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Thus, for Dirichlet boundary conditions we demand: Gn(x,x) = 0 for x' on S (1.43)
Then the first term in the surface integral in (1.42) vanishes and the solution is q}(
x) = | 2006/s ×) or -4. f ox)*da (144) P 477 JS 0n' For Neumann boundary ...
(9.64) With this condition on p it can readily be seen that the integral in (9,63)
over the hemisphere S, vanishes inversely as the hemisphere radius as that
radius goes to infinity. Then we obtain the Kirchhoff integral for p(x) in region II: 1
ik R x) ...
(GE))] dor' (9.74) y From the expansion, V × V × A = V(V - A) — W*A, it is evident
that the volume integral vanishes identically.” With the surface integral of the first
three terms in (9.72) identically zero, the remaining three terms give an ...
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Introduction to Electrostatics
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