By Professor Dr. Sc. Sergey Ivanovich Molokovsky, Professor Dr. Sc. Aleksandr Danilovich Sushkov (auth.)

**Intense Ion and Electron Beams** treats severe charged-particle beams utilized in vacuum tubes, particle beam know-how and experimental installations similar to unfastened electron lasers and accelerators. It addresses, between different issues, the physics and easy conception of excessive charged-particle beams; computation and layout of charged-particle weapons and focusing platforms; multiple-beam charged-particle structures; and experimental equipment for investigating excessive particle beams. The assurance is thoroughly balanced among the physics of excessive charged-particle beams and the layout of optical platforms for his or her formation and focusing. it may be prompt to all scientists learning or making use of vacuum electronics and charged-particle beam know-how, together with scholars, engineers and researchers.

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**Extra info for Intense Electron and Ion Beams**

**Sample text**

4). U = U0 + Um cos Solution of Cauchy Problem for Poisson Equation In the cylindrical coordinate system r, θ, z for the case of axial symmetry the Poisson equation is written as: 1 ∂ r ∂r r ∂U ∂r + ∂2U ρ (r, z) =− . 2 ∂z ε0 The problem is set up in the following way. 3 Calculation of Electrostatic Field: Cauchy Problem 35 as well as the space-charge density distributions ρ (r, z) are supposed to be known. A solution of this problem can be found by power series expansion of the potential U (r, z) and space-charge density ρ (r, z): U (r, z) = f0 (z) + f2 (z) r2 + f4 (z) r4 + .

The solution of this equation can be made by numerical methods with use of computers (see Sect. 4). 4 Method of Finite Diﬀerence The essence of the method consists of replacement of a partial diﬀerential equation by a ﬁnite diﬀerence equation that is obtained by replacement of partial derivatives with their approximate ﬁnite diﬀerence expressions. Let us consider this procedure on an example of a two-dimensional Poisson equation: ∂2U ρ ∂2U + =− . 10) The second derivatives including in this equation can be expressed at a point 0 through the values of the ﬁrst derivatives at the neighboring points a, b, c, d (Fig.

At the cylindrical boundary r = R the potential is changed by a jump at z = 0 from U1 = 0 to U2 , that is, U = U1 at z < 0 and U = U2 at z > 0. This potential distribution can be expressed in terms of Heaviside’s unit function changed by a jump from 0 to 1 at z = 0, U (R, z) = U2 · H (z). Heaviside’s function is analytically expressed through a Fourier integral: 1 1 H (z) = + 2 π ∞ 0 sin kz dk . k The potential at the cylindrical boundary r = R can be presented as: U2 U2 U (R, z) = U2 · H (z) = + 2 π ∞ 0 sin kz dk .