Question 3.3: Two infinite grounded metal plates lie parallel to the xz pl...

The configuration is independent of z, so this is really a two-imensional problem.In mathematical terms, we must solve Laplace’s equation,

\frac{\delta ^{2}V}{\delta x^{2}}+\frac{\delta ^{2}V}{\delta y^{2}}=0.      (3.20)

subject to the boundary conditions

\begin{cases}(i) V = 0 when y = 0,\\ii) V = 0 when y = a,\\(iii) V = V0(y) when x = 0,\\ (iv) V → 0 as x →∞.\end{cases}           (3.21)

(The latter, although not explicitly stated in the problem, is necessary on physical grounds: as you get farther and farther away from the “hot” strip at x = 0, the potential should drop to zero.) Since the potential is specified on all boundaries, the answer is uniquely determined. The first step is to look for solutions in the form of products:
V(x, y) = X(x)Y (y).           (3.22)
On the face of it, this is an absurd restriction—the overwhelming majority of solutions to Laplace’s equation do not have such a form. For example, V(x, y) =(5x + 6y) satisfies Eq. 3.20, but you can’t express it as the product of a function x times a function y. Obviously, we’re only going to get a tiny subset of all possible solutions by this means, and it would be a miracle if one of them happened to fit the boundary conditions of our problem . . . But hang on, because the solutions we do get are very special, and it turns out that by pasting them together we can construct the general solution. Anyway, putting Eq. 3.22 into Eq. 3.20, we obtain

The next step is to “separate the variables” (that is, collect all the x-dependence into one term and all the y-dependence into another). Typically, this is accomplished by dividing through by V:

\frac{1}{X} \frac{d^{2}X}{d^{2}x^{2}}+\frac{1}{Y} \frac{d^{2}Y}{dy^{2}}=0 .                         (3.23)

Here the first term depends only on x and the second only on y; in other words, we have an equation of the form

f (x) + g(y) = 0.                     (3.24)

Now, there’s only one way this could possibly be true: f and g must both be constant. For what if f (x) changed, as you vary x—then if we held y fixed and fiddled with x, the sum f (x) + g(y) would change, in violation of Eq. 3.24, which says it’s always zero. (That’s a simple but somehow rather elusive argument; don’t accept it without due thought, because the whole method rides on it.) It follows from Eq. 3.23, then, that

\frac{1}{X} \frac{d^{2}X}{d^{2}x^{2}}=C_{1} and \frac{1}{Y} \frac{d^{2}Y}{dy^{2}}=C_{2} with C_{1}+C_{2}=0.   (3.25)

One of these constants is positive, the other negative (or perhaps both are zero). In general, one must investigate all the possibilities; however, in our particular problem we need C_{1} positive and C_{2} negative, for reasons that will appear in a moment. Thus

\frac{d^{2}X}{d^{2}x^{2}}=k^{2}X, \frac{1}{Y} \frac{d^{2}Y}{dy^{2}}=-k^{2}Y .         (3.26)

Notice what has happened: A partial differential equation (3.20) has been converted into two ordinary differential equations (3.26). The advantage of this is obvious—ordinary differential equations are a lot easier to solve. Indeed:

so

V(x,y)=(Ae^{kx}+Be^{-kx}) (C\sin ky + D\cos ky),                 (3.27)

This is the appropriate separable solution to Laplace’s equation; it remains to impose the boundary conditions, and see what they tell us about the constants. To begin at the end, condition (iv) requires that A equal zero.^{8} Absorbing B into C and D, we are left with

Condition (i) now demands that D equal zero, so

V(x,y)=Ce^{-kx} sin ky ,                  (3.28)

Meanwhile (ii) yields sin ka = 0, from which it follows that

k=\frac{n \pi }{a} , (n=1,2,3,….)              (3.29)

