CHAPTER 7 :POISSON''''S AND LAPLACE''''S EQUATIONSA doc

CHAPTER 7 POISSON'S AND LAPLACE'S EQUATIONS A study of the previous chapter shows that several of the analogies used to obtain experimental field maps involved demonstrating that the analogous quan- tity satisfies Laplace's equation. This is true for small deflections of an elastic membrane, and we might have proved the current analogy by showing that the direct-current density in a conducting medium also satisfies Laplace's equation. It appears that this is a fundamental equation in more than one field of science, and, perhaps without knowing it, we have spent the last chapter obtaining solu- tions for Laplace's equation by experimental, graphical, and numerical methods. Now we are ready to obtain this equation formally and discuss several methods by which it may be solved analytically. It may seem that this material properly belongs before that of the previous chapter; as long as we are solving one equation by so many methods, would it not be fitting to see the equation first? The disadvantage of this more logical order lies in the fact that solving Laplace's equation is an exercise in mathe- matics, and unless we have the physical problem well in mind, we may easily miss the physical significance of what we are doing. A rough curvilinear map can tell us much about a field and then may be used later to check our mathematical solutions for gross errors or to indicate certain peculiar regions in the field which require special treatment. With this explanation let us finally obtain the equations of Laplace and Poisson. 195 | | | | ▲ ▲ e-Text Main Menu Textbook Table of Contents 7.1 POISSON'S AND LAPLACE'S EQUATIONS Obtaining Poisson's equation is exceedingly simple, for from the point form of Gauss's law, rÁ D   v 1 the definition of D, D  E 2 and the gradient relationship, E ÀrV 3 by substitution we have rÁ D rÁ EÀrÁ rV v or rÁrV À  v  4 for a homogeneous region in which  is constant. Equation (4) is Poisson's equation, but the ``double r'' operation must be interpreted and expanded, at least in cartesian coordinates, before the equation can be useful. In cartesian coordinates, rÁA  @A x @x  @A y @y  @A z @z rV  @V @x a x  @V @y a y  @V @z a z and therefore rÁrV  @ @x @V @x   @ @y @V @y   @ @z @V @z   @ 2 V @x 2  @ 2 V @y 2  @ 2 V @z 2 5 Usually the operation rÁ r is abbreviated r 2 (and pronounced ``del squared''), a good reminder of the second-order partial derivatives appearing in (5), and we have 196 ENGINEERING ELECTROMAGNETICS | | | | ▲ ▲ e-Text Main Menu Textbook Table of Contents r 2 V  @ 2 V @x 2  @ 2 V @y 2  @ 2 V @z 2 À  v  6 in cartesian coordinates. If  v  0, indicating zero volume charge density, but allowing point charges, line charge, and surface charge density to exist at singular locations as sources of the field, then r 2 V  0 7 which is Laplace's equation. The r 2 operation is called the Laplacian of V. In cartesian coordinates Laplace's equation is r 2 V  @ 2 V @x 2  @ 2 V @y 2  @ 2 V @z 2  0 cartesian8 and the form of r 2 V in cylindrical and spherical coordinates may be obtained by using the expressions for the divergence and gradient already obtained in those coordinate systems. For reference, the Laplacian in cylindrical coordinates is r 2 V  1  @ @  @V @   1  2 @ 2 V @ 2   @ 2 V @z 2 cylindrical9 and in spherical coordinates is r 2 V  1 r 2 @ @r r 2 @V @r   1 r 2 sin  @ @ sin  @V @   1 r 2 sin 2  @ 2 V @ 2 spherical10 These equations may be expanded by taking the indicated partial derivatives, but it is usually more helpful to have them in the forms given above; furthermore, it is much easier to expand them later if necessary than it is to put the broken pieces back together again. Laplace's equation is all-embracing, for, applying as it does wherever volume charge density is zero, it states that every conceivable configuration of electrodes or conductors produces a field for which r 2 V  0. All these fields are different, with different potential values and different spatial rates of change, yet for each of them r 2 V  0. Since every field (if  v  0 satisfies Laplace's equa- tion, how can we expect