Classical Electrodynamics |
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Page 19
... in (1.42) vanishes and the solution is on da (144) on' - For Neumann boundary
conditions we must be more careful. ... fox) 0GN (x, x') = 0 for x' on S 0n' since that
makes the second term in the surface integral in (1.42) vanish, as desired.
... in (1.42) vanishes and the solution is on da (144) on' - For Neumann boundary
conditions we must be more careful. ... fox) 0GN (x, x') = 0 for x' on S 0n' 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: ...
(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: ...
Page 284
(GE)] doz (9.74) y From the expansion, V × V x 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
alternative ...
(GE)] doz (9.74) y From the expansion, V × V x 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
alternative ...
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Contents
Introduction to Electrostatics | 1 |
References and suggested reading | 23 |
Multipoles Electrostatics of Macroscopic Media | 98 |
Copyright | |
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