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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 = - 1% c 0t (6.53) B = W x A 6.6 Green's Function for the Time-
Dependent Wave Equation The wave equations (6.37), (6.38), and (6.52) all
have the basic ...
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Introduction to Electrostatics
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