to reverse the procedure and use Laplace's equation to find one specific field in which we happen to have an interest? Obviously, more POISSON'S AND LAPLACE'S EQUATIONS 197 | | | | ▲ ▲ e-Text Main Menu Textbook Table of Contents information is required, and we shall find that we must solve Laplace's equation subject to certain boundary conditions. Every physical problem must contain at least one conducting boundary and usually contains two or more. The potentials on these boundaries are assigned values, perhaps V 0 , V 1 ; , or perhaps numerical values. These definite equi- potential surfaces will provide the boundary conditions for the type of problem to be solved in this chapter. In other types of problems, the boundary conditions take the form of specified values of E on an enclosing surface, or a mixture of known values of V and E: Before using Laplace's equation or Poisson's equation in several examples, we must pause to show that if our answer satisfies Laplace's equation and also satisfies the boundary conditions, then it is the only possible answer. It would be very distressing to work a problem by solving Laplace's equation with two different approved methods and then to obtain two different answers. We shall show that the two answers must be identical. \ D7.1. Calculate numerical values for V and  v at point P in free space if: a V  4yz x 2  1 ,atP1; 2; 3; b V  5 2 cos 2,atP  3;  3 ; z  2; c V  2 cos r 2 ,atPr  0:5;  458,   608: Ans.12V,À106:2 pC/m 3 ; 22.5 V, 0; 4 V, À141:7 pC/m 3 7.2 UNIQUENESS THEOREM Let us assume that we have two solutions of Laplace's equation, V 1 and V 2 , both general functions of the coordinates used. Therefore r 2 V 1  0 and r 2 V 2  0 from which r 2 V 1 À V 2 0 Each solution must also satisfy the boundary conditions, and if we repre- sent the given potential values on the boundaries by V b , then the value of V 1 on the boundary V 1b and the value of V 2 on the boundary V 2b must both be identical to V b ; V 1b  V 2b  V b or V 1b À V 2b  0 198 ENGINEERING ELECTROMAGNETICS | | | | ▲ ▲ e-Text Main Menu Textbook Table of Contents In Sec. 4.8, Eq. (44), we made use of a vector identity, rÁVDVr ÁDD ÁrV which holds for any scalar V and any vector D. For the present application we shall select V 1 À V 2 as the scalar and rV 1 À V 2  as the vector, giving rÁV 1 À V 2 rV 1 À V 2   V 1 À V 2 r Á rV 1 À V 2  rV 1 À V 2 ÁrV 1 À V 2  which we shall integrate throughout the volume enclosed by the boundary surfaces specified:  vol rÁV 1 À V 2 rV 1 À V 2 dv   vol V 1 À V 2 r Á rV 1 À V 2 dv   vol rV 1 À V 2  2 dv 11 The divergence theorem allows us to replace the volume integral on the left side of the equation by the closed surface integral over the surface surrounding the volume. This surface consists of the boundaries already specified on which V 1b  V 2b , and therefore  vol rÁV 1 À V 2 rV 1 À V 2 dv   S V 1b À V 2b rV 1b À V 2b  ÁdS  0 One of the factors of the first integral on the right side of (11) is rÁ rV 1 À V 2 ,orr 2 V 1 À V 2 , which is zero by hypothesis, and therefore that integral is zero. Hence the remaining volume integral must be zero:  vol rV 1 À V 2  2 dv  0 There are two reasons why an integral may be zero: either the integrand (the quantity under the integral sign) is everywhere zero, or the integrand is positive in some regions and negative in others, and the contributions cancel algebraically. In this case the first reason must hold because rV 1 À V 2  2 can- not be negative. Therefore rV 1 À V 2  2  0 and rV 1 À V 2 0 Finally, if the gradient of V 1 À V 2 is everywhere zero, then V 1 À V 2 cannot change with any coordinates and V 1 À V 2  constant If we can show that this constant is zero, we shall have accomplished our proof. The constant is easily evaluated by considering a point on the boundary. Here POISSON'S AND LAPLACE'S EQUATIONS 199 | | | | ▲ ▲ e-Text Main Menu Textbook Table of Contents V 1 À V 2  V 1b À V 2b  0, and we see that the constant is indeed zero, and there- fore V 1  V 2 giving two identical solutions. The uniqueness theorem also applies to Poisson's equation, for if r 2 V 1  À v = and r 2 V 2 À v =, then r 2 V 1 