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Understanding Electric Fields due to various charges


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.

Let’s look at the mutual repulsion of two positively charged bodies A and B. Suppose B has charge , and let be the electric force of A on B. One way to think about this force is as an “action at-a-distance” force—that is, as a force that acts across empty space without needing physical contact between A and B. But a more fruitful way to visualize the repulsion between A and B is as a two-stage process. We first envision that body A, as a result of the charge that it carries, somehow modifies the properties of the space around it. Then body B, as a result of the charge that it carries, senses how space has been modified at its position. The response of body B is to experience the force .

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.

If a point charge is then placed at point P, it experiences the force . We take the point of view that this force is exerted on by the field at P. Thus the electric field is the intermediary through which A communicates its presence to .

Because the point charge 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 produces an electric field in the space around it and that this electric field exerts the force on body A. For each force (the force of A on and the force of 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.

We define the electric field at a point as the electric force experienced by a test charge at the point, divided by the charge . That is, the electric field at a certain point is equal to the electric force per unit charge experienced by a charge at that point:

In SI units, in which the unit of force is 1 N and the unit of charge is 1 C, the unit of electric field magnitude is 1 N/C. If the field at a certain point is known, rearranging the equation gives the force experienced by a point charge placed at that point: (force exerted on a point charge by an electric field ). The charge can be either positive or negative. If is positive, the force experienced by the charge is in the same direction as if is negative, and are in opposite directions.
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 (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.
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

If the source distribution is a point charge , it is easy to find the electric field that it produces. We call the location of the charge the source point, and we call the point P where we are determining the field the field point. If we place a small test charge at the field point P, at a distance r from the source point, the magnitude of the force is given by Coulomb’s law,

Substituting the magnitude of the electric field at P is

Let be a unit vector that points along the line from source point to field point. This unit vector is equal to the displacement vector from the source point to the field point, divided by the distance between these two points; that is, . Using the unit vector , we can write a vector equation that gives both the magnitude and direction of the electric field :

where is the electric field due to a point charge, is the electric constant, is the value of the point charge, and is the distance from the point charge to where the field is measured. By definition, the electric field of a point charge always points away from a positive charge (in the same direction as ) but toward a negative charge (in the direction opposite ; ).

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.

can vary from point to point, it is not a single vector quantity but rather an infinite set of vector quantities, one associated with each point in space. This is an example of a vector field. If we use a rectangular coordinate system, each component of at any point is in general a function of the coordinates of the point.

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 nC?

Solution. We are given the magnitude of the charge and the distance from the charge to the field point, so we use to calculate the field magnitude .

(Coulomb’s constant )

Our result E = 5.0 N/C means that if we placed a 1.0-C charge at a point 3.0 m from , it would experience a 5.0 N force. The force on a 2.0-C charge at that point would be , and so on.

Example: Electric field vector for a point charge

A point charge is located at the origin. Find the electric field vector at the field point .

Solution. We must find the electric-field vector due to a point charge. To do this, first we must find the distance from the source point (the position of the charge , which in this example is at the origin) to the field point , and then we must obtain an expression for the unit vector that points from to .

The distance from to is

The unit vector is then

The electric field vector is

Since is negative, points from the field point to the charge (the source point), in the direction opposite to (hence the negative component). The magnitude of is

and the direction in degrees is

Charge Distribution and Superposition of Electric Fields

In most realistic situations that involve electric fields and forces, we encounter charge that is distributed over space. To find the field caused by a charge distribution, we imagine the distribution to be made up of many point charges At any given point , each point charge produces its own electric field , so a test charge placed at experiences a force from charge , a force from charge , and so on. From the principle of superposition of forces, the total force that the charge distribution exerts on is the vector sum of these individual forces:

The combined effect of all the charges in the distribution is described by the total electric field at point :

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.

When charge is distributed along a line, over a surface, or through a volume, a few additional terms are useful. For a line charge distribution (such as a long, thin, charged plastic rod), we conventionally use to represent the linear charge density (charge per unit length, measured in C/m). When charge is distributed over a surface (such as the surface of the imaging drum of a laser printer), we use to represent the surface charge density (charge per unit area, measured in ). And when charge is distributed through a volume, we use to represent the volume charge density (charge per unit volume, ).

Example: Uniformly charged rod

An insulating rod of length has charge uniformly distributed along its length.

(a) What is the linear charge density of the rod?
(b) Find the electric field at point a distance from the end of the rod.
(c) If were very far from the rod compared to , 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) The linear charge density is the charge per unit length, given by the total charge divided by the total length.

Since the rod is uniformly charged, is a constant.

(b) We use .

First, we must obtain an expression for the infinitesimal amount of charge on the infinitesimal length of the rod:

Assuming P is at (0,0), we obtain an expression for the electric field by integrating with respect to , through the whole length of the rod, from to .

The in the result confirms that the direction of the field at points towards the rod since it has a negative charge.

(c) When , . becomes very small compared to . Hence we can write and the expression for the electric field becomes

which is the electric field of a point charge.

Example: Ring of charge
Charge is uniformly distributed around a conducting ring of radius . Find the electric field at a point on the ring axis at a distance from its center.

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 linear charge density in this case is

The ring is uniformly charged, hence is a constant.

The charge in a segment of length is

Consider two identical segments, one as shown in the figure at and another halfway around the ring at . From the superposition of electric force vectors, we see that the net force these segments exert on a point test charge at , and thus their net field , is directed along the x-axis. The same is true for any such pair of opposite segments around the ring, so the net field at is along the x-axis:
To calculate , note that the square of the distance from a single ring segment to the point is . Hence the magnitude of this segment’s contribution to the electric field at is

The x-component of this field is . We know and the figure shows that

Substituting and ,

Substituting ,

To find we integrate this expression over the entire ring, that is from to . The integrand is constant for all points on the ring, so it can be taken outside the integral. Hence we get

Substituting ,

The result shows that at the center of the ring . This makes sense; charges on opposite sides of the ring push in opposite directions on a test charge at the center, and the vector sum of each such pair of forces is zero.
When the field point
is much farther from the ring than the ring’s radius, we have and the denominator becomes , so . In this limit the electric field at is

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

A nonconducting disk of radius has a uniform positive surface charge density . Find the electric field at a point along the axis of the disk a distance x from its center, assuming that is positive.

We represent the charge distribution as a collection of concentric rings of charge . 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.

A typical ring has charge , inner radius , and outer radius . Its area is approximately equal to its width times its circumference , or

Since a disk has an area, we use the charge per unit area . The disk is uniformly charged, hence is a constant. The charge per unit area is

so the charge of the ring is

Since the charge of the ring is dependent on the variable , we use in place of in the result of the previous example, and replace the ring radius with. Then the field component at point due to this ring is
To find the total field due to all the rings, we integrate over through the entire disk, from to .

Let .

When , and when , .

So the integral becomes

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.

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