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From (1.31) we know that V2(1/K) = -4ird(\ - x'), so that (1.35) becomes If the point
x lies within the volume V, we obtain: = f^W + J Jr R 4 If x lies outside the surface
S, the left-hand side of (1.36) is zero. [Note that this is consistent with the ...
Rg. 2.13 Hollow, rectangular box with five sides at zero potential, while the sixth (
z = c) has the specified potential fl> = V(x, y). directions. All surfaces of the box
are kept at zero potential, except the surface z = c, which is at a potential V(x, y).
D = [47r]E (8.1)* in order to give zero electric field inside the perfect conductor.
Similarly, for time-varying magnetic fields, the surface charges move in response
to the tangential magnetic field to produce always the correct surface current K:
n x ...
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Introduction to Electrostatics
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