## Classical Electrodynamics |

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...

For Neumann boundary conditions we ... x') seems to be 3GN (x, x') = 0 for x' on S

0n' since that makes the second term in the surface integral in (1.42)

...

**vanishes**and the solution is q}(x) = | 2006/s ×) or -4. f ox)*da (144) P 477 JS 0n'For Neumann boundary conditions we ... x') seems to be 3GN (x, x') = 0 for x' on S

0n' since that makes the second term in the surface integral in (1.42)

**vanish**, ...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: 1

ik R x) ...

(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: 1

ik R x) ...

Page 284

(GE))] dor' (9.74) y From the expansion, V × V × 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 ...

(GE))] dor' (9.74) y From the expansion, V × V × 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 ...

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### Contents

Introduction to Electrostatics | 1 |

Nē 3 | 3 |

Greens theorem | 14 |

Copyright | |

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