## Classical Electrodynamics |

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Page 19

Thus, for Dirichlet boundary conditions we demand : GD(\, x') = 0 for x' on S (1.43)

Then the first term in the surface integral in (1.42)

x) = f p(x')GD(x, x') dV - f (£ (D(x') ^ da' (1.44) Jv \TT Js on For Neumann ...

Thus, for Dirichlet boundary conditions we demand : GD(\, x') = 0 for x' on S (1.43)

Then the first term in the surface integral in (1.42)

**vanishes**and the solution is (D(x) = f p(x')GD(x, x') dV - f (£ (D(x') ^ da' (1.44) Jv \TT Js on For Neumann ...

Page 282

This means that the fields, and therefore y(x), will satisfy the radiation condition, C

-, r y (9-64) With this condition on y it can readily be seen that the integral in (9.63

) over the hemisphere S2

This means that the fields, and therefore y(x), will satisfy the radiation condition, C

-, r y (9-64) With this condition on y it can readily be seen that the integral in (9.63

) over the hemisphere S2

**vanishes**inversely as the hemisphere radius as that ...Page 284

... E) x V'G - Gn x (V' x E) (9.72) 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),

... E) x V'G - Gn x (V' x E) (9.72) 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 |

Scalar potential | 7 |

Greens theorem | 14 |

Copyright | |

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