Energy and Potential in Engineering Electromagnetics

Engineering Electromagnetics Lecture 6 Dr.-Ing. Erwin Sitompul President University http://zitompul.wordpress.com 2 0 1 5 President University Erwin Sitompul EEM 6/1

Chapter 4 Energy and Potential The Potential Field of a System of Charges: Conservative Property We will now prove, that for a system of charges, the potential is also independent of the path taken. Continuing the discussion, the potential field at the point r due to a single point charge Q1 located at r1 is given by: Q V r r = ( ) r 1 4 0 1 The field is linear with respect to charge so that superposition is applicable. Thus, the potential arising from n point charges is: ( ) 4 4 r r r r Q Q Q = + + + r V n 1 2 r r 4 0 1 0 2 0 n Q n = ( ) r m r r V 4 = 1 m 0 m President University Erwin Sitompul EEM 6/2

Chapter 4 Energy and Potential The Potential Field of a System of Charges: Conservative Property If each point charge is now represented as a small element of continuous volume charge distribution v v, then: ( ) ( ) ( ) 4 4 r r r r r r ( ) n r v v v = + + + r 1 1 2 2 V v v v n r r 4 0 1 0 2 0 n As the number of elements approach infinity, we obtain the integral expression: r r ( ) r dv = ( ) r V v 4 vol 0 If the charge distribution takes from of a line charge or a surface charge, ( ) ( ) 4 r r r r r dL = r V L 0 ( ) r dS = ( ) r V S 4 S 0 President University Erwin Sitompul EEM 6/3

Chapter 4 Energy and Potential The Potential Field of a System of Charges: Conservative Property As illustration, let us find V on the z axis for a uniform line charge L in the form of a ring, = a, in the z = 0 plane. ( ) ( ) 4 r r r dL = r V L 0 + Lad a 2 = La a 2 2 0 4 z 0 = + 2 2 2 z 0 The potential arising from point charges or continuous charge distribution can be seen as the summation of potential arising from each charge or each differential charge. It is independent of the path chosen. President University Erwin Sitompul EEM 6/4

Chapter 4 Energy and Potential The Potential Field of a System of Charges: Conservative Property With zero reference at , the expression for potential can be taken generally as: A E = L V d A Or, for potential difference: A E = = L V V V d AB A B B Both expressions above are not dependent on the path chosen for the line integral, regardless of the source of the E field. Potential conservation in a simple dc-circuit problem in the form of Kirchhoff s voltage law For static fields, no work is done in carrying the unit charge around any closed path. E d = L 0 President University Erwin Sitompul EEM 6/5

Chapter 4 Energy and Potential Potential Gradient We have discussed two methods of determining potential: directly from the electric field intensity by means of a line integral, or from the basic charge distribution itself by a volume integral. E or vknown V asked In practical problems, however, we rarely know E or v. Preliminary information is much more likely to consist a description of two equipotential surface, and the goal is to find the electric field intensity. V known E asked President University Erwin Sitompul EEM 6/6

Chapter 4 Energy and Potential Potential Gradient The general line-integral relationship between V and E is: V d = E dV d = E L For a very short element of length L, E is essentially constant: L E E L L cos V V V L cos E Assuming a conservative field, for a given reference and starting point, the result of the integration is a function of the end point (x,y,z). We may pass to the limit and obtain: dV dL = cos E President University Erwin Sitompul EEM 6/7

Chapter 4 Energy and Potential Potential Gradient From the last equation, the maximum positive increment of potential, vmax, will occur when cos = 1, or L points in the direction opposite to E. dV E dL = max We can now conclude two characteristics of the relationship between E and V at any point: 1. The magnitude of E is given by the maximum value of the rate of change of V with distance L. 2. This maximum value of V is obtained when the direction of the distance increment is opposite to E. President University Erwin Sitompul EEM 6/8

Chapter 4 Energy and Potential Potential Gradient For the equipotential surfaces below, find the direction of E at P. E dV dL = , max 180 President University Erwin Sitompul EEM 6/9

Chapter 4 Energy and Potential Potential Gradient : Electric field lines : Equipotential lines President University Erwin Sitompul EEM 6/10

Chapter 4 Energy and Potential Potential Gradient Since the potential field information is more likely to be determined first, let us describe the direction of L (which leads to a maximum increase in potential) in term of potential field. Let aN be a unit vector normal to the equipotential surface and directed toward the higher potential. The electric field intensity is then expressed in terms of the potential as: dV dL The maximum magnitude occurs when L is in the aN direction. Thus we define dN as incremental length in aN direction, E = a N max dV dL dV dN = max dV dN E = a N President University Erwin Sitompul EEM 6/11

Chapter 4 Energy and Potential Potential Gradient We know that the mathematical operation to find the rate of change in a certain direction is called gradient. Now, the gradient of a scalar field T is defined as: dT dN = = a Gradient of grad T T N Using the new term, dV dN grad V E = a = N President University Erwin Sitompul EEM 6/12

