2.4 Electric Field of Charge Distributions from Office of Academic Technologies on Vimeo.

- Example 1: Electric field of a charged rod along its Axis
- Example 2: Electric field of a charged ring along its axis
- Example 3: Electric field of a charged disc along its axis
- Example 4: Electric field of a charged infinitely long rod.
- Example 5: Electric field of a finite length rod along its bisector.

**2.4 Electric Field of Charge Distributions**

So far we have considered point charges, we studied the forces that the point charges exert to one another. We also studied the electric field generated by the charge or system of the charges. Now, we’re going to study a general case, we will look at the electric field generated by the charge distributions.

While we were studying the electric field of a point charge, we calculated the electric field that it generates, some R distance away, the electric field at point P, for a positive point charge is radially outward direction. And, we calculated the magnitude of this electric field by applying Coulomb’s law, which was experimental law. And, it simply, was equal to Coulomb constant 1 over 4 Pi Epsilon 0 times the magnitude of the charge divided by the square of the distance to the point of interest. And, this was the case of a point charge.

Well, whenever we go from a point charge to a charge distribution, through which the charge is distributed either along the length or through the surface, or through the volume of the object, let’s assume that it is a positively charge, then, we’re interested, again, in the electric field that it generates at a specific point in space.

What we’re going to do is choose an incremental segment of the distribution, and then treat that segment like a point charge, with a charge of DQ. And, assuming that this incremental segment is so small, it is behaving like a point charge, therefore, the electric field that it generates at the point of interest, R distance away, can be expressed as an incremental electric field associated with this element, DE.

And, directly, by applying Coulomb’s law, therefore, DE becomes equal to 1 over 4 Pi Epsilon 0 times the magnitude of this incremental charge. And, that is the DQ divided by the square of the distance between this incremental charge and the point of interest.

Now, once we determine the electric field associated with this element, now, we can go and choose a next incremental element and treat that one also, like a point charge and calculate it’s electric field at the point of interest, which will generate it’s own unique incremental electric field. And, after that, apply the same procedure to another incremental charge within the distribution, and so on, so forth.

So, as a result of this we will get the electric field generated by all the incremental charges which eventually makes the whole distribution at the point of interest. Once all these electric field vectors are obtained, then we add all these vectors, vectorialy of course, to be able to get the total electric field generated by the charge distribution. The addition process over here is integration.

And, let’s introduce a unit of radial vector, which represents the direction of each one of these vectors. Therefore, vector sum of all these incremental vectors, which is the integration of all these incremental electric field vectors, will eventually give us the total electric field at the point of interest.

So, according to this procedure, then, by adopting the Coulomb’s law, in terms of the electric field for a point charge, to our charge distribution, we’re assuming that the distribution consists of the collection of incremental charges, which can be treated like a point charge, then as we go from point charge to a charge distribution, the charge Q becomes incremental charge DQ and the corresponding electric field vector becomes incremental electric field vector.

Now, of course, to be able to apply this procedure, one has to put the integrant in an integral form. In other words, that the incremental charge has to be represented in terms of the total charge of the distribution. And, furthermore, the associated distances between the incremental charge and the point of interest, also, should be represented properly, so that one can take this integral.

When we look at the charge distributions in nature, we observe them, they can be in three different forms. The first one is the case that the charge can be distributed along the length of an object, which we call these distributions as linear charge distributions.

Let’s assume that we have a straight rod, like a plastic rod, such that the charges distributed along the length of this rod, with a magnitude of Q Coulombs, and the rod has the length of L. So L is the length of distribution.

If we are interested with the electric field that this charge distribution generates at a specific point in space, we’re going to choose an incremental segment at an arbitrary location of this rod and call it, length of this segment, as DL. Therefore, DL, represents the length of incremental charge element which we’re going to be treating this element like a point charge. Therefore, the charge on this element is going to be equal to DQ.