(At this point you can see why I chose C_{1} positive and C_{2} negative: If X were sinusoidal, we could never arrange for it to go to zero at infinity, and if Y were exponential we could not make it vanish at both 0 and a. Incidentally, n = 0 is no good, for in that case the potential vanishes everywhere. And we have already excluded negative n’s.) That’s as far as we can go, using separable solutions, and unless V_{0}(y) just happens to have the form sin(n\pi y/a) for some integer n, we simply can’t fit the final boundary condition at x = 0. But now comes the crucial step that redeems the method: Separation of variables has given us an infinite family of solutions (one for each n), and whereas none of them by itself satisfies the final boundary condition, it is possible to combine them in a way that does.  aplace’s equation is linear, in the sense that if V_{1}, V_{2}, V_{3}, . . . satisfy it, so does any linear combination,
V = α_{1}V_{1} + α_{2}V_{2} + α_{3}V_{3} + . . . , where α_{1}, α_{2}, . . . are arbitrary constants.
For

Exploiting this fact, we can patch together the separable solutions (Eq. 3.28) to construct a much more general solution:

V(x,y)=\sum\limits_{n=1}^{\infty }{C_{n}e^{-n\pi x/a}\sin (n \pi y/a)}.     (3.30)

This still satisfies three of the boundary conditions; the question is, can we (by astute choice of the coefficients C_{n}) fit the final boundary condition (iii)?

V(0,y)=\sum\limits_{n=1}^{\infty }{C_{n}\sin (n \pi y/a)}=V_{0}(y).       (3.31)

Well, you may recognize this sum—it’s a Fourier sine series. And Dirichlet’s theorem^{9} guarantees that virtually any function V_{0}(y)—it can even have a finite number of discontinuities—can be expanded in such a series. But how do we actually determine the coefficients C_{n}, buried as they are in that infinite sum? The device for accomplishing this is so lovely it deserves a name—I call it Fourier’s trick, though it seems Euler had used essentially the same idea somewhat earlier. Here’s how it goes: Multiply Eq. 3.31 by sin(\acute{n} πy/a) (where \acute{n} is a positive integer), and integrate from 0 to a:

\sum\limits_{n=1}^{\infty }{C_{n}\int_{0}^{a}{} \sin (n \pi y/a)\sin (\acute{n }\pi y/a})dy=\int_{0}^{a}{V_{0}(y)\sin (\acute{n}\pi y/a)dy} .                   (3.32)

You can work out the integral on the left for yourself; the answer is

\int_{0}^{a}{} \sin (n \pi y/a)\sin (\acute{n }\pi y/a)dy=\begin{cases} 0, if \acute{n}\neq n, \\ \frac{a}{2}, if \acute{n}\neq n\end{cases} .         (3.33)

Thus all the terms in the series drop out, save only the one where n =\acute{n}, and the left side of Eq. 3.32, reduces to (a/2)C_{\acute{n}} . Conclusion:10

C_{n}\frac{2}{a}\int_{0}^{a}{V_{0}(y)\sin (n\pi y/a)dy}.    (3.34)

That does it: Eq. 3.30 is the solution, with coefficients given by  Eq.3.34. As a concrete example, suppose the strip at x = 0 is a metal plate with constant potential V_{0} (remember, it’s insulated from the grounded plates at y = 0 and y = a). Then

C_{n}=\frac{2V_{0}}{a} \int_{0}^{a}{} \sin (n \pi y/a)=\frac{2V_{0}}{n\pi }(1-\cos n\pi ) =\begin{cases} 0, if n is even, \\ \frac{a}{2}, if n is odd\end{cases}.       (3.35)

V(x,y)=\frac{4V_{0}}{\pi } \sum\limits_{n=1.3.5..}^{}{\frac{1}{n}e^{-n\pi x/a} }\sin(n\pi y/a).       (3.36)

Figure 3.18 is a plot of this potential; Fig. 3.19 shows how the first few terms in the Fourier series combine to make a better and better approximation to the constant V_{0}: (a) is n = 1 only, (b) includes n up to 5, (c) is the sum of the first 10 terms, and (d) is the sum of the first 100 terms.

Incidentally, the infinite series in Eq. 3.36 can be summed explicitly (try your hand at it, if you like); the result is

V(x,y)=\frac{4V_{0}}{\pi }\tan^{-1}\left(\frac{\sin (\pi )}{\sinh (\pi x/a)} \right) .         (3.37)

In this form, it is easy to check that Laplace’s equation is obeyed and the four boundary conditions (Eq. 3.21) are satisfied.