Example 1- Electric field of a concentric solid spherical and conducting spherical shell charge distribution
Now let’s consider another example. A sphere of radius a, and charge +q uniformly distributed throughout its volume. It is concentric with a spherical conducting shell of inner radius b and outer radius c. In our system, we have an outer spherical shell — let’s exaggerate the thickness — and it is concentric to a spherical charged distribution which has the radius a. The inner radius of the spherical shell is b. The outer radius is c.
When we read the problem carefully, it says that the sphere of radius a, which has a charge +q and that charge is uniformly distributed through its volume. So we have a +q charge here, and that charge is uniformly distributed throughout the volume of this 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 of 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 of the charge will move to the surface. Since we are having a volume charge distribution, that is indicating that the inner sphere is an insulator.
The outer sphere information is already given in the problem and it says that we are dealing with a conducting shell of inner radius b and outer radius c. This shell, let’s say, has a net charge of –q. Let’s add that over here. The shell has a net charge of –q. Therefore, we have a total charge of –q within the outer shell.
We’d like to figure out the electric field generated by this distribution — and let me indicate this outer shell with these yellow lines — and we’d like to figure out the electric field generated by this distribution at different regions. Let’s start first, E is the question mark for the point inside the inner sphere. In other words, for r is less than a region. This is identical to a problem we did earlier as an example, a charged solid spherical charge. The charge is distributed uniformly throughout the volume. We calculated the electric field inside and outside.
We can do that one more time over here. Taking 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 S1. Gauss’s law states that the integral of an electrical field totaled with incremental surface area vector integrated over all in this closed surface S1 is equal to q-enclosed over ε0.
As in the case of all spherical 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 π r2, which is equal to q-enclosed over ε0. At this point we are interested in the net charge inside of the region surrounded by the Gaussian sphere. That is the region we’re talking about, this shaded area.
Of course, to be able to get that, we will first define the volume charge density. That is total charge of the distribution. For that part, the total charge is +q, for the inner sphere divided by the total volume of the inner sphere. That is 4 over 3, π times its radius cubed, a3. Therefore, q-enclosed is going to be equal to ρ times volume of the region surrounded by the Gaussian sphere and volume of the region surrounded by surface S1.
Therefore, q-enclosed is going to be equal to ρ — in explicit form that is q over 4 over 3 π a3. Volume of the region surrounded by surface S1 is 4 over 3 π r3. The radius of S1 is r. Here we can cancel four-third’s and π’s. Then the q-enclosed will turn out to be q r3 over a3. Substituting this to the right hand side of Gauss’s law, we will have: E times 4 π r2 is equal to q-enclosed (which is q r3 over a3 divided by ε0). r2 and r3 will cancel from both sides. Leaving the electric field alone, we will have q over 4 π ε0 a3 times r.
Since the charge is positive, the electric field in radially outward direction, we can represent this expression in vector form by multiplying this magnitude with the unit vector r̂ 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 are 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, 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 charged as concentrated at their center. Therefore, the problem induces into a form that we have a positive charge sitting over here.
We are r distance away, trying to figure out the electric field that it generates. Of course the electrical field will be radially outward. The magnitude is going to be q over 4 π ε0 r2 from Coulomb’s law. So this spherical charge will behave as if all of its charge are concentrated at its center for all the exterior points and our 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 π ε0 r2.
Using that fact, we can say that the inner sphere behaves like a point charge, as its whole charge is concentrated at its center for all the exterior points. Therefore, the electric field is going to be equal to q over 4 π ε0 r2 as if we are having a positive q sitting over here and calculating the electric field r distance away from that point charge. It is going to be in radial direction so we, again, multiply this magnitude by the unit vector r̂ in radial direction to express it in vector notation.
Okay. So far, all the parts that we did were identical to the example that we studied earlier. Now we are 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. We know that the electric field inside of a conducting medium is always 0. Knowing that fact, we would expect the electric field at any point inside of this region should be equal to 0.
Let’s verify that. Assume that this is our point of interest over here. We will place our Gaussian surface passing through this point in the form of a spherical surface. Let’s call this surface S3 since this one was S2. This is part 3. Now we’re interested in the electric field for the region such that r is between c and b.
