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Thus, for Dirichlet boundary conditions we demand : GD(\, x') = 0 for x' on S (1.43)
Then the first term in the surface integral in (1.42) vanishes and the solution is <D
(x) = f P(x')GD(x, x') d3x' - -1 <£ O(x') ^ da' (1.44) Jv 4n Js on For Neumann ...
This means that the fields, and therefore y(x), will satisfy the radiation condition, C
-, r y (9-64) With this condition on y it can readily be seen that the integral in (9.63
) over the hemisphere S2 vanishes inversely as the hemisphere radius as that ...
... E) x V'G - Gn x (V' x E) (9.72) While it may not appear very fruitful to transform
the two terms in (9.68) into six terms, we will now show that the surface integral of
the first three terms in (9.72), involving the product (GE), vanishes identically.
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Introduction to Electrostatics
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