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... x ' ) = 0 for x ' on S ( 1.43 ) Then the first term in the surface integral in ( 1.42 ) vanishes and the solution is 1 ag ♡ Φ ' ( 1.44 ) дn ' For ...
... 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 .
... we will now show that the surface integral of the first three terms in ( 9.72 ) , involving the product ( GE ) , vanishes identically .
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Introduction to Electrostatics
BoundaryValue Problems in Electrostatics I
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