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Electromagnetic Field Theory: A Problem Solving Approach Part 75

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and capacitance and charge distributions, 204-208 conservation theorem, 199 and current distributions, 454 density, electric field, 208-209. Exponential transmission line, 649 External inductance, 456-457 Fair weather electric field, 195 Farad, 175. and divergence theorem, 26-28 and Gauss's law, 74-75 and magnetic field, 338. between point charge and dielectric boundary, 165. between point charge and grounded plane, 108. between point charge...

Electromagnetic Field Theory: A Problem Solving Approach Part 1

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ar r •O r sin 0 a O. is = cos 0 cos oi, +cos 0 sin 4i,. sin Oi,+cos Oi,. cos Oi, -sin Oi,. r sin 0 cos 4. sin 0 cos i. cos Oi,- sin Oie. sin Oi, +cos ie. cos Oi, -sin 0iO Cylindrical. Heavy reli- ance on vector and integral calculus can obscure physical phenomena so...

Electromagnetic Field Theory: A Problem Solving Approach Part 2

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1.2.4 The Dot (Scalar)Product 1.2.5 The Cross (Vector) Product 1.3 THE GRADIENT AND THE DEL. OPERA TOR 1.3.1 The Gradient. (b) Spherical 1.3.3 The Line Integral 1.4 FLUX AND DIVERGENCE. Chapter 2-THE ELECTRIC FIELD 2.1 ELECTRIC CHARGE. 2.1.3 Faraday's"Ice-Pail"Experiment 2.2 THE COULOMB FORCE LAW. 2.2.3 The Electric Field 56. 2.3.2 The Electric Field Due to a Charge Dis-. 2.5 THE...

Electromagnetic Field Theory: A Problem Solving Approach Part 3

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7.7.1 Propagationat an ArbitraryAngle 529 7.7.2 The Complex PropagationConstant 530 7.7.3 Nonuniform Plane Waves 532 7.8 OBLIQUE INCIDENCE ONTO A PER-. 8.2.3 Approach to the dc Steady State 585 8.2.4 Inductors and Capacitorsas Quasi-static. 8.3.1 Solutions to the TransmissionLine Equa-. (a) Load Impedance Reflected Back to the. 8.5.1 Use of the Smith Chart for Admittance. 9.1.1 Nonhomogeneous Wave Equations 664...

Electromagnetic Field Theory: A Problem Solving Approach Part 4

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(b) The direction of the unit vectors i, and i, vary with the angle 46. Table 1-2 Geometric relations between coordinates and unit vectors for Cartesian, cylindrical, and spherical coordinate systems*. sin 0 cos •i, +cos 0 cos Oio -sin )i,#. from the coordinate (r, 0, 46) now depends on the angle 0 and the radial position r as shown...

Electromagnetic Field Theory: A Problem Solving Approach Part 5

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Find the unit vector i,, perpendicular in the right-hand sense to the vectors shown in Figure 1-10.. 16 Review of Vector Analysis. 1-3 THE GRADIENT AND THE DEL OPERATOR 1-3-1 The Gradient. i)dl ax ay az. where the spatial derivative terms in brackets are defined as the gradient of f:. The symbol V with the gradient term is introduced as...

Electromagnetic Field Theory: A Problem Solving Approach Part 6

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Figure 1-16 Infinitesimal volumes used to define the divergence of a vector in (a) cylindrical and (b) spherical geometries.. 26 Review of Vector Analysis. lets each of the bracketed terms become a partial derivative as the differential lengths approach zero and (8) becomes an exact relation. The divergence is then. r 9 sin 0 Ar AO A4 in Figure 1-16b...

Electromagnetic Field Theory: A Problem Solving Approach Part 7

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The Curl and Stokes' Theorem 35. (19) The 4 component of the curl is found using contour c:. VxA= I (A.sin 0)- i,. 1-5-3 Stokes' Theorem. We now piece together many incremental line contours of the type used in Figures 1-19-1-21 to form a macroscopic surface S like those shown in Figure 1-23. so that the total circulation is obtained...

