tailieunhanh - Electromagnetic Field Theory: A Problem Solving Approach Part 31

Electromagnetic Field Theory: A Problem Solving Approach Part 31. Electromagnetic field theory is often the least popular course in the electrical engineering curriculum. Heavy reliance on vector and integral calculus can obscure physical phenomena so that the student becomes bogged down in the mathematics and loses sight of the applications. This book instills problem solving confidence by teaching through the use of a large number of worked problems. To keep the subject exciting, many of these problems are based on physical processes, devices, and models. This text is an introductory treatment on the junior level for a two-semester electrical engineering. | Separation of Variables in Cylindrical Geometry 275 The other time-dependent amplitudes A t and C t are found from the following additional boundary conditions i the potential is continuous at r a which is the same as requiring continuity of the tangential component of E V r a V r a- E r a- E r a Aa Ba Cla 17 ii charge must be conserved on the interface day Jr r - a -Jr r _ 0 Ol criEr r a - r2 r r a_ ei r r - a -e2-Er r a- 0 9t 18 In the steady state 18 reduces to 11 for the continuity of normal current while for t 0 the time derivative must be noninfinite so af is continuous and thus zero as given by 10 . Using 17 in 18 we obtain a single equation in C t dC ZCi o- a2 I 1 IC - I 0 o-i rz e i - 2 at Ei e2 ei e2 19 Since Eo is a step function in time the last term on the right-hand side is an impulse function which imposes the initial condition C i 0 o2 Eo 20 1 s2 so that the total solution to 19 is ei e2 Ti cr2 . . 21 The interfacial surface charge is oy r a t ei r r a -e2 r r a_ Ei B cos 4 d o dt -it ri o ie2 o 2 i _ T C t a ol 7 7 7 -e X Ti iTo liTi T fl ol Ifii ei E2 o ei e2 5 costp a J _2 a2ei-aie2 --------- ----Eo 1 - e cos p 22 Ti cr2 276 Electric Field Boundary Value Problems The upper part of the cylinder tt 2 tt 2 is charged of one sign while the lower half rr 2 tw is charged with the opposite sign the net charge on the cylinder being zero. The cylinder is uncharged at each point on its surface if the relaxation times in each medium are the same i Ti The solution for the electric field at t 0 is E t 0 Eo 2fii o __- 2fi o 1 - cos 0ir - sin 0i 1 El íỉ 1 4- 2 _2 _ . a 2 1 . F 2 ------- I cos pir r 14 2 a 2 1 I 2 sin0i d r 14- 2 J r a 23 r a The field inside the cylinder is in the same direction as the applied field and is reduced in amplitude if 2 i and increased in amplitude if 2 1 up to a limiting factor of two as 1 becomes large compared to 2. If 2 i the solution reduces to the uniform applied field everywhere. The de steady-state solution is .