this tutorial we performed quite tedious calculations for the electric field. In the next tutorial, we will discuss Gauss’s Law and apply this general law for the electric field at any point along a closed surface. Gauss’s law allows the evaluation of the amount of enclosed charge by mapping the field on a surface outside the charge distribution. It is an essential tool since it simplifies the calculation of the electric field for geometries of sufficient symmetry.

Understanding Electric Fields due to various charges

Electric Field of a Point Charge

Charge Distribution and Superposition of Electric Fields

Authored By

Susie Maestre

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Understanding Electric Fields due to various charges

Electric Field of a Point Charge

Charge Distribution and Superposition of Electric Fields

Authored By

Susie Maestre

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