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Capacitance

Chapter 6. Dielectrics 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 .

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Capacitance

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  1. Chapter 6 Dielectrics and Capacitance Capacitance • Now let us consider two conductors embedded in a homogenous dielectric. • Conductor M2 carries a total positive charge Q, and M2 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

  2. Chapter 6 Dielectrics and Capacitance Capacitance • The electric flux is directed from M2 to M1, thus M2 is at the more positive potential. • Works must be done to carry a positive charge from M1 to M2. • Let us assign V0 as the potential difference between M2 and M1. • 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.

  3. Chapter 6 Dielectrics and Capacitance 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. • 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.

  4. Chapter 6 Dielectrics and Capacitance 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,

  5. Chapter 6 Dielectrics and Capacitance Capacitance • The potential difference between lower and upper planes is: • 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:

  6. Chapter 6 Dielectrics and Capacitance Capacitance • Example • Calculate the capacitance of a parallel-plate capacitor having a mica dielectric, εr = 6, a plate area of 10 in2, and a separation of 0.01 in.

  7. Chapter 6 Dielectrics and Capacitance Capacitance • The total energy stored in the capacitor is:

  8. Chapter 6 Dielectrics and Capacitance 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: • Next, consider a spherical capacitor formed of two concentric spherical conducting shells of radius a and b, b>a.

  9. Chapter 6 Dielectrics and Capacitance Several Capacitance Examples • If we allow the outer sphere to become infinitely large, we obtain the capacitance of an isolated spherical conductor: • 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 = ε1, extending from r = a to r = r1,

  10. Chapter 6 Dielectrics and Capacitance Several Capacitance Examples • While the potential difference is: • Therefore,

  11. Chapter 6 Dielectrics 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 ε1S/d, using a dielectric of permittivity ε1. • Now, let us replace a part of this dielectric by another of permittivity ε2, placing the boundary between the two dielectrics parallel to the plates. • Assuming a charge Q on one plate, ρS = Q/S, while DN1 = DN2, since D is only normal to the boundary. • E1 = D1/ε1 = Q/(ε1S),E2 = D2/ε2 = Q/(ε2S). • V1 = E1d1, V2 = E2d2.

  12. Chapter 6 Dielectrics 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 S1 and S2. • Assuming a charge Q on one plate, Q = ρS1S1 + ρS2S2. • ρS1 = D1 = ε1E1,ρS2 = D2 = ε2E2. • V0 = E1d = E2d.

  13. Chapter 6 Dielectrics and Capacitance Homework 8 • D6.4 • D6.5 • Deadline: 19.06.2012, at 08:00.

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