UPB / ETTI O.DROSU Electrical Engineering 2

1. UPB / ETTIO.DROSUElectrical Engineering 2 Lecture 4: Electrostatics: Electrostatic Shielding; Poisson’s and Laplace’s Equations; Capacitance; Dielectric Materials and Permittivity 1

2. Lecture 4 Objectives • To continue our study of electrostatics with electrostatic shielding; • Poisson’s and Laplace’s equations; • capacitance; • dielectric materials and permittivity. 2

3. b a Q Ungrounded Spherical Metallic Shell • Consider a point charge at the center of a spherical metallic shell: Electrically neutral 3

4. Ungrounded Spherical Metallic Shell • The applied electric field is given by 4

5. Ungrounded Spherical Metallic Shell • The total electric field can be obtained using Gauss’s law together with our knowledge of how fields behave in a conductor. (1) Assume from symmetry the form of the field (2) Construct a family of Gaussian surfaces spheres of radius r where 5

6. a b Ungrounded Spherical Metallic Shell • Here, we shall need to treat separately 3 sub-families of Gaussian surfaces: 1) 2) 3) 6

7. Ungrounded Spherical Metallic Shell (3) Evaluate the total charge within the volume enclosed by each Gaussian surface 7

8. Ungrounded Spherical Metallic Shell Gaussian surfaces for which Gaussian surfaces for which Gaussian surfaces for which 8

9. Ungrounded Spherical Metallic Shell • For • For Shell is electrically neutral: The net charge carried by shell is zero. 9

10. Ungrounded Spherical Metallic Shell • For since the electric field is zero inside conductor. • A surface charge must exist on the inner surface and be given by 10

11. Ungrounded Spherical Metallic Shell • Since the conducting shell is initially neutral, a surface charge must also exist on the outer surface and be given by 11

12. Ungrounded Spherical Metallic Shell (4) For each Gaussian surface, evaluate the integral surface area of Gaussian surface. magnitude of D on Gaussian surface. 12

13. Ungrounded Spherical Metallic Shell (5) Solve for D on each Gaussian surface (6) Evaluate E as 13

14. Ungrounded Spherical Metallic Shell 14

15. Ungrounded Spherical Metallic Shell • The induced field is given by 15

16. Ungrounded Spherical Metallic Shell E total electric field Eapp r a b Eind 16

17. Ungrounded Spherical Metallic Shell • The electrostatic potential is obtained by taking the line integral of E. To do this correctly, we must start at infinity (the reference point or ground) and “move in” back toward the point charge. • For r > b 17

18. Ungrounded Spherical Metallic Shell • Since the conductor is an equipotential body (and potential is a continuous function), we have for 18

19. Ungrounded Spherical Metallic Shell • For 19

20. Ungrounded Spherical Metallic Shell V No metallic shell r b a 20

21. Grounded Spherical Metallic Shell • When the conducting sphere is grounded, we can consider it and ground to be one huge conducting body at ground (zero) potential. • Electrons migrate from the ground, so that the conducting sphere now has an excess charge exactly equal to -Q. This charge appears in the form of a surface charge density on the inner surface of the sphere. 21

22. - - - b - - a - - Q - - - Grounded Spherical Metallic Shell • There is no longer a surface charge on the outer surface of the sphere. • The total field outside the sphere is zero. • The electrostatic potential of the sphere is zero. 22

23. Grounded Spherical Metallic Shell E total electric field Eapp R a b Eind 23

24. Grounded Spherical Metallic Shell V Grounded metallic shell acts as a shield. R a b 24

25. The Need for Poisson’s and Laplace’s Equations • So far, we have studied two approaches for finding the electric field and electrostatic potential due to a given charge distribution. 25

26. The Need for Poisson’s and Laplace’s Equations • Method 1: given the position of all the charges, find the electric field and electrostatic potential using • (A) 26