Chapter 4 Energy and Potential Potential Gradient Since V is a function of x, y, and z, the total differential is: V V V dV dx dy dz x y z = + + But also, = = E L dV d E dx E dy E dz x y z Both expression are true for any dx, dy, and dz. Thus: V E x V E y V E z = V x V y V z x = + + E a a a x y z = y V x V y V z = + + a a a grad V = x y z z Note: Gradient of a scalar is a vector. President University Erwin Sitompul EEM 6/13

Chapter 4 Energy and Potential Potential Gradient Introducing the vector operator for gradient: = + + a a a x y z x y z We now can relate E and V as: = E V V x V y V z Rectangular = + + a a a V x y z 1 V V V z Cylindrical = + + a a a V z 1 r 1 V r V V Spherical = + + a a a V r sin r President University Erwin Sitompul EEM 6/14

Chapter 4 Energy and Potential Potential Gradient Example Given the potential field, V = 2x2y 5z, and a point P( 4,3,6), find V, E, direction of E, D, and v. V = 2( 4) (3) 5(6) a 2 = 66 V 2 P 5 V x V y V z a = = + = + + 2 E a a a V a a a 4 2 5 xy x x y z x y z = + 4( 4)(3) = + a a a E a 48 32 5 V m 2( 4) x y z P x y z =E a P , E P E P = + = 3 a a a D E 425 283.3 44.27 pC m 0 P P x y z v At , = = 35.42 pC m y = = )( 4 ) y 3 D E 12 div P = div (8.854 10 0 106.26 pC m = 3 3 35.42(3) pC m v President University Erwin Sitompul EEM 6/15

Chapter 4 Energy and Potential The Dipole The dipole fields form the basis for the behavior of dielectric materials in electric field. The dipole will be discussed now and will serve as an illustration about the importance of the potential concept presented previously. An electric dipole, or simply a dipole, is the name given to two point charges of equal magnitude and opposite sign, separated by a distance which is small compared to the distance to the point P at which we want to know the electric and potential fields. President University Erwin Sitompul EEM 6/16

Chapter 4 Energy and Potential The Dipole The distant point P is described by the spherical coordinates r, , = 90 . The positive and negative point charges have separation d and described in rectangular coordinates (0,0, 0.5d) and (0,0, 0.5d). President University Erwin Sitompul EEM 6/17

Chapter 4 Energy and Potential The Dipole The total potential at P can be written as: 1 1 4 R R The plane z = 0 is the locus of points for which R1 = R2 The potential there is zero (as also all points at ). Q R R Q = = V 2 R R 1 4 0 1 2 0 1 2 President University Erwin Sitompul EEM 6/18

Chapter 4 Energy and Potential The Dipole For a distant point, R1 R2 r, R2 R1 dcos cos 4 r Qd = V 2 0 Using the gradient in spherical coordinates, = E 1 r r r V 1 V V V = + + a a a sin r cos sin Qd Qd = E a a r 3 3 2 4 r r 0 0 Qd ( ) = E a a 2cos sin r 3 4 r 0 President University Erwin Sitompul EEM 6/19

Chapter 4 Energy and Potential The Dipole To obtain a plot of the potential field, we choose Qd/(4 0) = 1 and thus cos = Vr2. The colored lines in the figure below indicate equipotentials for V = 0, +0.2, +0.4, +0.6, +0.8, and +1. r = 2.236 r = 1.880 Plane at zero potential 45 cos Qd = V 2 4 r 0 President University Erwin Sitompul EEM 6/20

Chapter 4 Energy and Potential The Dipole The potential field of the dipole may be simplified by making use of the dipole moment. If the vector length directed from Q to +Q is identified as d, then the dipole moment is defined as Qd and is assigned the symbol p. Q = p d Since d ar = dcos , we then have: p a = V r r 2 4 0 r r r r 1 r r 1 r 1 r Dipole charges: = p E , V V 2 3 2 4 0 1 r 1 r E , V Point charge: 2 President University Erwin Sitompul EEM 6/21

Chapter 4 Energy and Potential Exercise Problems 1. A charge in amount of 13.33 nC is uniformly distributed in a circular disk form, with the radius of 2 m. Determine the potential at a point on the axis, 2 m from the disk. Compare this potential with that which results if all the charge is at the center of the disk. (Sch.S62.E3) Answer: 49.7 V, 60 V. 2. For the point P(3,60 ,2) in cylindrical coordinates and the potential field V = 10( +1)z2cos V in free space, find at P: (a) V; (b) E; (c) D; (d) v; (e) dV/dN; (f) aN. 708az pC/m2; (d) 334 pC/m3;(e) 94.5 V/m; (f) 0.212a 0.489a +0.846az (Hay.E5.S112.23) Answer: (a) 80 V; (b) 20a + 46.2a 80az V/m; (c) 177.1a + 409a 3. Two opposite charges are located on the z-axis and centered on the origin, configuring as an electric dipole is located at origin. The distance between the two charges, each with magnitude of 1 nC, is given by 0.1 nm. The electric potential at A(0,1,1) nm is known to be 2 V. Find out the electric potential at B(1,1,1) nm. Hint: Do not assume that A and B are distant points. Determine E due to both charges first, then calculate the potential difference. (EEM.Pur5.N4)Answer: 1.855 V. President University Erwin Sitompul EEM 6/22