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CALCULATION OF THE ELECTRIC FIELD DISTORTION IN FREE SPACE DUE TO APPEARANCE OF THE OBJECT WITH CONDUCTIVE SUREACE Gennady Jatchevitch, Ph.D Saint-Petersburg State University of Aerospace Instrumentation, Saint-Petersburg, Russia Abstract For the development alarm systems, using the electrostatic field, it is necessary to evaluate the electric field distortion due to appearance of the object with conductive surface. We introduce the concept of the electric field distortion as the deflection of the electric field intensity vector E at each point of the check space due to appearance of the object with conductive surface [1]. We examine four cases: 1. Free space with the uniform electric field, the object is right circular cylinder with the conductive side surface of infinite extent. 2. Free space with the point electric charge, the object is plane conductive surface of infinite extent. 3. Free space with the electric dipole, the object is plane conductive surface of infinite extent. The axis of the dipole is paralleled to the surface. 4. Free space with the electric dipole, the object is plane conductive surface of infinite extent. The axis of the dipole is perpendicular to the surface. On the first case we solve the Laplace’s equation in cylindrical coordinates and use the boundary conditions on the conductive surface of the cylinder. For the second, the third and the fourth cases we use the method of images. 1 1 0 z z , (1.1) where – is potential at the check point of the space. Fig.1.1 Uniform external electric field and the object As potential doesn’t depend on Z coordinate, we have 1 2 1 2 0 2 2 2 Let us assume that potential is equal I. FREE SPACE WITH THE UNIFORM EXTERNAL ELECTRIC FIELD AND THE OBJECT IS RIGHT CIRCULAR CYLINDER WITH THE CONDUCTIVE SIDE SURFACE OF INFINITE EXTENT In order to calculate electric field strength vector it is necessary to solve Laplace’s equation in cylindrical coordinates [2], [3] fig (1.1) Laplace’s equation is f 1 f 2 f1 f2 , (1.3) where f1 depends on coordinate , and f2 depend on coordinate . Substituting (1.3) into (1.2), we obtain 1 f 2 f 1 2 f f 2 1 f 2 21 f1 2 22 0 (1.4) After separating variables we have f1 2 2 f1 2 2 f1 f1 20 (1.2) 1 2 f2 p2 , f 2 2 (1.5) where p – is constant. In our case f2 (1.6) Consequently 1 (1.7) d e d ¸ d d e df 1 d d d2 f 1 2 e d (1.8) E e d2 f 1 df 1 e 2 e d (1.10) d A1 e 2 A1 A2 e ¸ (1.11) (1.17) a2 1 sin 2 (1.18) E0 E0 for the E 1 E E0 cos (1.19) E0 sin (1.20) E=E .e+ E .e (1.21) relatively vector E0 E0=E1 .e+ E1 .e A2 . (1.12) A2 A 1 cos (1.22) Angle of deflection is given by the following formula [2] Unknown potential equals a2 1 cos 2 E0 And distortion of the electric field is the deflection of electric vector and function f1 become f1 (1.16) free space are 1 Common integral of this equation f1 (1.15) The components of electric field (1.9) 1 sin . Then we have df 1 A2 For cylinder surface =a The components of electric field are given by the following formulas E e ¸ A1 A1 E0 Let us introduce the new variable such that 1 cylinder doesn’t affect upon For the electric field, therefore cos p2 E The results of calculation with the help of Matlab 7.0 are illustrated on fig. 2.1, where x=/a, y~ (1.13) Therefore the components of electric field are E A1 A2 cos 2 (1.14) 21 z Ez q a 2 3 4 a a 2 z 2 a a (2.3) After appearance of the object potential at the point M(,z) of system of point charge is equal to the sum of the potential created at a given point by each of the charges [2] The potential at the point M(,z) is given by 1 z Fig.1.2 Angle of deflection Z due to X, Y II. FREE SPACE WITH THE POSITIVE POINT CHARGE, THE OBJECT IS PLANE SURFACE OF INFINITE EXTENT q 1 4 Ea z2 2 2a z2 1 2 (2.4) And the components of electric field will be This case is illustrated on fig. 2.1 E 1 q 2 4 a a 3 3 2 z 2 2 z 2 2 a a a a 2 a a (2.5) z z q a a Ez 2 2 2 2 3 2 3 4 a a z z 2 a a a a 1 Fig. 2.1 Positive point charge and the object in free space The potential of the paint M (,z) fig 2.1 in free space q z 4 Еa Using Eq.