27. The Need for Poisson’s and Laplace’s Equations • (B)  Method 1 is valid only for charges in free space. 27

28. The Need for Poisson’s and Laplace’s Equations • Method 2: Find the electric field and electrostatic potential using Gauss’s Law  Method 2 works only for symmetric charge distributions, but we can have materials other than free space present. 28

29. The Need for Poisson’s and Laplace’s Equations • Consider the following problem:  What are E and V in the region? Conducting bodies Neither Method 1 nor Method 2 can be used! 29

30. The Need for Poisson’s and Laplace’s Equations • Poisson’s equation is a differential equation for the electrostatic potential V. Poisson’s equation and the boundary conditions applicable to the particular geometry form a boundary-value problem that can be solved either analytically for some geometries or numerically for any geometry. • After the electrostatic potential is evaluated, the electric field is obtained using 30

31. Derivation of Poisson’s Equation • For now, we shall assume the only materials present are free space and conductors on which the electrostatic potential is specified. However, Poisson’s equation can be generalized for other materials (dielectric and magnetic as well). 31

32. Derivation of Poisson’s Equation 32

33. Derivation of Poisson’s Equation Poisson’s equation  2 is the Laplacian operator. The Laplacian of a scalar function is a scalar function equal to the divergence of the gradient of the original scalar function. 33

34. Laplacian Operator in Cartesian, Cylindrical, and Spherical Coordinates 34

35. Laplace’s Equation • Laplace’s equation is the homogeneous form of Poisson’s equation. • We use Laplace’s equation to solve problems where potentials are specified on conducting bodies, but no charge exists in the free space region. Laplace’s equation 35

36. Uniqueness Theorem • A solution to Poisson’s or Laplace’s equation that satisfies the given boundary conditions is the unique (i.e., the one and only correct) solution to the problem. 36

37. y + a V0 x b Potential Between Coaxial Cylinders Using Laplace’s Equation • Two conducting coaxial cylinders exist such that 37

38. Potential Between Coaxial Cylinders Using Laplace’s Equation • Assume from symmetry that 38

39. Potential Between Coaxial Cylinders Using Laplace’s Equation • Two successive integrations yield • The two constants are obtained from the two BCs: 39

40. Potential Between Coaxial Cylinders Using Laplace’s Equation • Solving for C1 and C2, we obtain: • The potential is 40

41. Potential Between Coaxial Cylinders Using Laplace’s Equation • The electric field between the plates is given by: • The surface charge densities on the inner and outer conductors are given by 41

42. + - V2 V1 + V12 = V2-V1 Capacitance of a Two Conductor System • The capacitance of a two conductor system is the ratio of the total charge on one of the conductors to the potential difference between that conductor and the other conductor. 42

43. Capacitance of a Two Conductor System • Capacitance is a positive quantity measured in units of F = Farads. • Capacitanceis a measure of the ability of a conductor configuration to store charge. 43

44. Capacitance of a Two Conductor System • The capacitance of an isolated conductor can be considered to be equal to the capacitance of a two conductor system where the second conductor is an infinite distance away from the first and at ground potential. 44

45. Capacitors • A capacitoris an electrical device consisting of two conductors separated by free space or another conducting medium. • To evaluate the capacitance of a two conductor system, we must find either the charge on each conductor in terms of an assumed potential difference between the conductors, or the potential difference between the conductors for an assumed charge on the conductors. 45

46. Capacitors • The former method is the more general but requires solution of Laplace’s equation. • The latter method is useful in cases where the symmetry of the problem allows us to use Gauss’s law to find the electric field from a given charge distribution. 46

47. Parallel-Plate Capacitor • Determine an approximate expression for the capacitance of a parallel-plate capacitor by neglecting fringing. Conductor 2 d A Conductor 1 47

48. Parallel-Plate Capacitor • “Neglecting fringing” means to assume that the field that exists in the real problem is the same as for the infinite problem. z z = d V = V12 z = 0 V = 0 48

49. Parallel-Plate Capacitor • Determine the potential between the plates by solving Laplace’s equation. 49

50. Parallel-Plate Capacitor 50