- Example 1: Electric field of a concentric solid spherical and conducting spherical shell charge distribution
- Example 2: Electric field of an infinite conducting sheet charge
3.2 Conducting charge Distributions
Let’s revisit the concept of insulators and conductors one more time. As we have seen earlier, insulators were the mediums, such that they didn’t provide easy motion for the charges. On the other hand, conductors were the mediums such that the charges can easily move around.
We gave the example for the insulators as a piece of wood for example, or a plastic, rubber, Styrofoam. For the conductors, a very good example was the all metals. The metals were good electrical conductors, because when we look at their atomic structure, we observe that the outermost electron of a metal atom is not bound to the nucleus of the atom, so it is free to move under the influence of any force. We call those types of electrons as free electrons. The metals have abundance of free electrons.
In that sense, if we compare an insulator and a conductor, first when we look at an insulator, which is a medium that the charges cannot move easily around, if we just place an excess amount of, let’s say a positive charge into an insulating medium, the charges basically will stay at the locations that we will place them. In other words, they are not going to be able to move easily around in this medium, although they will observe repulsive Coulomb force to one another, but since the medium does not provide them to move around, they basically remain in their original locations. So through this way, we can charge the object throughout its volume, either uniformly or non-uniformly.
On the other hand, if you consider a conducting object, a piece of metal for example, like aluminum, and if you do the similar type of process to this medium by placing some excess amount of charge into the region of this conducting medium, this charges will naturally immediately exert Coulomb force to one another and since they are likely charge, the nature of the force will be a repulsive one.
Under the influence of this repulsive Coulomb, force, the charges will immediately move to the surface of the conductor. In other words, they will go as far as possible away from each other and finally, they will reach the boundary of the medium, which they cannot go any further. Therefore, then they will be uniformly distributed along the surface of this object.
In other words, we will have these charges only at the surface of the conducting medium. Any charge that we place inside of the conducting medium, we will end up with immediate movement of the charges to the surface of that medium. So we will never have an enclosed charge inside of a conducting medium.
If we make a note of this then, we can say any excess charge put into a conducting medium will immediately move to the surface of the conducting medium. No excess charge can be enclosed inside of a conducting medium. Therefore, q enclosed [inaudible 05:42] will always be zero inside of a conducting medium, inside a conductor.
Since the source of electric field is charged, and since we cannot enclose any charge inside of a conductor, therefore electric field inside of a charged conductor is always zero in an electrostatic case. This is a case, if we recall, we assume that charges are at rest or they are moving at very low velocities relative to one another. These are important features which differ the insulators from the conductors.
Now let’s consider another example. A sphere of radius a and charge plus q uniformly distributed throughout its volume and it is concentric with a spherical shell. Let’s say with a spherical conducting shell of inner radius b and outer radius c.
So in our system, let me use different colors over here, we have an outer spherical shell. Let’s exaggerate the thickness. It is concentric to a spherical charge distribution, which has the radius a. The inner radius of the spherical shell is b, outer radius is c.
When we read the problem carefully, it says the sphere of radius a, which has a charge plus q, and that charge is uniformly distributed through its volume. We have a plus q charge here and that charge is uniformly distributed throughout the volume of this inner sphere. This volume distribution immediately indicates that the inner sphere is an insulator because we know now that whenever we place an excess charge inside a conducting medium, it immediately moves to the surface of the medium. Therefore we cannot have any charge enclosed inside of a conducting medium. All the charge will move to the surface.
Since we have a volume charge distribution, that is indicated that the inner sphere is an insulator. The outer sphere information is already given in the problem and it says that we’re dealing with a conducting shell of inner radius b and outer radius c.
This shell, let’s say, has a net charge of minus q. Let’s add that over here. This shell has a net charge of minus q. Therefore we have a total charge of minus q with the outer shell. We’d like to figure out the electric field generated by this distribution and let me indicate this outer shell with this yellow lines. We’d like to figure out the electric field generated by this distribution at different regions. Let’s start first, here is the question mark, for the point inside of the inner sphere, in other words, for r is less than a region.