Gauss’s law is simply stating that E dot dA is integrated over the surface S3 is equal to q-enclosed over ε0. In this case, we will just directly look at q-enclosed before we start to deal with the left hand of Gauss’s law. Q-enclosed, as you recall, is the net charge inside of the region surrounded by Gaussian surface. When we look at that Gaussian surface that we are dealing with is S3 and the region inside of that is this region.
In that region, the inner sphere is completely enclosed so we have +q over here. Furthermore, we have only this segment of the outer shell that is inside of the region that we’re interested in. The amount of charge in the inner shell is already given, that is +q. We don’t have to worry about that. We have to figure out have how much charge that 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 medium, this conducting medium.
As a result of this attractive Coulomb force, a charge magnitude of –q, or the charge magnitude with q and sign minus, will move and get collected along this inner surface. In other words, the +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. They will be distributed uniformly along this inner surface. The magnitude, the amount of that charge, will be just –q.
As you recall, the total charge of the system of the outer shell is given as –q. Since these minus charges are going to move from the outer regions and 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 +q charge along the outer surface. We also have negative q of total charge of this outer shell. That +q and this negative q will, therefore, cancel and we’re going to end up with 0 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.
If we didn’t have this inner sphere with positive q, then once we have the total charge of –q along that spherical shell of distribution, all that –q says they’re going to be continuously pulling one another will be distributed along the outer surface of the spherical distribution. In that case, we will have ended up with –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 get positively charged and that +q will cancel with this –q and therefore we will end up with no charge at all along the outer surface as a result of this distribution.
From that point of view, when we look at the q enclosed inside of the surface S3, we have +q due to the inner sphere and we also have –q induced along this inner surface of the shell. The total charge will be +q plus –q and those two charges will cancel one another. As a result of that, q-enclosed, which is +q plus –q will be equal to 0. Since q-enclosed is 0, then the electric field will be 0 for that specific region, in other words, when we are inside of this conducting medium.
If that shell were an insulator, of course then the story was going to be completely different. If it were an insulator and a total charge of –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 S3, 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 electric 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 dA integrated now over, let’s call this surface S4, is equal to q-enclosed over ε0.
Again, for a spherical geometry, as is the case of previous examples, we will end up with E times 4 π r2 on the left-hand side. On the right-hand side is q-enclosed over ε0. Here q-enclosed, basically, since the whole volume encloses the net charge of this whole distribution, then the outer shell has a net charge of –q, the inner shell has the net charge of +q. We will end up with 0 net charge in this case. As a result of that, the electric field outside of this distribution will also be equal to 0.
If the inner charge was +2q, for example, and the outer charge is – q then we would end up with the net charge of +q. Or, if the shell had -3q for example, and inner shell had +q, then the net-charge would have been -2q. In these specific cases both of the spheres, spherical shell and inner sphere both have the same magnitude, charged with opposite signs, then the total charge will be 0.
Let’s consider the case where the outer shell is an insulator, rather than a conductor. Let’s say that let outer shell be an insulator with total charge –q distributed uniformly throughout its volume. If we redraw the picture over here, we have our outer spherical shell, and we have our inner sphere with radius a, inner radius of shell was b and outer radius was c. These are our charge distributions. The net charge in the inner sphere is +q distributed throughout its volume and the net change on the outer spherical shell is –q, that too also distributed throughout the volume of the spherical shell.
We are interested with the electrical field inside of the shell, so r is between c and b. If our point of interest is somewhere over here then we place our Gaussian sphere passing through this point, like this. The distance of the point of interest to the center is r. Gauss’s law, which is E dot dA, integrated over this closed surface S is equal to the net charge enclosed in the volume surrounded by the Gaussian sphere divided by the permittivity of free space.
Again, the left-hand side, by following the exact same steps as earlier, will give us E times the surface area of the Gaussian sphere, which is 4 times π r2. One the right-hand side we have q-enclosed over ε0. We need to determine the q-enclosed and q-enclosed is the net charge inside of the region surrounded by the Gaussian sphere. When we look at that region we see that the inner sphere is fully inside of that region, so it has a charge of +q. From the shell we have only this much fraction of the outer shell that remains inside of the region of interest.