Electromagnetic Field Theory: A Problem Solving Approach Part 8

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Verify Stokes' theorem for the rectangular bounding contour in the xy plane with a vector field. Check the result for (a) a flat rectangular surface in the xy plane, and (b) for the rectangular cylinder.. Show that the order of differentiation for the mixed second derivative. Some of the unit vectors in cylindrical and spherical coordinates change direction in space...

Electromagnetic Field Theory: A Problem Solving Approach Part 9

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He discovered that the force between two small charges q, and q 2 (idealized as point charges of zero size) is pro- portional to their magnitudes and inversely proportional to the square of the distance r 1 2 between them, as illustrated in Figure 2-6. The force acts along the line joining the charges in the same or opposite direction...

Electromagnetic Field Theory: A Problem Solving Approach Part 10

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Figure 2-11 An infinitely long uniform distribution of line charge only has a radially directed electric field because the z components of the electric field are canceled out by symmetrically located incremental charge elements as also shown in Figure 2-8a.. 2-3-4 Field Due to Infinite Sheets of Surface Charge (a) Single Sheet. A surface charge sheet of infinite extent in...

Electromagnetic Field Theory: A Problem Solving Approach Part 11

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We can now use the divergence theorem to convert the surface integral to a volume integral:. fsE.dS= 4• V. When the point charge q is outside the surface every point in the volume has a nonzero value of rQp. Then, using (6) with rQp # 0, we see that the net flux of E through the surface is zero.. This...

Electromagnetic Field Theory: A Problem Solving Approach Part 12

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The Electric Potential 85. The minus sign in front of the integral is necessary because the quantity W represents the work we must exert on the test charge in opposition to the coulombic force between charges.. The dot product in (1) tells us that it takes no work to move the test charge perpendicular to the electric field, which in...

Electromagnetic Field Theory: A Problem Solving Approach Part 13

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96 The Electric Field. For the field given by (5), the equation for the lines tangent to the electric field is. where K 2 is a constant determined by specifying a single coordinate (xo, yo) along the field line of interest. The field lines are also circles of radius a/sin K 2 with centers at x = 0, y =...

Electromagnetic Field Theory: A Problem Solving Approach Part 14

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We ignore the blower= D solution with q'= -q since the image charge must always be outside sphere with valuregion of interest. If we allowed the this solution, the net charge at the position of the inducing charge is zero, contrary to our statement that the net charge is q.. The image charge distance b obeys a similar relation as...

Electromagnetic Field Theory: A Problem Solving Approach Part 15

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Find the total charge in each of the following dis- tributions where a is a constant parameter:. (b) A spherically symmetric volume charge distributed over all space. (c) An infinite sheet of surface charge with density. A point charge q with mass M in a gravity field g is released from rest a distance xo above a sheet of surface...

Electromagnetic Field Theory: A Problem Solving Approach Part 16

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An infinitely long line charge A is a distance D from the center of a conducting cylinder of radius R that carries a total charge per unit length Ac. What is the force per unit length on. An infinitely long sheet of surface charge of width d and uniform charge density ao 0 is placed in the yz plane.. The...

Electromagnetic Field Theory: A Problem Solving Approach Part 17

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The presence of matter modifies the electric field because even though the material is usually charge neutral, the field within the material can cause charge motion, called conduc- tion, or small charge displacements, called polarization.. Because of the large number of atoms present, 6.02 x 1023 per gram molecular weight (Avogadro's number), slight imbalances in the distribution have large effects...

Electromagnetic Field Theory: A Problem Solving Approach Part 18

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(b) The Local Electric Field. If this dipole were isolated, the local electric field would equal the applied macroscopic field. However, a large number density N of neighboring dipoles also contributes to the polarizing electric field. The electric field changes dras- tically from point to point within a small volume containing many dipoles, being equal to the superposition of fields...

Electromagnetic Field Theory: A Problem Solving Approach Part 19

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The net pressure force on the small rectangular volume shown in Figure 3-10 is. where we see that the pressure only exerts a net force on the volume if it is different on each opposite surface. As the volume shrinks to infinitesimal size, the pressure terms in (12) define partial derivatives so that the volume force density becomes. to the...