À V 2 0 as before. Boundary conditions still require that V 1b À V 2b  0, and the proof is identical from this point. This constitutes the proof of the uniqueness theorem. Viewed as the answer to a question, ``How do two solutions of Laplace's or Poisson's equation com- pare if they both satisfy the same boundary conditions?'' the uniqueness theorem should please us by its ensurance that the answers are identical. Once we can find any method of solving Laplace's or Poisson's equation subject to given boundary conditions, we have solved our problem once and for all. No other method can ever give a different answer. \ D7.2. Consider the two potential fields V 1  y and V 2  y  e x sin y. a Is r 2 V 1  0? b Is r 2 V 2  0? c Is V 1  0aty  0? d Is V 2  0aty  0? e Is V 1   at y  ? f  Is V 2   at y  ? g Are V 1 and V 2 identical? h Why does the uniqueness theorem not apply? Ans. Yes; yes; yes; yes; yes; yes; no; boundary conditions not given for a closed surface 7.3 EXAMPLES OF THE SOLUTION OF LAPLACE'S EQUATION Several methods have been developed for solving the second-order partial differ- ential equation known as Laplace's equation. The first and simplest method is that of direct integration, and we shall use this technique to work several exam- ples in various coordinate systems in this section. In Sec. 7.5 one other method will be used on a more difficult problem. Additional methods, requiring a more advanced mathematical knowledge, are described in the references given at the end of the chapter. The method of direct integration is applicable only to problems which are ``one-dimensional,'' or in which the potential field is a function of only one of the three coordinates. Since we are working with only three coordinate systems, it might seem, then, that there are nine problems to be solved, but a little reflection will show that a field which varies only with x is fundamentally the same as a field which varies only with y. Rotating the physical problem a quarter turn is no change. Actually, there are only five problems to be solved, one in cartesian coordinates, two in cylindrical, and two in spherical. We shall enjoy life to the fullest by solving them all. 200 ENGINEERING ELECTROMAGNETICS | | | | ▲ ▲ e-Text Main Menu Textbook Table of Contents h Example 7.1 Let us assume that V is a function only of x and worry later about which physical problem we are solving when we have a need for boundary conditions. Laplace's equa- tion reduces to @ 2 V @x 2  0 and the partial derivative may be replaced by an ordinary derivative, since V is not a function of y or z, d 2 V dx 2  0 We integrate twice, obtaining dV dx  A and V  Ax  B 12 where A and B are constants of integration. Equation (12) contains two such constants, as we should expect for a second-order differential equation. These constants can be determined only from the boundary conditions. What boundary conditions should we supply? They are our choice, since no physical problem has yet been specified, with the exception of the original hypothesis that the potential varied only with x. We should now attempt to visualize such a field. Most of us probably already have the answer, but it may be obtained by exact methods. Since the field varies only with x and is not a function of y and z, then V is a constant if x is a constant or, in other words, the equipotential surfaces are described by setting x constant. These surfaces are parallel planes normal to the x axis. The field is thus that of a parallel-plate capacitor, and as soon as we specify the potential on any two planes, we may evaluate our constants of integration. To be very general, let V  V 1 at x  x 1 and V  V 2 at x  x 2 . These values are then substituted into (12), giving V 1  Ax 1  B A  V 1 À V 2 x 1 À x 2 V 2  Ax 2  B B  V 2 x 1 À V 1 x 2 x 1 À x 2 and V  V 1 x À x 2 ÀV 2 x À x 1  x 1 À x 2 A simpler answer would have been obtained by choosing simpler boundary conditions. If we had fixed V  0atx  0andV  V 0 at x  d, then A  V 0 d B  0 POISSON'S AND LAPLACE'S EQUATIONS 201 | | | | ▲ ▲ e-Text Main Menu Textbook Table of Contents and V  V 0 x d 13 Suppose our primary aim is to