MIT OpenCourseWare http://ocw.mit.edu Electromagnetic Field Theory: A Problem Solving Approach For any use or distribution of this textbook, please cite as follows: Markus Zahn, Electromagnetic Field Theory: A Problem Solving Approach. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-NonCommercial-Share Alike. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.chapter 4electricfield boundaryvalue problems258 Electric Field Boundary Value ProblemsThe electric field distribution due to external sources isdisturbed by the addition of a conducting or dielectric bodybecause the resulting induced charges also contribute to thefield. The complete solution must now also satisfy boundaryconditions imposed by the materials.4-1 THE UNIQUENESS THEOREMConsider a linear dielectric material where the permittivitymay vary with position:D = e(r)E = -e(r)VV (1)The special case of different constant permittivity mediaseparated by an interface has e (r) as a step function. Using (1)in Gauss's law yieldsV -[(r)VV]=-pf (2)which reduces to Poisson's equation in regions where E (r) is aconstant. Let us call V, a solution to (2).The solution VL to the homogeneous equationV -[e(r)V VI= 0 (3)which reduces to Laplace's equation when e(r) is constant,can be added to Vp and still satisfy (2) because (2) is linear inthe potential:V -[e (r)V( Vp + VL)] = V [e (r)V VP] +V [e (r)V VL] = -Pf0 (4)Any linear physical problem must only have one solutionyet (3) and thus (2) have many solutions. We need to findwhat boundary conditions are necessary to uniquely specifythis solution. Our method is to consider two different solu-tions V1 and V2 for the same charge distributionV (eV Vi)= -P, V (eV V2) = -Pf (5)so that we can determine what boundary conditions forcethese solutions to be identical, V, = V2.___Boundary Value Problems in CartesianGeometries 259The difference of these two solutions VT = V, - V2 obeysthe homogeneous equationV* (eV Vr) = 0 (6)We examine the vector expansionV *(eVTVVT)= VTV (EVVT)+eVVT" VVT= eVVTI2 (7)0noting that the first term in the expansion is zero from (6) andthat the second term is never negative.We now integrate (7) over the volume of interest V, whichmay be of infinite extent and thus include all spaceV.V(eVTVVT)dV= eVTVVT-dS=I EIVVTI dV (8)The volume integral is converted to a surface integral overthe surface bounding the region using the divergencetheorem. Since the integrand in the last volume integral of (8)is never negative, the integral itself can only be zero if VT iszero at every point in the volume making the solution unique(VT= O0 V1 = V2). To force the volume integral to be zero,the surface integral term in (8) must be zero. This requiresthat on the surface S the two solutions must have the samevalue (VI = V2) or their normal derivatives must be equal[V V1 -n = V V2 n]. This last condition is equivalent torequiring that the normal components of the electric fields beequal (E = -V V).Thus, a problem is uniquely posed when in addition togiving the charge distribution, the potential or the normalcomponent of the electric field on the bounding surface sur-rounding the volume is specified. The bounding surface canbe taken in sections with some sections having the potentialspecified and other sections having the normal fieldcomponent specified.If a particular solution satisfies (2) but it does not satisfythe boundary conditions, additional homogeneous solutionswhere pf = 0, must be added so that the boundary conditionsare met. No matter how a solution is obtained, even ifguessed, if it satisfies (2) and all the boundary conditions, it isthe only solution.4-2 BOUNDARY VALUE PROBLEMS IN CARTESIAN GEOMETRIESFor most of the problems treated in Chapters 2 and 3 werestricted ourselves to one-dimensional problems where theelectric field points in a single direction and only depends onthat coordinate. For many cases, the volume is free of chargeso that the system is described by Laplace's equation. Surface260 Electric FieldBoundary Value Problemscharge is present only on interfacial boundaries separatingdissimilar conducting materials. We now consider suchvolume charge-free problems with two-and three dimen-sional variations.4-2-1 Separation of VariablesLet us assume that within a region of space of constantpermittivity with no volume charge, that solutions do notdepend on the z coordinate. Then Laplace's equation reducesto82V O2Vax2 +y2 = 0 (1)We try a solution that is a product of a function only of the xcoordinate and a function only of y:V(x, y) = X(x) Y(y) (2)This assumed solution is often convenient to use if the systemboundaries lay in constant x or constant y planes. Then alonga boundary, one of the functions in (2) is constant. When (2) issubstituted into (1) we have_d'2X d2Y 1 d2X 1 d2,YY-+X = 0 + (3)dx dy X dx2 Y dywhere the partial derivatives become total derivatives becauseeach function only depends on a single coordinate. Thesecond relation is obtained by dividing through by XY so thatthe first term is only a function of x while the second is only afunction of y.The only way the sum of these two terms can be zero for allvalues of x and y is if each term is separately equal to aconstant so that (3) separates into two equations,1 d2X 2 1 d2Y_kX k d----(4)where k2 is called the separation constant and in general canbe a complex number. These equations can then be rewrittenas the ordinary differential equations:d2X d2Sk-2X= O, ++k'Y=O2dx dyBoundary Value Problems in CartesianGeometries2614-2-2 Zero Separation Constant SolutionsWhen the separation constant is zero (A2 = 0) the solutionsto (5) areX = arx +bl,Y= cry+dlwhere a,, b1, cl, and dl are constants. The potential is given bythe product of these terms which is of the formV = a2+ b2x + C2y + d2xyThe linear and constant terms we have seen before, as thepotential distribution within a parallel plate capacitor with nofringing, so that the electric field is uniform. The last term wehave not seen previously.(a) Hyperbolic ElectrodesA hyperbolically shaped electrode whose surface shapeobeys the equation xy = ab is at potential Vo and is placedabove a grounded right-angle corner as in Figure 4-1. TheVo0525125Equipotential lines -- -Vo abField lines -y2 -X2 = const.Figure 4-1 The equipotential and field lines for a hyperbolically shaped electrode atpotential Vo above
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