## Classical Electrodynamics |

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Page 9

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

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 ...Page 10

The tangential component of electric field can be shown to be continuous across

a boundary

It is only necessary to take a rectangular path with negligible ends and one side ...

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 ...

Page 240

According to Ohm's law, there exists a current density J near 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

...

According to Ohm's law, there exists a current density J near the

**surface**of theconductor: 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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### Contents

Introduction to Electrostatics | 1 |

BoundaryValue Problems in Electrostatics I | 26 |

BoundaryValue Problems in Electrostatics II | 54 |

Copyright | |

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