We need to determine how much charge we have in this shaded area. To be able to do that, we are going to first express the charge density for the outer shell. That is total charge — denoted by ρS for the shell — that is equal to total charge of the distribution, which is, – q, divided by total volume of the distribution. We are talking about the volume of this outer spherical shell. That is going to be the volume of this outer sphere minus of the volume of the inner sphere. In other words, 4 over 3 π, that’s going to become a quantity c3 minus b3. That is the total volume of the outer-spherical shell.
q enclosed is equal to q′, the charge along the shaded region in the spherical shell, since the whole charge is inside the region of interest and that is q, we add q over there. q′ is going to be equal to ρS times volume of the region of the shell inside the surface S that we choose. If we express that — let’s write these in explicit form now — ρS is –q over 4 over 3 π c3 minus b3.
Now, we’re going to express this volume of the spherical shell, which is green shaded, 4 over 3 π times r3 minus b3. In this case, again, 4 over 3’s and π’s will cancel. q‘ will turn out to be minus q over c3 minus b3 times r3 minus b3. We can place this negative sign into the parentheses over here. That is going to make then q times b3 minus r3 over c3 minus b3. q-enclosed is q′ plus q. Therefore, in explicit form, that is going to be equal to q times b3 minus r3 divided by c3 minus b3 plus q.
The left-hand side of Gauss’s law was E times 4 times π r2 and the right-hand side is this q-enclosed divided by ε0. In q-enclosed, q is common quantity so let’s take into that parenthesis we have q and we have ε0 in the denominator and here we’re going to have, let’s say, times b3 minus r3 divided by c3 minus b3 plus 1. We can carry this to one more step by having a common denominator over here. q over ε0 times b3 minus r3 plus c3 minus b3 divided by c3 minus b3. The b3’s will cancel in the numerator. Solving for the electric field we will have q over 4 π ε0 r2 times c3 minus r3 divided by c3 minus b3.
This is going to be the magnitude of the electric field inside the spherical shell region. This is also in radial direction since c is greater than r and it’s going to be a positive value as well as in the denominator c is greater than b. So it is in positive radial direction. We can multiply this with the unit vector in radial direction. That will give us the electric field inside of the shell.
As you can see, if it is an insulator we have a net electric field inside of the shell. If it is a conducting shell, then the net electric field is going to be 0 inside of that region. With these charges +q through the volume of inner sphere and –q through the volume of the outer shell, when we look at the electric field outside, this is going to be identical to the previous case because the net charge inside of the region, surrounded by the Gaussian sphere for the outer part will –q plus +q which they will cancel one another and q enclosed will be 0. The electric field outside of this distribution will also be 0.
You need to be extremely careful whenever you are analyzing charge distribution problems in identifying whether the medium is a conductor or an insulator. Another interesting couple of hints, in terms of the induction of the charge through the conducting mediums. Let’s consider, again, a similar type of mechanism here that we have a shell and an inner charge and an outer charge. Let’s assume that we have placed +q in the center. This +q is going to immediately cause, if the outer shell is neutral, a charge separation throughout this conducting medium. We’re looking at this shell as a conductor. That +q will attract equal magnitude of –q to the inner surface. As these charges move towards the inner surface, the other outer surface will lack that much of charge. Therefore, it is going to immediately get charged positively.
This should be your starting point, always. If the outer shell has some net charge, if it were not originally neutral then for the outer region you’re going to add that charge. If the net charge, for example, +5q then the outer surface will get charged +5q and plus this positive q over here due to the charge separation so the net charge will be, in that case, +6q.
If the net charge of the outer shell is – 4q, for example, the positive charge will attract, again, –q into the inner side. The net charge was -4q because of this negative charge in the inner surface will cause the other surface to get an extra +q. If we add that to the total charge, which was -4q. In that case the net charge along the outer surface will be -3q.
Whenever we have a conducting medium and if we introduce a charge nearby, that charge is going to cause a charge separation in the conducting medium because whenever the charge that we are approaching through the end of that conducting medium will attract equal magnitude of opposite charge towards the surface near to that, which will cause the opposing surface to be charged oppositely. These are the points where you should be careful whenever you are analyzing these types of problems.