This is an identical problem we did earlier as an example. Solid spherical charged and the charge is distributed uniformly throughout the volume and we calculated the electric field inside and outside. If we do that one more time over here, taking the advantage of spherical symmetry, we will choose a Gaussian surface in the form of a sphere passing through the point of interest. Let’s call this surface first, surface s 1. The Gauss’s law states that the integral of electric field [inaudible 13:51] with incremental surface area vector, integrated over all this closed surface s q, is equal to q enclosed over Epsilon zero.
As in the case of all spherical symmetry examples that we did earlier, I’m not going to go through those steps one more time. The left-hand side of this equation, Gauss’s law, will give us electric field times the area of the Gaussian surface, 4 Pi r squared, which is equal to q enclosed over Epsilon zero.
At this point, we are interested with the net charge inside of the region surrounded by the Gaussian sphere and that is the region we are talking about, this shaded area. Of course, to be able to get that, we will first define the volume charge density and that is total charge of the distribution, and for that part, the total charge is plus q for the inner sphere and divided by the total volume of the inner sphere, that is 4 over 3 Pi times its radius cubed, a cubed and therefore q enclosed is going to be equal to Rho times volume of the region surrounded by the Gaussian sphere, the volume of the region surrounded by surface s 1. Therefore q enclosed is going to be equal to Rho, in explicit form that is q over 4 over 3 Pi a cubed, volume of the region surrounded by surface s 1 is 4 over 3 Pi r cubed, the radius of s 1 is the little r. Here we can cancel 4 thirds and Pi’s. Then the q enclosed will turn out to be q r cubed over a cubed.
Substituting this to the right-hand side of the Gauss’s law, over here, we will have e times 4 Pi r squared is equal to q enclosed, which is q r cubed over a cubed, divided by Epsilon zero. r squared and r cubed will cancel from both sides and leaving electric field alone, we will have q over 4 Pi epsilon zero a cubed time r.
Since the charge is positive, electric field is in radially outward direction, we can represent this expression in vector form by multiplying this magnitude with the unit vector r in radial direction. This is the identical result that we obtained earlier when we did this example in more detailed form.
Now, as a second region, we’re going to calculate the electric field when our point of interest is placed between the shell and the sphere. For that, if this is the location of our point of interest here, again we choose our Gaussian surface such that it passes through that point and choose that surface again in the form of a sphere because we are dealing with a spherical geometry.
For the second part, the electric field for the region that the r is between b and a, we can apply Gauss’s law like in the previous part and obtain the electric field. Or, if we recall, that we said if we’re dealing with spherical charge distributions, and if our point of interest is outside of the distribution, then the spherical distributions behave like a point charge. In other words, they behave as if they are all charges concentrated at their center and therefore the problem induces a form that we have a positive charge sitting over here and we’re r distance away trying to figure out the electric field that it generates. Of course the electric field will be radially outward and the magnitude is going to be q over 4 Pi Epsilon zero r squared from Coulomb’s law.
This spherical charge will behave as if it’s all charges concentrated at its center. For all the exterior points, and r point is an exterior point to that charge at this location therefore, the electric field that it generates is going to be radially out and it will have the magnitude of q over 4 Pi Epsilon zero r squared.
So using that fact, we can say that inner sphere behaves like a point charge as it’s all charge concentrated at its center for all the exterior points. Therefore, electric field is going to be equal to q over 4 Pi Epsilon zero r squared, as if we are having a positive q sitting over here and calculating the electric field r distance away from that point charge is going to be in radial direction so we multiply this magnitude by the unit vector r in radial direction to express it in vector notation.
Okay. So far, all the parts that we did are identical to the example that we studied earlier. Now, we’re going to come to a different region, which is the region inside of the outer shell. When we treat that region, we have to be careful. The outer shell is a conducting shell and we know that the electric field inside of a conducting medium is always zero. So knowing that fact, then we would expect electric field at any point inside of this region should be equal to zero.
Let’s verify that. Assume that this our point of interest over here, so we will place our Gaussian surface passing through this point in the form of a spherical surface and let’s call this surface s 3 since this one was s 2.
