5.11 Example from Office of Academic Technologies on Vimeo.

**5.11 Example**

Let’s do an example associated with the energy density concept. Let’s assume that we have a cylindrical capacitor with radii a and b. I would like to show that half the stored electrical potential energy lies within a cylinder whose radius, r, is equal to the square root of the inner radius times the outer radius.

Therefore, we have a cylindrical capacitor, and let’s see, this is the outer surface of the cylindrical capacitor, and this represents the inner surface of the cylindrical capacitor. And the inner radius of this capacitor is given as a, and the outer radius is b. The question is asking to find a cylindrical region inside of this cylindrical capacitor, such that the amount of stored electrical potential energy in that region is half of the total energy stored in the electric field of this capacitor.

Now, as you recall, the energy density is given as one half epsilon zero times the square of the electric field between the plates of the capacitor. And let’s assume that the cylindrical region that we’re interested in is this region, which has the radius of r.

Now, when we charge this capacitor, we know that if we charge the inner one positively and the outer one negatively, by connecting to the terminals of a power supply, we’re going to generate a potential difference between these two plates, and we will also generate an electric field originating from the positive plate and entering into the negative plate in a radially outward direction, filling the region between these plates.

And we can calculate the strength of this electric field, which we did earlier, by applying Gauss’ law, and that was e dot da, integrated over this Gaussian cylinder that we choose by using the symmetry of the problem, such that it’s side surface passes through the point of interest located between the plates, and we choose this location away from the ends of the capacitor, you know, to be able to apply Gauss’ law by neglecting the end effects that we did this earlier. And the right hand side of the Gauss’ law, as you recall, will be equal to q enclosed over epsilon zero, which is the net charge enclosed in the region surrounded by the Gaussian cylinder divided by epsilon zero.

And again, as you recall, we divided this closed surface integration into the sum of three integrals taken over the top, bottom and side surfaces of the cylinder. And when we add them, we ended up with the closed surface integral. So integral over the top surface of EdA, as we recall, here the dA was perpendicular to the top surface, so the angle between them, E and dA was 90 degrees, cosine of 90 from that product.

And integral over the bottom surface of EdA, again we have the similar type of situation, that dA’s pointing in a downward direction, the electric field is radially out, neglecting the end effects. So therefore the angle between them is again 90 degrees. Plus integral over the side surface, EdA, and for the side surface, as we have seen again earlier, e is radially out, and dA is also perpendicular to the surface. Therefore the angle for the side surface between e and dA is always zero degrees, and the right-hand side is again q enclosed over epsilon zero. Since cosine 90 is zero, there will not be any contribution from the top and bottom surface integration, beyond the contribution that will come from the side surface, since cosine of zero is 1, and as long as we are along the side surface of the cylinder, the electric field is constant, we can take it outside of the integral. And also we can easily see that since this Gaussian cylinder enclosed the inner charged surface completely, and if the total charge on that surface is q, therefore q enclosed is simply equal to q, and so the left-hand side, taking E outside, becomes E times integral of dA over the side surface, is equal to q over epsilon zero for the right-hand side.

Integral over the side surface, as we have seen earlier, is adding all these incremental areas to one another over the side surface, and will give us the side surface area of the cylinder. If we say that the height of our cylindrical capacitor is h and the radius of the cylinder is r, so we can express the side surface area as 2 pi r times h. Times the electric field is equal to q over epsilon zero, and from there the electric field turns out to be q over 2 pi epsilon zero h times r.

And as we can see, this electric field is varying with radial distance. And the energy density, by definition, is one half epsilon zero E squared, and therefore the energy density, the energy per unit volume of this cylindrical capacitor, becomes equal to one half epsilon zero E squared, and that is q squared over 4 pi squared epsilon zero squared h squared and r squared.

So now we can see that the energy density is varying with one over r squared. Energy density is energy per unit volume, and that density is not constant. If you recall the variable charged density problem, for example, that we did earlier, we’re going to approach this problem also in a similar way. Right now our interest is energy. So what we’re looking at is the amount of energy stored in this region, okay? And we’re saying that if the total energy stored between the plates of this capacitor is U, we’d like to figure out the radius of the cylinder, such that the energy stored inside of that cylinder, which is the region inside of this Gaussian cylinder, will be half of the total energy. That’s what we are after.

And since our energy density is varying with the radial distance, what we’re going to do is, in this case, we will choose an incremental cylindrical shell with a very small thickness along the radial direction. Something like this. And this shell is so thin that we will assume that the amount of energy stored along this spherical shell, and the volume of this spherical shell, let’s call that volume as db, volume of incremental cylindrical shell. And let’s say the radius of this shell is s and the thickness is ds, for example, and that this du is so small that we can assume this energy density, which is varying with the radial distance squared, remains constant along that distance. Then we can calculate the amount of energy stored inside of this cylindrical shell.

