A capacitor at high frequencies
Now we want to discuss in detail the behavior of a capacitor—a geometrically ideal capacitor—as the frequency gets larger and larger, so we can see the transition of its properties. (We prefer to use a capacitor instead of an inductance, because the geometry of a pair of plates is much less complicated than the geometry of a coil.) We consider the capacitor shown in Fig. 23–4(a), which consists of two parallel circular plates connected to an external generator by a pair of wires. If we charge the capacitor with dc, there will be a positive charge on one plate and a negative charge on the other; and there will be a uniform electric field between the plates. Now suppose that instead of dc, we put an ac of low frequency on the plates. (We will find out later what is “low” and what is “high”.) Say we connect the capacitor to a lower-frequency generator. As the voltage alternates, the positive charge on the top plate is taken off and negative charge is put on. While that is happening, the electric field disappears and then builds up in the opposite direction. As the charge sloshes back and forth slowly, the electric field follows. At each instant the electric field is uniform, as shown in Fig. 23–4(b), except for some edge effects which we are going to disregard. We can write the magnitude of the electric field as \begin{equation} \label{Eq:II:23:2} E=E_0e^{i\omega t}, \end{equation}  where $E_0$ is a constant. Now will that continue to be right as the frequency goes up? No, because as the electric field is going up and down, there is a flux of electric field through any loop like $\Gamma_1$ in Fig. 23–4(a). And, as you know, a changing electric field acts to produce a magnetic field. One of Maxwell’s equations says that when there is a varying electric field, as there is here, there has got to be a line integral of the magnetic field. The integral of the magnetic field around a closed ring, multiplied by $c^2$, is equal to the time rate-of-change of the electric flux through the area inside the ring (if there are no currents): \begin{equation} \label{Eq:II:23:3} c^2\oint_\Gamma\FLPB\cdot d\FLPs=\ddt{}{t}\kern{-1.7ex} \underset{\text{inside $\Gamma$}}{\int} \FLPE\cdot\FLPn\,da. \end{equation}  So how much magnetic field is there? That’s not very hard. Suppose that we take the loop $\Gamma_1$, which is a circle of radius $r$. We can see from symmetry that the magnetic field goes around as shown in the figure. Then the line integral of $\FLPB$ is $2\pi rB$. And, since the electric field is uniform, the flux of the electric field is simply $E$ multiplied by $\pi r^2$, the area of the circle: \begin{equation} \label{Eq:II:23:4} c^2B\cdot2\pi r=\ddp{}{t}\,E\cdot\pi r^2. \end{equation}  The derivative of $E$ with respect to time is, for our alternating field, simply $i\omega E_0e^{i\omega t}$. So we find that our capacitor has the magnetic field \begin{equation} \label{Eq:II:23:5} B=\frac{i\omega r}{2c^2}\,E_0e^{i\omega t}. \end{equation}  In other words, the magnetic field also oscillates and has a strength proportional to $r$. What is the effect of that? When there is a magnetic field that is varying, there will be induced electric fields and the capacitor will begin to act a little bit like an inductance. As the frequency goes up, the magnetic field gets stronger; it is proportional to the rate of change of $E$, and so to $\omega$. The impedance of the capacitor will no longer be simply $1/i\omega C$.  Let’s continue to raise the frequency and to analyze what happens more carefully. We have a magnetic field that goes sloshing back and forth. But then the electric field cannot be uniform, as we have assumed! When there is a varying magnetic field, there must be a line integral of the electric field—because of Faraday’s law. So if there is an appreciable magnetic field, as begins to happen at high frequencies, the electric field cannot be the same at all distances from the center. The electric field must change with $r$ so that the line integral of the electric field can equal the changing flux of the magnetic field. Let’s see if we can figure out the correct electric field. We can do that by computing a “correction” to the uniform field we originally assumed for low frequencies. Let’s call the uniform field $E_1$, which will still be $E_0e^{i\omega t}$, and write the correct field as \begin{equation*} E=E_1+E_2, \end{equation*}  where $E_2$ is the correction due to the changing magnetic field. For any $\omega$ we will write the field at the