Gauss’s law, let’s say this is part 3, now we are interested with the electric field for the region such that r is between c and b. The Gauss’s law is simply stating that e dot d a integrated over this surface s 3 is equal to q enclosed over Epsilon zero. In this case, we will directly look at q enclosed before we start to deal with the left-hand side of the Gauss’s law. q enclosed, as you recall, is the net charge inside of the region surrounded by the Gaussian surface. When we look at that, the Gaussian surface we are dealing with is s 3 and the region inside of that is this region. In that region, the inner sphere is completely enclosed, so we have plus q over here and furthermore, we have only this segment of the outer shell as inside of the region that we’re interested in.
The amount of charge in the inner shell is already given, that is plus q, so we don’t have to worry about that. We need to figure out how much charge we have in this region. Since the medium is a conducting medium, whenever we place this positive q charge at the center, that charge will immediately exert an attractive Coulomb force to the free electrons which are filling this metallic, conducting medium. As a result of this attractive Coulomb force, a charge magnitude of minus q, or the charge with magnitude of q and sign minus, will move and get collected along this inner surface. In other words, the plus q will attract equal magnitude of free electrons, which are negatively charged, towards itself. Those electrons will move and they will come to the boundary of the surface and they will be distributed uniformly along this inner surface. The magnitude or amount of that charge will be just minus q.
As you recall, the total charge of the system of the outer shell is given as minus q. Well, since these minus charges are going to move from the outer regions and come be collected and distributed throughout this inner surface of the spherical shell, the other surface will lack that much amount of negative charge. It means that it will automatically get charged positively. So we will have a plus q charge along the outer surface.
Well, we also have negative q of total charge of this outer shell. That plus q and this negative q will therefore cancel, and we’re going to end up with zero charge along the outer surface. In other words, it will be neutral. That’s how the charge is going to be distributed throughout the outer shell.
Now, if we didn’t have this inner sphere with positive q, then once we have the total charge of minus q along that spherical shell of distribution, all the minus q’s are going to be continuously repelling one another, will be distributed along the outer surface of the spherical distribution. In that case we would have ended up with minus q along the outer surface.
But having this positive q is going to attract equal magnitude of negative charge to the inner surface, which will leave the outer surface to naturally charge positively, and that plus q will cancel with this minus q and therefore we will end up with no charge at all along the outer surface as a result of this distribution.
Now, from that point of view, when we look at the q enclosed, inside of the surface s 3, we have plus q due to the inner sphere and we also have minus q induced along this inner surface of the shell. The total charge will be plus q plus minus q, and those two charges will cancel one another and as a result of that q enclosed, which is plus q plus minus q, will be equal to zero. Since q enclosed is zero, then the electric field will be zero for that specific region. In other words, when we are inside of this conducting medium.
Now if that shell were an insulator, of course then the story was going to be completely different. If it were insulator, with a total charge of minus q and if the charge is distributed uniformly throughout the volume of that shell, then we were going to calculate the net charge inside of this region in s 3 and plus the charge due to the inner sphere. To be able to get the amount of charge for this region, we needed to write down the charge density and then multiply that density by the volume of the region over here to be able to get the amount of charge in this shaded green area, which is inside of the Gaussian surface.
As a matter of fact, let me just do that after we calculate the electric field in the last region, which is the outside of all distribution, and that is for part 4. Electrical field is the question mark for r is larger than c. In that case of course, our point of interest is somewhere over here. Again, we choose our Gaussian sphere such that it is passing through the point of interest.
The left-hand side of the Gauss’s law will be identical to the previous cases. e dot d a integrated now over, let’s call this surface as s4, is equal to q enclosed over Epsilon zero. Again, for a spherical geometry, as in the case of previous examples, we will end up with e times 4 Pi r squared on the left-hand side and the right-hand side is q enclosed over Epsilon zero.
Here, q enclosed, since the whole volume encloses the net charge of this whole distribution, the outer shell has the net charge of minus q, inner shell has the net charge of plus q, we will end up with zero net charge in this case. As a result of that, the electric field outside of this distribution will also be equal to zero.
Now, if the inner charge was plus 2 q for example, and the outer charge is minus q, then we were going to end up with the net charge of plus q. Or if the shell had minus 3 q for example, and the inner one had plus q, then the net charge would have been minus 2 q. But in the specific case of both of these spherical shell and inner sphere has the same magnitude charge with opposite signs, then the total charge will be zero.
Here, let’s say as another part of this problem, let’s assume that the spherical shell is an insulator and the charge is distributed throughout.