Results 1-3 of 50
... (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.
(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 ...
... 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.
What people are saying - Write a review
LibraryThing ReviewUser Review - barriboy - LibraryThing
A soul crushing technical manual written by a sadist that has served as the right of passage for physics PhDs since the dawn of time. Every single one of my professors studied this book, and every ... Read full review
LibraryThing ReviewUser Review - aproustian - LibraryThing
"Jackson E&M is about learning how to approximate reliably...the entire book, with few exceptions, is a mathematical discussion on how to solve [the same] 4 problems for different boundary conditions." Read full review
Introduction to Electrostatics
BoundaryValue Problems in Electrostatics I
References and suggested reading
16 other sections not shown