By Robert Langlois
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Extra resources for Build an EDM, Electrical Discharge Machining, Removing Metal By Spark Erosion
85) V Expanding now V (r ) in Taylor series around the origin O, chosen at an arbitrary point of D, we obtain ∂2 V 1 + xi xk + ... 2 ∂xi ∂xk 0 0 ∂Ek 1 = V (0) − xi Ei (0) − xi xk − ... , 6 ∂xi 0 V (r ) = V (0) + xi ∂V ∂xi where we added the null term r 2 δik ∂Ek ∂xi = r 2 (∇ · E)0 = 0. 85), we find finally that 1 We = QV (0) − pi Ei (0) − pik 6 ∂Ek ∂xi ∞ + ... = 0 W (l) . 72)), with the dipole at the origin; k is the quadrupole potential energy of the charges, with the (c) W (2) = − 16 pik ∂E ∂xi 0 quadrupole at the origin, etc.
2 which can also be written as 1 1 = |r − r | r ∞ l=0 r r ∞ l Pl (cos θ) = l=0 (r )l Pl (cos θ). 137) Comparing this result with the analysis presented in Sect. 137) (multiplied by 4πq 0 ) stands for the monopole potential, the second – for dipole potential, the third – for quadrupole potential, etc. 34) from Appendix D, we write the Laplace equation in cylindrical coordinates as ∂V 1 ∂2 V 1 ∂ ∂2 V ρ + 2 + = 0. 138) 2 ρ ∂ρ ∂ρ ρ ∂ϕ ∂z2 We separate the variables and seek the solution in the form V (ρ, ϕ, z) = R(ρ)Φ(ϕ)Z(z).
Ln . The moment of such a system is p(n) = n! lim (ql1 l2 . . ln ) . 76) Here are a few examples of multipoles: (a) Monopole (zeroth-order multipole), with a single point charge; (b) Dipole (first-order multipole), of moment (see Sect. 10): p(1) = lim (ql) ; l→0 q→∞ (c) Quadrupole (second-order multipole), formed by two opposite, parallel dipoles, situated at small mutual distances, with charges at the corners of a parallelogram (Fig. 9). The quadrupole moment is p(2) = 2! 77) (d) Octupole (third-order multipole), composed of two opposite, parallel quadrupoles, at small distances, the charges being placed at the corners of a parallelepiped (Fig.
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