1. President UniversityErwin SitompulEEM 9/1 Dr.-Ing. Erwin Sitompul President University Lecture 9 Engineering Electromagnetics http://zitompul.wordpress.com

2. President UniversityErwin SitompulEEM 9/2 Chapter 6Dielectrics and Capacitance Capacitance Now let us consider two conductors embedded in a homogenous dielectric. Conductor M 2 carries a total positive charge Q, and M 2 carries an equal negative charge –Q. No other charges present  the total charge of the system is zero. The charge is carried on the surface as a surface charge density. The electric field is normal to the conductor surface. Each conductor is an equipotential surface

3. President UniversityErwin SitompulEEM 9/3 Capacitance The electric flux is directed from M 2 to M 1, thus M 2 is at the more positive potential. Works must be done to carry a positive charge from M 1 to M 2. Let us assign V 0 as the potential difference between M 2 and M 1. We may now define the capacitance of this two-conductor system as the ratio of the magnitude of the total charge on either conductor to the magnitude of the potential difference between the conductors. Chapter 6Dielectrics and Capacitance

4. President UniversityErwin SitompulEEM 9/4 Capacitance The capacitance is independent of the potential and total charge for their ratio is constant. If the charge density is increased by a factor, Gauss's law indicates that the electric flux density or electric field intensity also increases by the same factor, as does the potential difference. Chapter 6Dielectrics and Capacitance Capacitance is a function only of the physical dimensions of the system of conductors and of the permittivity of the homogenous dielectric. Capacitance is measured in farads (F), 1 F = 1 C/V.

5. President UniversityErwin SitompulEEM 9/5 Capacitance Chapter 6Dielectrics and Capacitance We will now apply the definition of capacitance to a simple two- conductor system, where the conductors are identical, infinite parallel planes, and separated a distance d to each other. The charge on the lower plane is positive, since D is upward. The charge on the upper plane is negative,

6. President UniversityErwin SitompulEEM 9/6 Capacitance The potential difference between lower and upper planes is: Chapter 6Dielectrics and Capacitance The total charge for an area S of either plane, both with linear dimensions much greater than their separation d, is: The capacitance of a portion of the infinite-plane arrangement, far from the edges, is:

7. President UniversityErwin SitompulEEM 9/7 Capacitance Chapter 6Dielectrics and Capacitance Example Calculate the capacitance of a parallel-plate capacitor having a mica dielectric, ε r = 6, a plate area of 10 in 2, and a separation of 0.01 in.

8. President UniversityErwin SitompulEEM 9/8 Capacitance The total energy stored in the capacitor is: Chapter 6Dielectrics and Capacitance

9. President UniversityErwin SitompulEEM 9/9 Several Capacitance Examples As first example, consider a coaxial cable or coaxial capacitor of inner radius a, outer radius b, and length L. The capacitance is given by: Chapter 6Dielectrics and Capacitance Next, consider a spherical capacitor formed of two concentric spherical conducting shells of radius a and b, b>a.

10. President UniversityErwin SitompulEEM 9/10 If we allow the outer sphere to become infinitely large, we obtain the capacitance of an isolated spherical conductor: Chapter 6Dielectrics and Capacitance Several Capacitance Examples A sphere about the size of a marble, with a diameter of 1 cm, will have: Coating this sphere with a different dielectric layer, for which ε = ε 1, extending from r = a to r = r 1,

11. President UniversityErwin SitompulEEM 9/11 Several Capacitance Examples While the potential difference is: Chapter 6Dielectrics and Capacitance Therefore,

12. President UniversityErwin SitompulEEM 9/12 Chapter 6Dielectrics and Capacitance Several Capacitance Examples A capacitor can be made up of several dielectrics. Consider a parallel-plate capacitor of area S and spacing d, d << linear dimension of S. The capacitance is ε 1 S/d, using a dielectric of permittivity ε 1. Now, let us replace a part of this dielectric by another of permittivity ε 2, placing the boundary between the two dielectrics parallel to the plates. Assuming a charge Q on one plate, ρ S = Q/S, while D N1 = D N2, since D is only normal to the boundary. E 1 = D 1 /ε 1 = Q/(ε 1 S), E 2 = D 2 /ε 2 = Q/(ε 2 S). V 1 = E 1 d 1, V 2 = E 2 d 2.

13. President UniversityErwin SitompulEEM 9/13 Chapter 6Dielectrics and Capacitance Several Capacitance Examples Another configuration is when the dielectric boundary were placed normal to the two conducting plates and the dielectrics occupied areas of S 1 and S 2. Assuming a charge Q on one plate, Q = ρ S1 S 1 + ρ S2 S 2. ρ S1 = D 1 = ε 1 E 1, ρ S2 = D 2 = ε 2 E 2. V 0 = E 1 d = E 2 d.

14. President UniversityErwin SitompulEEM 9/14 Homework 8 D6.4 D6.5 Deadline: 19.06.2012, at 08:00. Chapter 6Dielectrics and Capacitance