Classical Electrodynamics |
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Page 19
... (1.42) vanishes and the solution is q}(x) –s. p(x')Go(x, x') dor' — # fox) 3GD da'
1.44 on' (1.44) For Neumann boundary ... to be 39s G. x') = 0 for x' on S on' since
that makes the second term in the surface integral in (1.42) vanish, as desired.
... (1.42) vanishes and the solution is q}(x) –s. p(x')Go(x, x') dor' — # fox) 3GD da'
1.44 on' (1.44) For Neumann boundary ... to be 39s G. x') = 0 for x' on S on' since
that makes the second term in the surface integral in (1.42) vanish, as desired.
Page 282
(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:
irre 1 ...
(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:
irre 1 ...
Page 284
... three terms in (9.72), involving the product (GE), vanishes identically. To do this
we make use of the following easily proved identities connecting surface
integrals over a closed surface S to volume integrals over the interior of S: f A n
da -s v.
... three terms in (9.72), involving the product (GE), vanishes identically. To do this
we make use of the following easily proved identities connecting surface
integrals over a closed surface S to volume integrals over the interior of S: f A n
da -s v.
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Contents
Introduction to Electrostatics | 1 |
BoundaryValue Problems in Electrostatics I | 26 |
Multipoles Electrostatics of Macroscopic Media | 98 |
Copyright | |
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