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Engineering Electromagnetics Lecture 4 Dr.-Ing. Erwin Sitompul President University http://zitompul.wordpress.com President University Erwin Sitompul EEM 4/1 Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Application of Gauss’s Law: Differential Volume Element We are now going to apply the methods of Gauss’s law to a slightly different type of problem: a surface without symmetry. We have to choose such a very small closed surface that D is almost constant over the surface, and the small change in D may be adequately represented by using the first two terms of the Taylor’s-series expansion for D. The result will become more nearly correct as the volume enclosed by the gaussian surface decreases. We intend eventually to allow this volume to approach zero. President University Erwin Sitompul EEM 4/2 Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Taylor’s Series Expansion f ( x0 x) f ( x) f ( x0 ) x0 x x0 x A point near x0 f ( x) f ( x0 x) f ( x0 ) f ( x0 ) f ( x) f ( x0 ) x (x) 2 1! 2! f n ( x0 ) (x) n n! Only the linear terms are used for the linearization President University Erwin Sitompul EEM 4/3 Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Application of Gauss’s Law: Differential Volume Element Consider any point P, located by a rectangular coordinate system. The value of D at the point P may be expressed in rectangular components: D0 Dx0ax Dy 0a y Dz 0az We now choose as our closed surface, the small rectangular box, centered at P, having sides of lengths Δx, Δy, and Δz, and apply Gauss’s law: D dS Q S S D dS front President University back left right top bottom Erwin Sitompul EEM 4/4 Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Application of Gauss’s Law: Differential Volume Element We will now consider the front surface in detail. The surface element is very small, thus D is essentially constant over this surface (a portion of the entire closed surface): front Dfront Sfront Dfront yz ax Dx,front yz The front face is at a distance of Δx/2 from P, and therefore: Dx ,front x Dx 0 rate of change of Dx with x 2 x Dx Dx 0 2 x President University Erwin Sitompul EEM 4/5 Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Application of Gauss’s Law: Differential Volume Element We have now, for front surface: front x Dx Dx 0 yz 2 x In the same way, the integral over the back surface can be found as: back Dback S back Dback (yz a x ) Dx,back yz Dx ,back back x Dx Dx 0 2 x x Dx D x0 yz 2 x President University Erwin Sitompul EEM 4/6 Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Application of Gauss’s Law: Differential Volume Element If we combine the two integrals over the front and back surface, we have: front back Dx xyz x Repeating the same process to the remaining surfaces, we find: right top left bottom Dy y yxz Dz zxy z These results may be collected to yield: Dx Dy Dz S D dS x y z xyz Dx Dy Dz S D dS Q x y z v President University Erwin Sitompul EEM 4/7 Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Application of Gauss’s Law: Differential Volume Element The previous equation is an approximation, which becomes better as Δv becomes smaller, and in the following section the volume Δv will be let to approach zero. For the moment, we have applied Gauss’s law to the closed surface surrounding the volume element Δv. The result is the approximation stating that: Dx Dy Dz Charge enclosed in volume v v y z x President University Erwin Sitompul EEM 4/8 Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Application of Gauss’s Law: Differential Volume Element Example Let D = y2z3 ax + 2xyz3 ay + 3xy2z2 az nC/m2 in free space. (a) Find the total electric flux passing through the surface x = 3, 0 ≤ y ≤ 2, 0 ≤ z ≤ 1 in a direction away from the origin. (b) Find |E| at P(3,2,1). (c) Find the total charge contained in an incremental sphere having a radius of 2 mm centered at P(3,2,1). (a) ψ S DS dS 1 z 0 y z a 1 2 2 y 0 0 0 y 1 3 2 3 3 3 2 2 2 xyz a 3 xy z a z dydz a x x y x 3 y 2 z 3dydz 2 1 0 4 z 4 1 0 23 nC President University Erwin Sitompul EEM 4/9 Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Application of Gauss’s Law: Differential Volume Element (b) D = y2 z3ax 2xyz3a y 3xy2 z 2az DP = (2)2 (1)3 ax 2(3)(2)(1)3 a y 3(3)(2)2 (1)2 az = 4ax 12a y 36az nC m2 DP = DP (4) 2 (12) 2 (36) 2 38.158nC m2 EP DP 0 38.158 nC m 2 8.854 1012 4.31 kV m President University Erwin Sitompul EEM 4/10 Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Application of Gauss’s Law: Differential Volume Element (c) Dx Dy Dz Q v y z x Dx Dy Dz QP v y z P x 0 2 xz 3 6 xy 2 z x 3 nC m3 43 (2 103 )3 m3 y 2 z 1 4 3 2 0 2(3)(1) 6(3)(2) (1) 3 (2 103 )3 nC 2.611015 C President University Erwin Sitompul EEM 4/11 Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Divergence We shall now obtain an exact relationship, by allowing the volume element Δv to shrink to zero. Dx Dy Dz x y z D dS S v Dx Dy Dz lim y z v0 x Q v D dS S v Q lim v 0 v The last term is the volume charge density ρv, so that: Dx Dy Dz lim y z v0 x President University D dS S v v Erwin Sitompul EEM 4/12 Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Divergence Let us no consider one information that can be obtained from the last equation: Dx Dy Dz lim y z v0 x D dS S v This equation is valid not only for electric flux density D, but also to any vector field A to find the surface integral for a small closed surface. Ax Ay Az lim y z v0 x President University A dS S v Erwin Sitompul EEM 4/13 Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Divergence This operation received a descriptive name, divergence. The divergence of A is defined as: Divergence of A div A lim v 0 A dS S v “The divergence of the vector flux density A is the outflow of flux from a small closed surface per unit volume as the volume shrinks to zero.” A positive divergence of a vector quantity indicates a source of that vector quantity at that point. Similarly, a negative divergence indicates a sink. President University Erwin Sitompul EEM 4/14 Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Divergence Dx Dy Dz div D x y z Rectangular 1 1 D Dz div D ( D ) z Cylindrical 1 2 1 1 D div D 2 (r Dr ) (sin D ) r r r sin r sin Spherical President University EEM 4/15 Erwin Sitompul Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Divergence Example If D = e–xsiny ax – e–x cosy ay + 2z az, find div D at the origin and P(1,2,3). Dx Dy Dz e x sin y e x sin y 2 div D x y z 2 Regardless of location the divergence of D equals 2 C/m3. President University Erwin Sitompul EEM 4/16 Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Maxwell’s First Equation (Electrostatics) We may now rewrite the expressions developed until now: div D lim D dS S v 0 div D v Dy Dx D z x y z div D v Maxwell’s First Equation Point Form of Gauss’s Law This first of Maxwell’s four equations applies to electrostatics and steady magnetic field. Physically it states that the electric flux per unit volume leaving a vanishingly small volume unit is exactly equal to the volume charge density there. President University Erwin Sitompul EEM 4/17 Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence The Vector Operator and The Divergence Theorem Divergence is an operation on a vector yielding a scalar, just like the dot product. We define the del operator as a vector operator: ax a y az x y z Then, treating the del operator as an ordinary vector, we can write: D a x a y a z ( Dxa x Dy a y Dz a z ) y z x Dx Dy Dz D x y z Dx Dy Dz div D = D = x y z President University Erwin Sitompul EEM 4/18 Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence The Vector Operator and The Divergence Theorem The operator does not have a specific form in other coordinate systems than rectangular coordinate system. Nevertheless, 1 1 D Dz D ( D ) z Cylindrical 1 2 1 1 D D 2 (r Dr ) (sin D ) r r r sin r sin Spherical President University EEM 4/19 Erwin Sitompul Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence The Vector Operator and The Divergence Theorem We shall now give name to a theorem that we actually have obtained, the Divergence Theorem: S D dS Q v dv Ddv vol vol The first and last terms constitute the divergence theorem: S D dS D dv vol “The integral of the normal component of any vector field over a closed surface is equal to the integral of the divergence of this vector field throughout the volume enclosed by the closed surface.” President University Erwin Sitompul EEM 4/20 Electric Flux Density, Gauss’s Law, and DIvergence Chapter 3 The Vector Operator and The Divergence Theorem Example Evaluate both sides of the divergence theorem for the field D = 2xy ax + x2 ay C/m2 and the rectangular parallelepiped fomed by the planes x = 0 and 1, y = 0 and 2, and z = 0 and 3. S S D dS D dv Divergence Theorem vol 3 2 0 0 DS dS (D) (D) x 0 3 1 0 0 3 2 0 0 (dydz a x ) y 0 (dxdz a y (D) ) (D) x 1 3 1 0 0 (dydz a x ) y 2 (dxdz a y ) But ( Dx ) x0 0, ( Dy ) y0 ( Dy ) y2 D S S dS 3 0 President University 2 0 ( Dx ) x1 dydz 3 0 2 0 2ydydz 12 C Erwin Sitompul EEM 4/21 Electric Flux Density, Gauss’s Law, and DIvergence Chapter 3 The Vector Operator and The Divergence Theorem D = vol (2 xy) ( x 2 ) 2 y x y D dv 3 2 1 z 0 y 0 x 0 x0 y 1 2 2 0 (2 y)dxdydz 3 z0 12 C D dS D dv 12 C S vol President University Erwin Sitompul EEM 4/22 Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Homework 4 D3.6. D3.7. D3.9. All homework problems from Hayt and Buck, 7th Edition. Deadline: 8 May 2012, at 08:00. President University Erwin Sitompul EEM 4/23