And how can we calculate that? The energy, let’s call that d sub u, the energy stored in incremental cylindrical shell. And that is going to be equal to energy density, u, that is energy per unit volume, as you recall, by definition, times the volume of the region that we’re interested in, and that is the volume of incremental cylindrical shell. So that expression, therefore, will give us how much energy we have inside of that region. In explicit form, this is going to be equal to one half epsilon zero q squared over 4 pi squared epsilon zero squared h squared r squared. And that is basically the energy density, u, and let’s define this as dv, the volume of the incremental cylindrical shell.

And let’s calculate that volume. This is an incremental cylindrical shell, something like this, and we’re interested in the volume of this cylinder. It has the radius of s and the thickness of ds. Of course, since its cross-sectional area is constant along its length, then the volume of this shell, dv, is going to be equal to the surface area, and that is the area of this incremental ring, which is 2 pi r dr, times the height of this cylinder, and that is h. Since we’re using the s variable over here, let us replace these “r”s with ds, and as well as the r in the energy density with s. Okay, then the explicit form of the incremental potential energy stored inside of the volume of this cylindrical shell will be equal to one over 2 epsilon zero q squared over 4 pi squared epsilon zero squared h squared s squared times dv, the volume of this incremental cylindrical shell, which is 2 pi s ds times h.

So what we have over here is energy per unit volume times the volume that we’re interested in, so the volumes will cancel and we’re going to end up with the energy inside of that region, energy stored within that region. And here we can cancel one of these s, the numerator and the denominator, and we can cancel one of these h squared with the h in the numerator, and similarly, epsilon zero cancels with the epsilon zero squared, and pi square cancels with this pi, and we can cancel 2 with this 2. And we’re going to end up with the incremental potential energy stored inside of the volume of this incremental cylindrical shell as q squared divided by 4 pi epsilon zero h s ds.

Okay. Well, this is the amount of energy stored along the volume of this incremental cylindrical shell. I can just go ahead and calculate the energy stored in the next shell and in the next shell and so on and so forth, throughout the region that I’m interested in. And that region extends, the radius starts from a to this specific radius r, which I’m actually trying to calculate. Therefore, if I integrate that du from here, from a to r, then I will get the amount of energy stored inside of this purple cylinder.

Well, I want this energy to be half of the total energy stored in the electric field of this capacitor, and the total energy can be obtained by adding those “du”s, incremental energies stored in these incremental cylindrical shells, starting from this inner radius a to outer radius b. In other words, what we have over here is that the integral of du integrated from a to r, to the radius that we’re interested in, we want this energy to be equal to the half of the total energy. And the total energy is the integral of du, the amount of energy stored in these incremental cylindrical shells, integrated from a to b, that will give us the total energy, and I want this energy to be half of that energy. Therefore, I will divide that by 2.

Okay. Let’s calculate these integrals now. Integral from a to r of q squared over 4 pi epsilon zero h times s times ds will be equal to one half the integral from a to b of q squared 4 pi epsilon zero h s times ds. And here q squared 4 pi epsilon zero h, these terms are constant, we can take it outside of the integral for both sides. And after we take them outside, we can easily see that, since we have the same terms on both sides, dividing both sides of the equation by the same terms, then we can eliminate them.

So we’re going to end up with integral from a to r of ds over s is going to be equal to one half integral from a to b of ds over s. Moving on, integral of ds over s is ln of s, and this will be evaluated at a and r, will be equal to one half, again ln of S, now it is evaluated at a and b. Substituting the boundaries, we will have ln of r minus ln of a, will be equal to one half ln of b minus one half ln of a. If we leave ln of r alone on one side of the equation, moving ln of a to the other side, we will half one half ln of a will go as positive to the other side, and minus one half ln of a will give us one half ln of a plus one half ln of b.

We can rewrite this as ln of r is equal to ln of a in one half parentheses plus one half ln of b, which is equal to one half ln of a times b. ln of r will be equal to ln of a times b to the power of one half, and if we take the immersed logarithm of both sides, we will have r is equal to the square root of a times b. And this was the case that we were set to prove. Indeed, if the radius of the cylinder is equal to the square root of the inner radius times the outer radius, then the amount of energy for this capacitor, stored in this region, will be half of the potential energy stored between the plates of this cylinder. In other words, when we consider this region with that radius, such that r is equal to the square root of a times b, inside of this cylinder the amount of energy stored is simply equal to half of the total energy stored by this cylindrical capacitor.

Therefore, this example demonstrates the use of energy density, which we define as energy per unit volume. It also shows you how we approach, again, variable density cases. This is, in a way, a similar type of example to that variable charged density problem. But here, instead of variable charged density, we’re dealing with variable energy density. But the mathematical way of analyzing these types of problems is almost identical.