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In obtaining result (1.36)—not a solution—we chose the function p to be 1/|x – x',
it being the potential of a unit point charge, satisfying the equation: Vo2 s 1 ) |x —
x"| = —4trö(x — x') (1.31) The function 1/|x – x' is only one of a class of functions ...
To see that potentials can always be found to satisfy the Lorentz condition,
suppose that the potentials A, D which satisfy (6.32) and (6.33) do not satisfy (
6.36). Then let us make a gauge transformation to potentials A', 'b' and demand
that A', ...
Then b = 0, and A satisfies the homogeneous wave equation. The fields are
given by E ... Since the time is involved, the Green's function will depend on the
variables (x, x', t, t'), and will satisfy the equation, 2 (v. –4 #)06. t; x', t') = —4tt ö(x
— x') ...
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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
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Introduction to Electrostatics
BoundaryValue Problems in Electrostatics I
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