Electromagnetic Field Theory: A Problem Solving Approach Part 28 ppsx
... A capacitor has a moveable part that can rotate through the angle 0 so that the capacitance C(O) depends on 0. (a) What is the torque on the moveable part? (b) An electrostatic ... charge - Q at radius R 2 . 48. A capacitor C is charged to a voltage V 0 . At t = 0 another initially uncharged capacitor of equal capacitance C is Switc...
Ngày tải lên: 03/07/2014, 02:20
... the field plots. Markus Zahn Cartesian Coordinates (x, y, z) af. af af Vf = - , + O i, + i, ax ay Oz aA, aA, aA, V . A= a+ + ax ay az (LAX, A AaA\ .a, O aA. ay z az Ox ... ax ay V2f a& apos; +f + a& apos;f Ox jy az Cylindrical Coordinates (r, 4, z) Of. I af af. Vf = r 4 +• 1 ar r 04 az 1 a 1iA, aA, V * A= - -(rAr)+M +M rr rr a Oz a...
Ngày tải lên: 03/07/2014, 02:20
... becomes infinitesimally small, each of the bracketed terms defines a partial derivative aA 3A ýAMz lim = ( + + A V (4) ax-o \ax ay az where AV = Ax Ay Az is the volume ... Section 1-3-1 and the vector A: aAx aA, aA, div A = V - A = -+ + (5) ax ay az 1-4-3 Curvilinear Coordinates In cylindrical and spherical coordinates, the divergence o...
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Electromagnetic Field Theory: A Problem Solving Approach Part 10 ppsx
... incremental line charge alone has a radial field component as given by (5) that in Cartesian coordinates results in x and y components. Consider the line charge dA 1 , a distance ... dis- tance from the infinite sheet. (b) Parallel Sheets of Opposite Sign A capacitor is formed by two oppositely charged sheets of surface charge a distance...
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Electromagnetic Field Theory: A Problem Solving Approach Part 16 ppsx
... A point charge q is placed between two parallel grounded conducting planes a distance d apart. (a) The point charge q a distance a above the lower plane and a distance ... Try power-law solutions q, = AA", b, = Ba and find the characteristic values of A and a that satisfy the equations in (d). (f) Taking a linear combination...
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Electromagnetic Field Theory: A Problem Solving Approach Part 40 ppsx
... right-hand side in (12) that accounts for an iron object to be drawn towards a magnet. Magnetiz- able materials are attracted towards regions of higher H. 5-8-2 Force on a ... the magnetic dipole moment m = i Ax Ay iý: lim f = m [LB +-aB, + B,, (3) A O xL ax x y iy Ampere's and Gauss's law for the magnetic field relate the...
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Electromagnetic Field Theory: A Problem Solving Approach Part 57 ppsx
... e -a The solutions are now nonuniform plane waves, as discussed in Section 7-7. Complex angles of transmission are a valid mathematical concept. What has happened is that ... e-" ' 4, e-'ix" +4, e-i.,x 4, e-i'x (23) Again the phase factors must be equal so that (5) and (6) are again true. Snell's law and the angle o...
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Electromagnetic Field Theory: A Problem Solving Approach Part 62 ppsx
... (35) and (36) for two values of T/7 and see that the quasi-static and transmission line results approach each other as T/r.becomes small. When the roundtrip wave transit ... V~L~t) t>T (C) Figure 8-14 (a) A step voltage is applied to transmission lines loaded at z = 1 with a capacitor CL or inductor LL. The load voltage and current ar...
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Electromagnetic Field Theory: A Problem Solving Approach Part 72 ppsx
... L Secondary maxima then occur at nearby angles but at much smaller amplitudes compared to the main lobe at 0 = 7r/2. 9-4-3 Radiation Resistance The total time-average radiated power ... distribution and calculates the resultant (near and far) fields. If all boundary conditions along the antenna are satisfied, then the solution has been found. Unfor...
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Electromagnetic Field Theory: A Problem Solving Approach Part 75 pdf
... magnetic scalar potential, 365 spherical coordinates, 284 Laplacian of reciprocal distance, 73-74 Larmor angular velocity, 316 Laser, 517 Law of sines and cosines, 41 Leakage flux, ... VfxA+fVxA (V x A) x A = (A V )A - 'V (A . A) Vx (Vx A) = V(V - A) - V A INTEGRAL THEOREMS Line Integral of a Gradient Vf dlI =f(b) -f (a) Divergence...
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