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Page 19
Thus , for Dirichlet boundary conditions we demand : Gy ( 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 1 ( x , x ' ) d'x ' 0 ( x ' ) da ' ( 1.44 ) 47 Js an ...
Thus , for Dirichlet boundary conditions we demand : Gy ( 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 1 ( x , x ' ) d'x ' 0 ( x ' ) da ' ( 1.44 ) 47 Js an ...
Page 282
... f ( 0 , 0 ) 1 др ( 9.64 ) yar With this condition on y 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 .
... f ( 0 , 0 ) 1 др ( 9.64 ) yar With this condition on y 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 .
Page 284
... 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 .
... 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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Contents
Introduction to Electrostatics | 1 |
BoundaryValue Problems in Electrostatics I | 26 |
TimeVarying Fields Maxwells Equations Con | 169 |
Copyright | |
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