If our point of interest, P, is located over here, then the distance of this element to the point of interest, which is this length, will be what we call R. Once we calculate the electric field of this DQ at this location, then we go then do the same process for the next incremental charge element, and then the next one, and then the next one, and, so on, so forth. Obtain all these associated electric fields that they generate at the point of interest and then eventually, add all those electric fields vectorialy to be able to get the total electric field generated at this location.

Of course, as we do this, by applying the Coulomb’s law, which is 1 over 4 Pi Epsilon 0 integral of DQ over R square in magnitude form, we need to determine, or we need to express this incremental charge in terms of the total charge of the distribution. And, to do that, we will use the concept of charge density.

Here we’re going to define linear charge density. We will denote this quantity by Greek letter Lambda and we’re going to define it as total charge of the distribution divided by total length of the distribution. Then Lambda becomes equal to Q, which is the total charge of the distribution divided by the total length and that’s capital L. And, in SI unit system, the unit of charge is in Coulombs and the unit of length is in meters, therefore, Coulomb per meter is the unit of linear charge density, charge per unit length.

Now, if we take the product of this quantity, Lambda, with the length of the region that we’re interested, which is the length of charge DQ, charge per unit length times length will give us the amount of charge along that distance. Therefore, DQ is going to be equal to Lambda times DL, which is the length of incremental charge DQ. And, using this system, therefore, we’re going be able to express DQ in terms of the total charge of the system, or of the distribution.

The second type of distribution is the case that the charge might be distributed along the surface of an object. These type of distributions are called surface charge distributions. Therefore, in this case, charge is distributed along the surface of the object and, let’s assume, that again, our charge is positive and has the magnitude of Q Coulombs, and let’s say, that the A, capital A, represents the area of the distribution.

Again, like in the previous case, to be able to calculate the electric field of such a distribution at a specific point in space, again, we choose an incremental segment along this surface. And, in this case, of course, that element is going to be an incremental surface element and call the area of that element as DA and the amount of charge associated with that element as DQ.

Again, we calculate the electric field that such a surface element, incremental charge, generates at the point of interest by applying Coulomb’s law. And, after we do that, we go and choose another surface element, and then another one, and so on, so forth. And calculate the associated electric field generated by each one of these incremental charges. And, once the electric field vectors are obtained due to each one of these, at the point of interest, we add them vectorialy and obtain the total electric field generated by the whole distribution.

So, again, the task is to represent DQ in terms of the total charge of the system. Again, if we represent DA as the area of incremental charge element DQ to be able to express DQ in terms of the total charge of the distribution now we will introduce the concept of surface charge density. Which, we will denote this by Sigma, and it is defined as total charge of the distribution divided by total area of the distribution. Therefore, Sigma is going to be equal to Q over A. And, again, if we look at the units in SI unit system, it’s going to be equal to coulomb per meter square.

Knowing this quantity, then we can easily express DQ. DQ is going to be equal to charge per unit area, that is surface charge density times the area of the region of interest, which is the area of incremental charge DQ. And, this quantity, therefore, will give us the amount of charge associated with this incremental region.

The third and the last type of charge distribution is the volume charge distribution. In this case, the charge is distributed along the volume of an object. We choose, just, then an arbitrary shape object which occupies a volume in space.

And, again, if we are interested with the electric field that it generates, assuming that it is positively charged throughout it’s volume, with a charge of, total charge of Q, we choose, now, an incremental volume element with a charge of DQ and volume of DV.

If we call the total volume of the distribution as V, total volume of distribution, and DV as the volume of incremental charge, which we treat like a point of charge, then we will define volume charge density. And, we will denote this quantity by Greek letter Row, which is defined as the total charge of distribution divided by total volume of distribution.

In other words Row is going to be equal to Q over V, with units in SI unit system, as coulomb per meter cubed, then we can express the incremental charge DQ as volume charge density Row, which is charge per unit volume, times volume of the charge, or the region of interest, and that is the region where the DQ is occupying. Therefore, Row, times DV.