find the capacitance of a parallel-plate capacitor. We have solved Laplace's equation, obtaining (12) with the two constants A and B. Should they be evaluated or left alone? Presumably we are not interested in the potential field itself, but only in the capacitance, and we may continue successfully with A and B or we may simplify the algebra by a little foresight. Capacitance is given by the ratio of charge to potential difference, so we may choose now the potential difference as V 0 , which is equivalent to one boundary condition, and then choose whatever second boundary condition seems to help the form of the equation the most. This is the essence of the second set of boundary conditions which produced (13). The potential difference was fixed as V 0 by choosing the potential of one plate zero and the other V 0 ; the location of these plates was made as simple as possible by letting V  0atx  0: Using (13), then, we still need the total charge on either plate before the capacitance can be found. We should remember that when we first solved this capacitor problem in Chap. 5, the sheet of charge provided our starting point. We did not have to work very hard to find the charge, for all the fields were expressed in terms of it. The work then was spent in finding potential difference. Now the problem is reversed (and simplified). The necessary steps are these, after the choice of boundary conditions has been made: 1. Given V, use E ÀrV to find E: 2. Use D  E to find D: 3. Evaluate D at either capacitor plate, D  D S  D N a N : 4. Recognize that  S  D N : 5. Find Q by a surface integration over the capacitor plate, Q   S  S dS: Here we have V  V 0 x d E À V 0 d a x D À V 0 d a x 202 ENGINEERING ELECTROMAGNETICS | | | | ▲ ▲ e-Text Main Menu Textbook Table of Contents D S  D     x0 À V 0 d a x a N  a x D N À V 0 d   S Q   S ÀV 0 d dS À V 0 S d and the capacitance is C  jQj V 0  S d 14 We shall use this procedure several times in the examples to follow. h Example 7.2 Since no new problems are solved by choosing fields which vary only with y or with z in cartesian coordinates, we pass on to cylindrical coordinates for our next example. Variations with respect to z are again nothing new, and we next assume variation with respect to  only. Laplace's equation becomes 1  @ @  @V @   0 or 1  d d  dV d   0 Noting the  in the denominator, we exclude   0 from our solution and then multiply by  and integrate,  dV d  A rearrange, and integrate again, V  A ln  B 15 The equipotential surfaces are given by   constant and are cylinders, and the problem is that of the coaxial capacitor or coaxial transmission line. We choose a potential difference of V 0 by letting V  V 0 at   a, V  0at  b, b > a, and obtain V  V 0 lnb= lnb=a 16 POISSON'S AND LAPLACE'S EQUATIONS 203 | | | | ▲ ▲ e-Text Main Menu Textbook Table of Contents from which E  V 0  1 lnb=a a  D Na  V 0 a lnb=a Q  V 0 2aL a lnb=a C  2L lnb=a 17 which agrees with our results in Chap. 5. h Example 7.3 Now let us assume that V is a function only of  in cylindrical coordinates. We might look at the physical problem first for a change and see that equipotential surfaces are given by   constant. These are radial planes. Boundary conditions might be V  0at   0andV  V 0 at   , leading to the physical problem detailed in Fig. 7.1. Laplace's equation is now 1  2 @ 2 V @ 2  0 We exclude   0 and have d 2 V d 2  0 The solution is V  A  B 204 ENGINEERING ELECTROMAGNETICS FIGURE 7.1 Two infinite radial planes with an interior angle . An infinitesimal insulating gap exists at   0. The potential field may be found by applying Laplace's equation in cylindrical coordinates. | | | | ▲ ▲ e-Text Main Menu Textbook Table of Contents [...]... Er 3 0 when r 3 0, and also that V 3 0 as r 3 I …b† Now find V…r† by using Gauss's law and a line integral 7. 22 Let the volume charge density in Fig 7: 3a be given by v ˆ v0 …x=a†eÀjxja …a† Determine v;max and v;min and their locations …b† Find Ex and V…x† if V…0† ˆ 0 and Ex 3 0 as x 3 I …c† Use a development similar to that of Sec 7. 4 to show that C ˆ dQ=dV0 ˆ p 0 S=…4 2a† 7. 23 A rectangular... V1 ? 7. 