## Classical Electrodynamics |

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Page 19

... in (1.42)

conditions we must be more careful. ... fox) 0GN (x, x') = 0 for x' on S 0n' since that

makes the second term in the surface integral in (1.42)

... in (1.42)

**vanishes**and the solution is on da (144) on' - For Neumann boundaryconditions we must be more careful. ... fox) 0GN (x, x') = 0 for x' on S 0n' since that

makes the second term in the surface integral in (1.42)

**vanish**, as desired.Page 282

(9.64) With this condition on p it can readily be seen that the integral in (9.63)

over the hemisphere S,

radius goes to infinity. Then we obtain the Kirchhoff integral for p(x) in region II: ...

(9.64) With this condition on p it can readily be seen that the integral in (9.63)

over the hemisphere S,

**vanishes**inversely as the hemisphere radius as thatradius goes to infinity. Then we obtain the Kirchhoff integral for p(x) in region II: ...

Page 284

(GE)] doz (9.74) y From the expansion, V × V x A = V(V. A.) – W*A, it is evident

that the volume integral

three terms in (9.72) identically zero, the remaining three terms give an

alternative ...

(GE)] doz (9.74) y From the expansion, V × V x A = V(V. A.) – W*A, it is evident

that the volume integral

**vanishes**identically.* With the surface integral of the firstthree terms in (9.72) identically zero, the remaining three terms give an

alternative ...

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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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