## Classical ElectrodynamicsProblems after each chapter |

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Page 19

Thus , for Dirichlet boundary conditions we demand : G ( x , x ' ) = 0 for x ' on S ( 1.43 ) Then the first term in the surface integral in ( 1.42 )

Thus , for Dirichlet boundary conditions we demand : G ( x , x ' ) = 0 for x ' on S ( 1.43 ) Then the first term in the surface integral in ( 1.42 )

**vanishes**and the solution is 1 Φ ( x ) p ( x ' ) Gp ( x , x ' ) d'x ' ( 1.44 ) an ' For ...Page 282

... radiation condition , » +10,4 m ( ik - 1 ) ( 9.64 ) With this condition on y it can readily be seen that the integral in ( 9.63 ) over the hemisphere S ,

... radiation condition , » +10,4 m ( ik - 1 ) ( 9.64 ) With this condition on y it can readily be seen that the integral in ( 9.63 ) over the hemisphere S ,

**vanishes**inversely as the hemisphere radius as that radius goes to infinity .Page 284

... While it may not appear very fruitful to transform the two terms in ( 9.68 ) into six terms , we will now show that the surface integral of the first three terms in ( 9.72 ) , involving the product ( GE ) ,

... While it may not appear very fruitful to transform the two terms in ( 9.68 ) into six terms , we will now show that the surface integral of the first three terms in ( 9.72 ) , involving the product ( GE ) ,

**vanishes**identically .### What people are saying - Write a review

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### Contents

Introduction to Electrostatics | 1 |

References and suggested reading | 23 |

Multipoles Electrostatics of Macroscopic Media | 98 |

Copyright | |

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