1. President UniversityErwin SitompulEEM 6/1 Lecture 6 Engineering Electromagnetics Dr.-Ing. Erwin Sitompul President University http://zitompul.wordpress.com 2013

2. President UniversityErwin SitompulEEM 6/2 The Potential Field of a System of Charges: Conservative Property Chapter 4Energy and Potential 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 Q 1 located at r 1 is given by: The field is linear with respect to charge so that superposition is applicable. Thus, the potential arising from n point charges is:

3. President UniversityErwin SitompulEEM 6/3 Chapter 4Energy and Potential The Potential Field of a System of Charges: Conservative Property As the number of elements approach infinity, we obtain the integral expression: If each point charge is now represented as a small element of continuous volume charge distribution ρ v Δv, then: If the charge distribution takes from of a line charge or a surface charge,

4. President UniversityErwin SitompulEEM 6/4 Chapter 4Energy 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 ρ L in the form of a ring, ρ = a, in the z = 0 plane. 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.

5. President UniversityErwin SitompulEEM 6/5 Chapter 4Energy 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: Or, for potential difference: Both expressions above are not dependent on the path chosen for the line integral, regardless of the source of the E field. For static fields, no work is done in carrying the unit charge around any closed path. Potential conservation in a simple dc-circuit problem in the form of Kirchhoff’s voltage law

6. President UniversityErwin SitompulEEM 6/6 Chapter 4Energy 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. In practical problems, however, we rarely know E or ρ 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.

7. President UniversityErwin SitompulEEM 6/7 Chapter 4Energy and Potential Potential Gradient The general line-integral relationship between V and E is: For a very short element of length ΔL, E is essentially constant: 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:

8. President UniversityErwin SitompulEEM 6/8 Potential Gradient Chapter 4Energy and Potential From the last equation, the maximum positive increment of potential, Δv max, will occur when cosθ = –1, or ΔL points in the direction opposite to E. 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.

9. President UniversityErwin SitompulEEM 6/9 Potential Gradient Chapter 4Energy and Potential For the equipotential surfaces below, find the direction of E at P.

10. President UniversityErwin SitompulEEM 6/10 Potential Gradient Chapter 4Energy and Potential Since the potential field information is more likely to be determined first, let us describe the direction of ΔL (which leads to a maximum increase in potential) in term of potential field. Let a N 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: The maximum magnitude occurs when ΔL is in the a N direction. Thus we define dN as incremental length in a N direction,

11. President UniversityErwin SitompulEEM 6/11 Chapter 4Energy 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: Using the new term,

12. President UniversityErwin SitompulEEM 6/12 Chapter 4Energy and Potential Potential Gradient Since V is a function of x, y, and z, the total differential is: But also, Both expression are true for any dx, dy, and dz. Thus: Note: Gradient of a scalar is a vector.

13. President UniversityErwin SitompulEEM 6/13 Potential Gradient Chapter 4Energy and Potential Introducing the vector operator for gradient: We now can relate E and V as: Rectangular Cylindrical Spherical

14. President UniversityErwin SitompulEEM 6/14 Potential Gradient Chapter 4Energy and Potential Example Given the potential field, V = 2x 2 y–5z, and a point P(–4,3,6), find V, E, direction of E, D, and ρ v.

15. President UniversityErwin SitompulEEM 6/15 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. Chapter 4Energy and Potential 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.

16. President UniversityErwin SitompulEEM 6/16 The Dipole The distant point P is described by the spherical coordinates r, θ, Φ = 90°. The positive and negative point charges have separation d and described in rectangular coordinates (0,0, 0.5d) and (0,0,–0.5d). Chapter 4Energy and Potential

17. President UniversityErwin SitompulEEM 6/17 The Dipole The total potential at P can be written as: The plane z = 0 is the locus of points for which R 1 = R 2 ► The potential there is zero (as also all points at ∞). Chapter 4Energy and Potential

18. President UniversityErwin SitompulEEM 6/18 The Dipole For a distant point, R 1 ≈ R 2 ≈ r, R 2 –R 1 ≈ dcosθ Using the gradient in spherical coordinates, Chapter 4Energy and Potential

19. President UniversityErwin SitompulEEM 6/19 The Dipole To obtain a plot of the potential field, we choose Qd/(4πε 0 ) = 1 and thus cosθ = Vr 2. 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° Chapter 4Energy and Potential

20. President UniversityErwin SitompulEEM 6/20 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. Since d  a r = d cosθ, we then have: Dipole charges: Point charge: Chapter 4Energy and Potential

21. President UniversityErwin SitompulEEM 6/21 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) Chapter 4Energy and Potential Answer: 49.7 V, 60 V. 2.For the point P(3,60°,2) in cylindrical coordinates and the potential field V = 10(ρ +1)z 2 cosφ V in free space, find at P: (a) V; (b) E; (c) D; (d) ρ v ; (e) dV/dN; (f) a N. (Hay.E5.S112.23) Answer: (a) 80 V; (b) –20a ρ + 46.2a φ – 80a z V/m; (c) –177.1a ρ + 409a φ – 708a z pC/m 2 ; (d) –334 pC/m 3 ;(e) 94.5 V/m; (f) 0.212a ρ – 0.489a φ +0.846a z 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.