Results 1-3 of 53
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') dV - f (£ (D(x') ^ da' (1.44) Jv \TT 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.
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
19 other sections not shown