10 Conducting planes at z ˆ 2 cm and z ˆ 8 cm are held at potentials of À3 V and 9 V, respectively The region between the plates is filled with a perfect dielectric with  ˆ 50 Find and sketch: …a† V…z†; …b† Ez …z†; …c† Dz …z†: 7. 11 The conducting planes 2x ‡ 3y ˆ 12 and 2x ‡ 3y ˆ 18 are at potentials of 100 V and 0, respectively Let  ˆ 0 and find: …a† V at P…5; 2; 6†; …b† E at P: 7. 12 Conducting... center of the trough 7. 25 In Fig 7. 7 change the right side so that the potential varies linearly from 0 at the bottom of that side to 100 V at the top Solve for the potential at the center of the trough 7. 26 If X is a function of x and X HH ‡ …x À 1†X H À 2X ˆ 0, assume a solution in the form of an infinite power series and determine numerical values for a2 to a8 if a0 ˆ 1 and a1 ˆ À1: 7. 27 It is known that... 0:188 À 2, and then find S and Q on the lower surface of the lower plane …g† Find the total charge on the lower plane and the capacitance between the planes 7. 16 …a† Solve Laplace's equation for the potential field in the homogeneous region between two concentric conducting spheres with radii a and b; b > a, if V ˆ 0 at r ˆ b, and V ˆ V0 at r ˆ a …b† Find the capacitance between them 7. 17 Concentric... of the fields having the form given by (36) or ( 37) and thus satisfying the boundary conditions of a more practical problem We close this chapter with such an example The problem to be solved is that shown in Fig 7. 6 The boundary conditions shown are V ˆ 0 at x ˆ 0, y ˆ 0, and y ˆ b, and V ˆ V0 at x ˆ d for all y FIGURE 7. 5 Cross section of the V ˆ 0 and V ˆ V1 equipotential surfaces for the potential... conversion, and ion propulsion require its use in constructing satisfactory theories \ D7.4 In the neighborhood of a certain semiconductor junction the volume charge density is given by v ˆ 75 0 sech 106 x tanh x C=m3 The dielectric constant of the semiconductor material is 10 and the junction area is 2  10 7 m2 Find: …a† V0 ; …b† C; …c† E at the junction Ans 2 .70 V; 8.85 pF; 2 .70 MV/m \ p D7.5 Given... zero-potential plane and add this amount to our answer above Fringing, or nonuniform, fields in this region have been neglected and introduce an additional source of error \ D7.3 Find jEj at P…3; 1; 2† for the field of: …a† two coaxial conducting cylinders, V ˆ 50 V at  ˆ 2 m, and V ˆ 20 V at  ˆ 3 m; …b† two radial conducting planes, V ˆ 50 V at  ˆ 108, and V ˆ 20 V at  ˆ 308: Ans 23.4 V/m; 27. 2 V/m 7. 4 EXAMPLE... by four conducting planes located at x ˆ 0 and 8 cm and y ˆ 0 and 5 cm in air The surface at y ˆ 5 cm is at a potential of 100 V, the other three are at zero potential, and the necessary gaps are placed at two corners Find the potential at x ˆ 3 cm, y ˆ 4 cm 7. 24 The four sides of a square trough are held at potentials of 0, 20, À30, and 60 V; the highest and lowest potentials are on opposite sides... space where v ˆ 0 It is known that both Ex and V are zero at the origin Find f …x† and   V…x; y†: 2 7. 4 Given the potential field V ˆ A ln tan ‡ B: …a† show that r2 V ˆ 0; 2 …b† select A and B so that V ˆ 100 V and E ˆ 500 V/m at P…r ˆ 5,  ˆ 608,  ˆ 458†: | v v 220 | e-Text Main Menu | Textbook Table of Contents | POISSON'S AND LAPLACE'S EQUATIONS | v v 7. 5 Given the potential field V ˆ …A4 ‡... charge density on a conductor surface passing through the point B…2; 308; 1†: 20 4 7. 8 Let V1 …r; ; † ˆ and V2 …r; ; † ˆ ‡ 4 …a† State whether V1 and V2 r r satisfy Laplace's equation …b† Evaluate V1 and V2 on the closed surface r ˆ 4 …c† Conciliate your results with the uniqueness theorem 7. 9 The functions V1 …; ; z† and V2 …; ; z† both satisfy Laplace's equation in the region a <  < b, 0  . solution and then multiply by  and integrate,  dV d  A rearrange, and integrate again, V  A ln  B 15 The equipotential surfaces are given by   constant and are cylinders, and the problem. À141 :7 pC/m 3 7. 2 UNIQUENESS THEOREM Let us assume that we have two solutions of Laplace's equation, V 1 and V 2 , both general functions of the coordinates used. Therefore r 2 V 1  0 and r 2 V 2 . CHAPTER 7 POISSON'S AND LAPLACE'S EQUATIONS A study of the previous chapter shows that several of the analogies used to obtain