(2.2), (2.3), (2.5) and (2.6) we can write [2] 1 2 z2 , (2.1) where Ea – capacitivity of the medium. The components of electric field vector is equal to [2],[3],[4] E 22 q 4 a a2 (2.6) a 2 z 2 a a z arccos E E Ez Ez1 1 E Ez E1 Ez1 2 2 2 (2.7) After calculating with the help of Matlab 7.0 we have the results which are illustraited on fig 2.2, where x=/a, y=z/a 3 (2.2) 2 Fig2.2 Angle of deflection Z due to X,Y III. FREE SPACE WITH THE ELECTRIC DIPOLE, THE OBJECT IS PLANE CONDUCTIVE SURFACE OF INFINITE EXTENT WHICH IS PARALLELED TO THE AXIS OF THE DIPOLE. This case is illustrated on fig 3.1. Potential at the point M in free space [2] ( M) q 1 4 a 2 ( z l) 2 z 1 2 2 (3.1) Fig.3.1 Electric dipole in free space and the object The components of electrical field vector are equal to E Ez q l l 2 2 2 3 2 3 4 a l 2 z z 1 l l l l (3.2) q 2 4 a l (3.3) l 2 2 2 3 2 3 z z 1 l l l l z 1 l z After appearance of the object (fig 2.1) the components of electric field vector are equal to E 1 2 2 q l l l l 2 3 3 3 3 4 a l 2 2 z 2 2 z 2 2 z 2 z 2 2 2 1 1 l l l l l l l l (3.4) 23 Ez1 z z z z 1 1 l l l l 2 2 2 2 2 3 2 3 2 3 2 3 4 a l 2 z z z z 2 2 1 l l 1 l l l l l l q (3.5) Using Eq. (3.2), (3.3), (3.4) and (3.5) we can write [2] z E E arccos 1 Ez Ez1 E Ez E1 Ez1 2 2 2 2 (3.6) After calculating Eq. (3.6) with the help of Matlab 7.0 we have the results which are illustrated on fig. 3.2, where x=/a, y=z/a IV. FREE SPACE WITH ELECTRIC DIPOLE, THE OBJECT IS PLANE CONDUCTIVE SURFACE OF INFINITE EXTENT. THE AXIS OF THE DIPOLE IS PERPENDICULAR TO THE SURFACE Fig.3.2 Angle of deflection Z due to X,Y Fig.4.1 Electric dipole and the object This case is illustrated on fig 4.1. In free space the components of electrical field vector are equal to E Ez z q l l 2 3 3 2 2 2 2 4 a l z z 0.5 0.5 l l l l (4.1) z z 0.5 0.5 q l l 2 3 3 2 2 2 2 4 a l z z 0.5 l l 0.5 l l (4.2) After appearance of the object (fig 4.1) the components of electric field vector are equal to 24 E 1 Ez1 l l l l 2 2 2 2 2 3 2 3 2 3 2 3 4 a l 2 z z z z 2.5 0.5 1.5 l l 0.5 l l l l l l (4.3) q z z z 0.5 2.5 0.5 q l l l 2 3 3 2 2 2 2 2 3 4 a l 2 z z z 2.5 0.5 l l 0.5 l l l l (4.4) Using Eq (4.1), (4.2), (4.3) and (4.4) we can write z E E arccos 1 Ez Ez1 E Ez E1 Ez1 After calculating Eq (4.5) with the help of Matlab 7.0 we have the results which are illustrated on fig 4.2, where x=/a, y=z/a 2 2 2 2 (4.5) Preliminary results show that the value of distortion at the check point of the space depends on the distance from check point to the object, on orientation of the object, and position of check point in the free space. The received results prove the possibility of development alarm system using electrostatic field. We developed the transducer of electric field distortion and we investigated experimentally the deflection of the electric field intensity vector at the same check points, that performed theoretically, due to appearance of the object of finite sizes and we had the confirmation of calculation results. But these results are topics of the another articles. REFERENCES [1] Fig.4.2 Angle of deflection Z due to X,Y [2] V. CONCLUSION [3] The received results are preliminary and have conceptual meaning. [4] Яцевич Г.Б. Варжапетян А.Г. Устройство охранной сигнализации. Свидетельство на полезную модель №23511 от 20.06.2002г. Smyth W.R. Static and dynamic electricity. New York-Toronto-London 1950. Стрэттон Дж. А. Теория электромагнетизма. М.1949г. Matveev A.N. Electricity and Magnetism Mir,1986 . 25