center of the condenser as $E_0e^{i\omega t}$ (thereby defining $E_0$), so that we have no correction at the center; $E_2=0$ at $r=0$. To find $E_2$ we can use the integral form of Faraday’s law: \begin{equation*} \oint_\Gamma\FLPE\cdot d\FLPs=-\ddt{}{t}(\text{flux of $\FLPB$}). \end{equation*}  The integrals are simple if we take them for the curve $\Gamma_2$, shown in Fig. 23–4(b), which goes up along the axis, out radially the distance $r$ along the top plate, down vertically to the bottom plate, and back to the axis. The line integral of $E_1$ around this curve is, of course, zero; so only $E_2$ contributes, and its integral is just $-E_2(r)\cdot h$, where $h$ is the spacing between the plates. (We call $E$ positive if it points upward.) This is equal to minus the rate of change of the flux of $\FLPB$, which we have to get by an integral over the shaded area $S$ inside $\Gamma_2$ in Fig. 23–4(b). The flux through a vertical strip of width $dr$ is $B(r)h\,dr$, so the total flux is \begin{equation*} h\int B(r)\,dr. \end{equation*}  Setting $-\ddpl{}{t}$ of the flux equal to the line integral of $E_2$, we have \begin{equation} \label{Eq:II:23:6} E_2(r)=\ddp{}{t}\int B(r)\,dr. \end{equation}  Notice that the $h$ cancels out; the fields don’t depend on the separation of the plates. Using Eq. (23.5) for $B(r)$, we have \begin{equation*} E_2(r)=\ddp{}{t}\,\frac{i\omega r^2}{4c^2}\,E_0e^{i\omega t}. \end{equation*}  The time derivative just brings down another factor $i\omega$; we get \begin{equation} \label{Eq:II:23:7} E_2(r)=-\frac{\omega^2r^2}{4c^2}\,E_0e^{i\omega t}. \end{equation}  As we expect, the induced field tends to reduce the electric field farther out. The corrected field $E=E_1+E_2$ is then \begin{equation} \label{Eq:II:23:8} E=E_1+E_2=\biggl(1-\frac{1}{4}\,\frac{\omega^2r^2}{c^2} \biggr)E_0e^{i\omega t}. \end{equation}   The electric field in the capacitor is no longer uniform; it has the parabolic shape shown by the broken line in Fig. 23–5. You see that our simple capacitor is getting slightly complicated. We could now use our results to calculate the impedance of the capacitor at high frequencies. Knowing the electric field, we could compute the charges on the plates and find out how the current through the capacitor depends on the frequency $\omega$, but we are not interested in that problem for the moment. We are more interested in seeing what happens as we continue to go up with the frequency—to see what happens at even higher frequencies. Aren’t we already finished?  No, because we have corrected the electric field, which means that the magnetic field we have calculated is no longer right. The magnetic field of Eq. (23.5) is approximately right, but it is only a first approximation. So let’s call it $B_1$. We should then rewrite Eq. (23.5) as \begin{equation} \label{Eq:II:23:9} B_1=\frac{i\omega r}{2c^2}\,E_0e^{i\omega t}. \end{equation}  You will remember that this field was produced by the variation of $E_1$. Now the correct magnetic field will be that produced by the total electric field $E_1+E_2$. If we write the magnetic field as $B=B_1+B_2$, the second term is just the additional field produced by $E_2$. To find $B_2$ we can go through the same arguments we have used to find $B_1$; the line integral of $B_2$ around the curve $\Gamma_1$ is equal to the rate of change of the flux of $E_2$ through $\Gamma_1$. We will just have Eq. (23.4) again with $B$ replaced by $B_2$ and $E$ replaced by $E_2$: \begin{equation*} c^2B_2\cdot2\pi r=\ddt{}{t}(\text{flux of $E_2$ through $\Gamma_1$}). \end{equation*}  Since $E_2$ varies with radius, to obtain its flux we must integrate over the circular surface inside $\Gamma_1$. Using $2\pi r\,dr$ as the element of area, this integral is \begin{equation} \int_0^rE_2(r)\cdot2\pi r\,dr.\notag \end{equation}  So we get for $B_2(r)$ \begin{equation} \label{Eq:II:23:10} B_2(r)=\frac{1}{rc^2}\,\ddp{}{t} \int E_2(r)r\,dr. \end{equation}  Using $E_2(r)$ from Eq. (23.7), we need the integral of $r^3\,dr$, which is, of course, $r^4/4$. Our correction to the magnetic field becomes \begin{equation} \label{Eq:II:23:11} B_2(r)=-\frac{i\omega^3r^3}{16c^4}\,E_0e^{i\omega t}. \end{equation}  But we are still not finished! If the magnetic field $B$ is not the same as we first thought, then we have incorrectly computed $E_2$. We must make a further correction to $E$, which comes from the extra magnetic field $B_2$. Let’s call this additional correction to the electric field $E_3$. It is related to the magnetic field $B_2$ in the