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If the path is closed, the line integral is zero, # e.a. = 0 (1.21) a result that can also
be obtained directly from Coulomb's law. Then application of Stokes's theorem [if
A(x) is a vector field, S is an open surface, and C is the closed curve bounding ...
The tangential component of electric field can be shown to be continuous across
a boundary surface by using (1.21) for the line integral of E around a closed path.
It is only necessary to take a rectangular path with negligible ends and one side ...
According to Ohm's law, there exists a current density J near the surface of the
conductor: puoo 8m. J = ok. - (1 – 1)(n x Hu)e^*** (8.13) The time-average rate of
dissipation of energy per unit volume in ohmic losses is J. E* = (1/20) |J|*, so that
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Introduction to Electrostatics
BoundaryValue Problems in Electrostatics I
BoundaryValue Problems in Electrostatics II
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