## Classical Electrodynamics |

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Page 19

... (1.42)

1.44 on' (1.44) For Neumann boundary ... to be 39s G. x') = 0 for x' on S on' since

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

... (1.42)

**vanishes**and the solution is q}(x) –s. p(x')Go(x, x') dor' — # fox) 3GD da'1.44 on' (1.44) For Neumann boundary ... to be 39s G. x') = 0 for x' on S on' 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:

irre 1 ...

(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:

irre 1 ...

Page 284

... three terms in (9.72), involving the product (GE),

we make use of the following easily proved identities connecting surface

integrals over a closed surface S to volume integrals over the interior of S: f A n

da -s v.

... three terms in (9.72), involving the product (GE),

**vanishes**identically. To do thiswe make use of the following easily proved identities connecting surface

integrals over a closed surface S to volume integrals over the interior of S: f A n

da -s v.

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### Contents

Introduction to Electrostatics | 1 |

BoundaryValue Problems in Electrostatics I | 26 |

Multipoles Electrostatics of Macroscopic Media | 98 |

Copyright | |

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