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r dr'G(r, r')(Er(r') - 1)k E(r') + ESelf(r) Jv- vo
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(2.1.41)
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(2.1.42) is the self field. The source is an infinitesimal volume Vb of relative permittivity E = EplE (i.e. relative to the background E), and with internal field r E(r) inside Vb. On the other hand, the first two terms on the right-hand side of (2.1.41) is the contribution to E(r) from the incident field and its surrounding medium. Since Vb -----+ 0, the self field Eself(r) can be interpreted in terms of electrostatic field created by free charge and polarization charge densities and surface charge densities. On the other hand, if the vector potential and scalar potential are used, we have the volume integral equation of (2.1.37). The subject of singularity of the dyadic Green's function are discussed in [van Bladel, 1961; Livesay and Chen, 1974; Yaghjian, 1980; Tsang et. al. 1985; Chew, 1990]. We note that by comparing (2.1.37) and (2.1.39) that the term due to electric current
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2 INTEGRAL EQUATION FORMULATIONS AND NUMERICAL METHODS
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] is benign. The singularity comes from the electrostatic part of electric field due to charges that are in the third term of (2.1.37) and (2.1.39). Thus we shall use electrostatics and examine how electric charges and polarization charges produce electric field in the source region. In the electrostatic limit
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g(r, r') = 41fr-r 1- _'I
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The electrostatic equations are
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(2.1.43)
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The electrostatic potential is <I> such that
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(2.1.44) (2.1.45) (2.1.46)
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E=-\7<I>
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D = EE+P P = (Ep(r) - E)E
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(2.1.48) (2.1.49) (2.1.50)
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PP = - \7 . P = - \7 . [(Ep(r) - E)E]
(2.1.51)
is the polarization charge density. The surface polarization charge is
(2.1.52) The electrostatic potential can be calculated from the superposition integral of these charges
-n (Pout -
Pin) = (jp
<I>(r) =
J +~J
dr'g(r, r') [pj(r')
+ pp(r')] + (jp(r')]
(2.1.53)
dS' g(r, r') [(js(r')
For the case of an infinitesimal volume V8, we can assume that it is of homogeneous permittivity Ep inside V8 (Fig. 2.1.4). The field inside is E. Pin = (Ep Pout = 0
(2.1.54)
(2.1.55)
1.3 Dyadic Green's Function Singularity and Electrostatics
Figure 2.1.4 Field produced by
If inside Vo on itself.
Note that Pin = constant inside V8 so that Pp = - \7 . Pin = O. For dielectric problem, PI = (js = O. Thus we only have
(jp =
-n (Pout =
Pin)
n' (Er -
I)EE
(2.1.56)
Thus for r inside V8, from (2.1.53) and (2.1.56),
q)self(r)
dS' g(r, r')(jp(r')
41r r - r
JdS'
1_1 _'I n''(E r
-l)E(r')
(2.1.57)
The electric field is E se1f = -\7q)self(r). Let H = (r - r')1 R with R = Ir - r'l and k = - k Then - \7 (1 I R) = HI R 2 . Also V8 is small so that E is constant inside V8.
Eself(r)
Js odS' 41r l~n'_'12' (E r r - r
I)E(r') = -(E r
1) . E
(2.1.58)
where (2.1.59)
-, r.
Note that r is at the center of V8 and
is a unit vector pointing from r to
o and 0 -----+ 0,
Example 1: Without loss of generality, let r = O. If V8 is a sphere of radius then k = n' = r = sin () cos x + sin () sin fJ + cos ()z. We have
L = = -1
d() sin ()
(2.1.60a)
2 INTEGRAL EQUATION FORMULATIONS AND NUMERICAL METHODS
rr =
dD(sin ecos <px + sin esin <py + cos ez)
. (sin ecos <px + sin esin <py + cos ez) 41r= = -1 (2.1.60b) 3 We have used the result that on integration, the cross terms in (2.1.60) vanish. Hence
L = xx + yy + zz
(2.1.61)
Example 2: Let V6 be a rectangular parallelepiped with sides equal to Ox, Oy and oz, respectively. Also let Ox = ao, Oy = M and Oz = cO and 0 -----+ O. Thus a, band e ar~ finite numbers and the ratios among them are important to determine L. In this case there are six faces for the surface integral of (2.1.60). The sum of the contributions from the top and bottom faces to L
~ lim j6 /2 dx j6,/2 dy z(xx + yy + (Oz/2)z) - z(xx + yy - (Oz/2)z) 41r 6--->0 -6 x /2 -6,/2 (x 2 + y2 + (Oz/2)2)3/2