1. Conductors and Dielectrics UNIT II 1B.Hemalath AP-ECE

2. Capacitance 1.To store charge 2.To store energy 3.To control variation time scales in a circuit 2B.Hemalath AP-ECE

3. When a capacitor is charged, its plates have charges of equal magnitudes but opposite signs: q+ and q-. However, we refer to the charge of a capacitor as being q, the absolute value of these charges on the plates. The charge q and the potential difference V for a capacitor are proportional to each other: The proportionality constant C is called the capacitance of the capacitor. Its value depends only on the geometry of the plates and not on their charge or potential difference. The SI unit is called the farad (F): 1 farad (1 F)= 1 coulomb per volt =1 C/V. 3B.Hemalath AP-ECE

4. Charging a Capacitor The circuit shown is incomplete because switch S is open; that is, the switch does not electrically connect the wires attached to it. When the switch is closed, electrically connecting those wires, the circuit is complete and charge can then flow through the switch and the wires. As the plates become oppositely charged, that potential difference increases until it equals the potential difference V between the terminals of the battery. With the electric field zero, there is no further drive of electrons. The capacitor is then said to be fully charged, with a potential difference V and charge q. 4B.Hemalath AP-ECE

5. Calculating the Capacitance: 5B.Hemalath AP-ECE

6. Calculating the Capacitance; A Cylindrical Capacitor : As a Gaussian surface, we choose a cylinder of length L and radius r, closed by end caps and placed as is shown. It is coaxial with the cylinders and encloses the central cylinder and thus also the charge q on that cylinder. 6B.Hemalath AP-ECE

7. Calculating the Capacitance; A Spherical Capacitor: 7B.Hemalath AP-ECE

8. *We can assign a capacitance to a single isolated spherical conductor of radius R by assuming that the “missing plate” is a conducting sphere of infinite radius. *The field lines that leave the surface of a positively charged isolated conductor must end somewhere; the walls of the room in which the conductor is housed can serve effectively as our sphere of infinite radius. *To find the capacitance of the conductor, we first rewrite the capacitance as: Now letting b→∞, and substituting R for a, Calculating the Capacitance; An Isolated Sphere: 8B.Hemalath AP-ECE

9. Example, Charging the Plates in a Parallel-Plate Capacitor: 9B.Hemalath AP-ECE

10. Capacitors in Parallel 10B.Hemalath AP-ECE

11. Capacitors in Series 11B.Hemalath AP-ECE

12. Example, Capacitors in Parallel and in Series: 12B.Hemalath AP-ECE

13. Example, Capacitors in Parallel and in Series: 13B.Hemalath AP-ECE

14. Example, One Capacitor Charging up Another Capacitor: 14B.Hemalath AP-ECE

15. Energy Stored in an Electric Field: 15B.Hemalath AP-ECE

16. Energy Density: 16B.Hemalath AP-ECE

17. Potential Energy & Energy Density of an Electric Field: Example 17B.Hemalath AP-ECE

18. Example Work and Energy when a Dielectric is inserted inside a Capacitor: 18B.Hemalath AP-ECE

19. Dielectrics, an Atomic View: 1.Polar dielectrics. The molecules of some dielectrics, like water, have permanent electric dipole moments. In such materials (called polar dielectrics), the electric dipoles tend to line up with an external electric field as in Fig. 25-14. Since the molecules are continuously jostling each other as a result of their random thermal motion, this alignment is not complete, but it becomes more complete as the magnitude of the applied field is increased (or as the temperature, and thus the jostling, are decreased).The alignment of the electric dipoles produces an electric field that is directed opposite the applied field and is smaller in magnitude. 2.Nonpolar dielectrics. Regardless of whether they have permanent electric dipole moments, molecules acquire dipole moments by induction when placed in an external electric field. This occurs because the external field tends to “stretch” the molecules, slightly separating the centers of negative and positive charge. 19B.Hemalath AP-ECE

20. Dielectrics and Gauss’ Law: A dielectric, is an insulating material such as mineral oil or plastic, and is characterized by a numerical factor , called the dielectric constant of the material. 20B.Hemalath AP-ECE

21. Dielectrics and Gauss’ Law: ( Q ) 21B.Hemalath AP-ECE

22. Dielectrics and Gauss’ Law: 1.The flux integral now involves  E, not just E. The vector (  0  E) is sometimes called the electric displacement, D. The above equation can be written as: 1.The charge q enclosed by the Gaussian surface is now taken to be the free charge only. The induced surface charge is deliberately ignored on the right side of the above equation, having been taken fully into account by introducing the dielectric constant  on the left side.  0 gets replaced by  0. We keep  inside the integral of the above equation to allow for cases in which  is not constant over the entire Gaussian surface. 22B.Hemalath AP-ECE

23. Capacitor with a Dielectric: The introduction of a dielectric limits the potential difference that can be applied between the plates to a certain value V max, called the breakdown potential. Every dielectric material has a characteristic dielectric strength, which is the maximum value of the electric field that it can tolerate without breakdown. It actually can increase the capacitance of the device. Recall that 23B.Hemalath AP-ECE