same way that $E_2$ was related to $B_1$. We can use Eq. (23.6) all over again just by changing the subscripts: \begin{equation} \label{Eq:II:23:12} E_3(r)=\ddp{}{t}\int B_2(r)\,dr. \end{equation}  Using our result, Eq. (23.11), for $B_2$, the new correction to the electric field is \begin{equation} \label{Eq:II:23:13} E_3(r)=+\frac{\omega^4r^4}{64c^4}\,E_0e^{i\omega t}. \end{equation}  Writing our doubly corrected electric field as $E=E_1+E_2+E_3$, we get \begin{equation} \label{Eq:II:23:14} E=E_0e^{i\omega t}\biggl[ 1-\frac{1}{2^2}\biggl(\frac{\omega r}{c}\biggr)^2\kern{-1.25ex}+ \frac{1}{2^2\cdot4^2}\biggl(\frac{\omega r}{c}\biggr)^4 \biggr]. \end{equation}  The variation of the electric field with radius is no longer the simple parabola we drew in Fig. 23–5, but at large radii lies slightly above the curve $(E_1+E_2)$. We are not quite through yet. The new electric field produces a new correction to the magnetic field, and the newly corrected magnetic field will produce a further correction to the electric field, and on and on. However, we already have all the formulas that we need. For $B_3$ we can use Eq. (23.10), changing the subscripts of $B$ and $E$ from $2$ to $3$. The next correction to the electric field is \begin{equation*} E_4=-\frac{1}{2^2\cdot4^2\cdot6^2} \biggl(\frac{\omega r}{c}\biggr)^6 E_0e^{i\omega t}. \end{equation*}  So to this order we have that the complete electric field is given by  \begin{equation} \label{Eq:II:23:15} E=E_0e^{i\omega t}\biggl[ 1-\frac{1}{(1!)^2}\biggl(\frac{\omega r}{2c}\biggr)^2\kern{-1.25ex}+ \frac{1}{(2!)^2}\biggl(\frac{\omega r}{2c}\biggr)^4\kern{-1.25ex}- \frac{1}{(3!)^2}\biggl(\frac{\omega r}{2c}\biggr)^6\pm\dotsb \biggr]. \end{equation} 
 \begin{align} \label{Eq:II:23:15} E=E_0&e^{i\omega t}\biggl[ 1-\frac{1}{(1!)^2}\biggl(\frac{\omega r}{2c}\biggr)^2\kern{-1ex}+\\[1ex] &\frac{1}{(2!)^2}\biggl(\frac{\omega r}{2c}\biggr)^4\kern{-1ex}- \frac{1}{(3!)^2}\biggl(\frac{\omega r}{2c}\biggr)^6\kern{-1ex}\pm\dotsb \biggr].\notag \end{align}  where we have written the numerical coefficients in such a way that it is obvious how the series is to be continued. Our final result is that the electric field between the plates of the capacitor, for any frequency, is given by $E_0e^{i\omega t}$ times the infinite series which contains only the variable $\omega r/c$. If we wish, we can define a special function, which we will call $J_0(x)$, as the infinite series that appears in the brackets of Eq. (23.15):  \begin{equation} \label{Eq:II:23:16} J_0(x)=1-\frac{1}{(1!)^2}\biggl(\frac{x}{2}\biggr)^2\kern{-1.25ex}+ \frac{1}{(2!)^2}\biggl(\frac{x}{2}\biggr)^4\kern{-1.25ex}- \frac{1}{(3!)^2}\biggl(\frac{x}{2}\biggr)^6\pm\dotsb \end{equation} 
 \begin{align} \label{Eq:II:23:16} J_0(x)=&\;1-\frac{1}{(1!)^2}\biggl(\frac{x}{2}\biggr)^2\kern{-1ex}+\\[1ex] &\frac{1}{(2!)^2}\biggl(\frac{x}{2}\biggr)^4\kern{-1ex}- \frac{1}{(3!)^2}\biggl(\frac{x}{2}\biggr)^6\kern{-1ex}\pm\dotsb\notag \end{align}  Then we can write our solution as $E_0e^{i\omega t}$ times this function, with $x=\omega r/c$: \begin{equation} \label{Eq:II:23:17} E=E_0e^{i\omega t}J_0\biggl(\frac{\omega r}{c}\biggr). \end{equation}  The reason we have called our special function $J_0$ is that, naturally, this is not the first time anyone has ever worked out a problem with oscillations in a cylinder. The function has come up before and is usually called $J_0$. It always comes up whenever you solve a problem about waves with cylindrical symmetry. The function $J_0$ is to cylindrical waves what the cosine function is to waves on a straight line. So it is an important function, invented a long time ago. Then a man named Bessel got his name attached to it. The subscript zero means that Bessel invented a whole lot of different functions and this is just the first of them. The other functions of Bessel—$J_1$, $J_2$, and so on—have to do with cylindrical waves which have a variation of their strength with the angle around the axis of the cylinder.  The completely corrected electric field between the plates of our circular capacitor, given by Eq. (23.17), is plotted as the solid line in Fig. 23–5. For frequencies that are not too high, our second approximation was already quite good. The third approximation was even better—so good, in fact, that if we had plotted it, you would not have been able to see the difference between it and the solid curve. You will see in the next section, however, that the complete series is needed to get an accurate description for large radii, or for high frequencies.