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Solving boundary-value electrostatics problems using Green’s reciprocity theorem
1.John David Jackson, Classical Electrodynamics (Wiley, New York, 1999), pp. 35–40.
2.Walter Greiner, Classical Electrodynamics (Springer, Berlin, 1998), pp. 48–50.
3.Richard S. Elliot, Electromagnetics: History and Applications (IEEE, New York, 1993), pp. 172–175.
4.David J. Griffiths, Introduction to Electrodynamics (Prentice-Hall, Englewoods Cliffs, NJ, 1999), pp. 157 and 158.
5.Reference 1, p. 52;
5.Ref. 3, p. 191.
6.The usual minus sign in relationship between the field and the derivative of the potential is absent because is defined to be positive pointing out of V into
7.See, e.g., Ref. 1, pp. 37 and 38;
7.Ref. 4, pp. 116–120.
8.See, e.g., Ref. 1, pp. 35–37;
8.Ref. 4, p. 56.
9.Note that a unit vector normal to a point on the surface is the negative of the unit vector normal to S. This accounts for change of sign of the terms in the square parentheses in Eq. (11).
10.This argument assumes S is a connected surface. If the surface is composed of distinct disconnected sections, one might argue that the net positive and negative charges reside on different sections, and it is impossible to find a path through from one section to the other; e.g., the case where V is the volume bounded by spherical surfaces and is the spherical surface of radius In this case, however, since in V, all the field lines would have to start and terminate on the same surface, which is impossible if there is only purely positive or negative charge on that surface. Finally, one might argue that there are situations where it is impossible find a path through V from the disconnected surfaces; e.g., the case where V is given by and (where ). In this case, the entire argument is moot because V consists of distinct disconnected volumes, and each disconnected volume can each be treated as a separate boundary-value problem without regard to the others.
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