24. Example, Dielectric Partially Filling a Gap in a Capacitor: 24B.Hemalath AP-ECE

25. 25B.Hemalath AP-ECE

26. Why are boundary conditions important ? When a free-space electromagnetic wave is incident upon a medium secondary waves are transmitted wave reflected wave The transmitted wave is due to the E and H fields at the boundary as seen from the incident side The reflected wave is due to the E and H fields at the boundary as seen from the transmitted side To calculate the transmitted and reflected fields we need to know the fields at the boundary These are determined by the boundary conditions 26B.Hemalath AP-ECE

27. At a boundary between two media, m r,e r s are different on either side. An abrupt change in these values changes the characteristic impedance experienced by propagating waves Discontinuities results in partial reflection and transmission of EM waves The characteristics of the reflected and transmitted waves can be determined from a solution of Maxwells equations along the boundary 27  2,  2  2  1,  1  1 B.Hemalath AP-ECE

28. Boundary conditions The tangential component of E is continuous at a surface of discontinuity E 1t, = E 2t Except for a perfect conductor, the tangential component of H is continuous at a surface of discontinuity H 1t, = H 2t 28 E 1t, H 1t  2,  2  2  1,  1  1 E 2t, H 2t  2,  2  2  1,  1  1 D 2n, B 2n The normal component of D is continuous at the surface of a discontinuity if there is no surface charge density. If there is surface charge density D is discontinuous by an amount equal to the surface charge density. – D 1n, = D 2n +r s The normal component of B is continuous at the surface of discontinuity – B 1n, = B 2n B.Hemalath AP-ECE

29. Proof of boundary conditions - continuity of E t Integral form of Faraday’s law: 0 0 00 0 That is, the tangential component of E is continuous  2,  2  2  1,  1  1 29B.Hemalath AP-ECE

30. Ampere’s law 30 That is, the tangential component of H is continuous 0 0 0 0 0  2,  2  2  1,  1  1 B.Hemalath AP-ECE

31. The integral form of Gauss’ law for electrostatics is: 31 Proof of boundary conditions - D n applied to the box gives hence The change in the normal component of D at a boundary is equal to the surface charge density  2,  2  2  1,  1  1 B.Hemalath AP-ECE

32. Poisson’s and Laplace Equations A useful approach to the calculation of electric potentials Relates potential to the charge density. The electric field is related to the charge density by the divergence relationship The electric field is related to the electric potential by a gradient relationship Therefore the potential is related to the charge density by Poisson's equation In a charge-free region of space, this becomes Laplace's equation 32B.Hemalath AP-ECE

33. Potential of a Uniform Sphere of Charge outside inside 33B.Hemalath AP-ECE

34. Poisson’s and Laplace Equations Poisson’s Equation 34B.Hemalath AP-ECE

35. Poisson’s and Laplace Equations 35B.Hemalath AP-ECE

36. Examples of the Solution of Laplace’s Equation D7.1 36B.Hemalath AP-ECE

37. Uniqueness Theorem Given is a volume V with a closed surface S. The function V(x,y,z) is completely determined on the surface S. There is only one function V(x,y,z) with given values on S (the boundary values) that satisfies the Laplace equation. Application: The theorem of uniqueness allows to make statements about the potential in a region that is free of charges if the potential on the surface of this region is known. The Laplace equation applies to a region of space that is free of charges. Thus, if a region of space is enclosed by a surface of known potential values, then there is only one possible potential function that satisfies both the Laplace equation and the boundary conditions. Example: A piece of metal has a fixed potential, for example, V = 0 V. Consider an empty hole in this piece of metal. On the boundary S of this hole, the value of V(x,y,z) is the potential value of the metal, i.e., V(S) = 0 V. V(x,y,z) = 0 satisfies the Laplace equation (check it!). Because of the theorem of uniqueness, V(x,y,z) = 0 describes also the potential inside the hole 37B.Hemalath AP-ECE

38. Examples of the Solution of Laplace’s Equation Assume V is a function only of x – solve Laplace’s equation 38B.Hemalath AP-ECE

39. Examples of the Solution of Laplace’s Equation Steps 1 – Given V, use E = - DelV to find E 2 – Use D =  E to find D 3 - Evaluate D at either capacitor plate, D = Ds = Dn an 4 – Recognize that  s = Dn 5 – Find Q by a surface integration over the capacitor plate Finding the capacitance of a parallel-plate capacitor 39B.Hemalath AP-ECE

40. Examples of the Solution of Laplace’s Equation Cylindrical 40B.Hemalath AP-ECE

41. Examples of the Solution of Laplace’s Equation 41B.Hemalath AP-ECE

42. Examples of the Solution of Laplace’s Equation (spherical coordinates) 42B.Hemalath AP-ECE

43. Examples of the Solution of Laplace’s Equation 43B.Hemalath AP-ECE

44. Examples of the Solution of Poisson’s Equation charge density E field intensity potential 44B.Hemalath AP-ECE

45. Product Solution Of Laplace's Equation 45B.Hemalath AP-ECE

46. Product Solution Of Laplace's Equation 46B.Hemalath AP-ECE

47. Product Solution Of Laplace's Equation 47B.Hemalath AP-ECE

48. Product Solution Of Laplace's Equation 48B.Hemalath AP-ECE

49. Product Solution Of Laplace's Equation 49B.Hemalath AP-ECE

50. Product Solution Of Laplace's Equation 50B.Hemalath AP-ECE