Chapter 24 Capacitance and Dielectrics

Transcription

1 Chapter 24 Capacitance and Dielectrics 1 Capacitors and Capacitance A capacitor is a device that stores electric potential energy and electric charge. The simplest construction of a capacitor is two parallel plates with a dielectric (insulating) material between the two plates. When placed in a circuit, equal and opposite charges are stored on the two plates providing a source of energy for the circuit. We will see that electric potential energy can be stored in a capacitor in the form of an electric field. The idea that the electric field contains energy forms our fundamental understanding of electromagnetic waves and the nature of light. Any two conductors separated by an insulator (or a vacuum) forma a capacitor. Figure 1: This figure shows the simplest form of a capacitor two charged conductors separated by air (or vacuum or an insulating dielectric). The capacitance is defined as the ratio of charge to potential difference, and this is a property of the capacitor due to its physical construction. 1

2 C = Q V ab (1) The SI units of capacitance is called one farad, in honor of the 19th-century English physicist Michael Faraday. 1 F = 1 farad = 1 C/V = 1 coulomb/volt 1.1 Calculating Capacitance: Capacitors in Vacuum Figure 2: This figure shows the arrangement of capacitor plates in a parallel-plate capacitor. The electric field produced by the accumulation of charge on the plates is also shown. We can write the E-field between two capacitor plates as: E = σ ɛ o = Q A ɛ o (2) 2

3 From the previous chapter we saw that the voltage between two capacitor plates is: V ab = E d = 1 ɛ o Qd A (3) Using our definition of capacitance (Eq. 1) we can write the capacitance as: C = Q V = ɛ o A d (4) Units of Capacitance Also, 1 F = 1 C 2 /N m = 1 C 2 /J ɛ o = F/m 1.2 A Cylindrical Capacitor Figure 3: This figure shows a cylindrical capacitor with linear charge density +λ on the inner conductor and λ on the outer conductor with an air gap between the two conductors. We saw from the previous chapter that voltage for an infinitely-long line charge was: V = λ 2πɛ o ln ( ro ) r 3

4 In order to calculate the capacitance of the two conductors, we must determine the voltage between the two conductors, V ab. V ab = λ 2πɛ o ln ( rb Using the electrical definition of capacitance we find: r a ) C = Q V ab = λl λ 2πɛ o ln ( r b r a ) = 2πɛ o L ln (r b /r a ) The Capacitance per unit Length C L = 2πɛ o ln (r b /r a ) = 55.6 pf/m ln (r b /r a ) (5) The capacitance per unit length of a coaxial cable is determined entirely by its dimensions. Most coaxial cables have an insulating material instead of vacuum between the conductors. A typical cable used for connecting a television has a capacitance per unit length of 69 pf/m. Ex A spherical capacitor contains a charge of 3.3 nc when connected to a potential difference of 220 V. If its plates are separated by vacuum and the inner radius of the outer shell is 4.00 cm, calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere. C = Q V ab = 4πɛ o r a r b r b r a 2 Capacitors in Series and Parallel Capacitors are produced with standard values of capacitance; however, they may not be the values you want. You can obtain the desired values by combining ca- 4

5 pacitors where the simplest combinations are series connections and parallel connections. 2.1 Capacitors in Series Figure 4: This figure shows two capacitors in series (i.e., they share the same charge Q on both capacitors. Also shown is the process for finding the equivalent capacitor that holds the same charge Q as the individual capacitors C 1 and C 2. The equivalent capacitor for capacitors in series is: 1 C eq = 1 C C 2 + (6) Ex In Fig. 24.8a, let C 1 = 3.00µF, C 2 = 5.00 µf, and V ab = V. Calculate (a) the charge on each capacitor and (b) the potential difference across each capacitor. 5

6 2.2 Capacitors in Parallel Figure 5: This figure shows two capacitors in parallel (i.e, they share the same voltage difference). Also shown is the process for finding the equivalent capacitance having charge Q = Q 1 + Q 2. The equivalent capacitor for capacitors in parallel is: C eq = C 1 + C 2 + (7) 2.3 Capacitors in Series and Parallel Figure 6: This figure shows the procedure for reducing a circuit containing capacitors in series and parallel into an equivalent circuit with only one capacitor. 6

7 Ex In Fig. E24.20, C 1 = 6.00µF, C 2 = 3.00µF, and C 3 = 5.00µF. The capacitor network is connected to an appliced potential V ab. After the charges on the capacitors have reached their final values, the charge on C 2 is 30.0µC. (a) What are the charges on capacitors C 1 and C 3? (b) What is the applied voltage V ab? 3 Energy Storage in Capacitors and Electric-Field Energy Most of the important applications of capacitors depend on their ability to store energy. In order to calculate the amount of energy stored in a capacitor we have must calculate how much work is required to charge a capacitor for a given voltage. If a battery with voltage V is connected to a capacitor, the total work W is calculated by summing up the differential works dw for every dq of charge placed on the capacitor plate. This is illustrated by the following figure. In the above picture, it takes an increasing amount of work to move the same charge dq on to the capacitor plate because the voltage V across the capacitor is continuously increasing proportional to the increasing amount charge q deposited on it. The amount work required to move a differential charge dq on to a capacitor plate already containing a charge q is: 7

8 Figure 7: This figure shows how a continuous stream of dq s are added to the capacitor until it reaches its total charge Q = V C. dw = dq V where V is the instantaneous voltage on the capacitor due to the instantaneous charge already on the capacitor: V = q/c. dw = V dq = q dq C (8) W = W 0 dw = 1 C Q 0 q dq = Q2 2C (work to charge a capacitor) (9) Because Q = V C, we can write the energy stored in a capacitor U using one of the three following expressions: Ex. 28: U = Q2 2C = 1 2 C V 2 = 1 QV (10) 2 For the capacitor network shown in Fig. E24.28, the potential difference across ab is 48 V. Find (a) the total charge stored in this network; (b) the charge on each capacitor; (c) the total energy stored in the network; (d) the energy stored in each capacitor; (e) the potential differences across each capacitor. 8

9 3.1 Electric-Field Energy Figure 8: This figure is used in problem As we charge a capacitor, electrons move from one plate to the other, thus creating an electric field between the plates. The volume between the plates is Ad where d is the separation distance between the plates and A is the area. where V = Ed. u = Energy density = 1 2 C V 2 A d However, we know the capacitance of the parallel-plate capacitor is: C = ɛ o A/d, and we can write the following for the energy density: (11) u = 1 2 ɛ oe 2 [ joules/m 3 ] (12) 4 Dielectrics When constructing capacitors, it is important to keep the separation distance between the plates d very uniform. This is accomplished by introducing an insulating material of uniform thickness, known as a dielectric. While introducing a fix to the construction tends to detract from its performance, in the case of a capacitor, the dielectric actually enhances its performance by increasing its capacitance. The new capacitance C in terms of its vacuum capacitance C o can be written as: C = K C o 9

10 where K is called the dielectric constant having a value K > 1. Some of the benefits of having a dielectric can be readily seen from the improved performance characteristics of the capacitor: The charge stored on the capacitor is: Q = V C = K (V C o ) = KQ o The energy density between the plates is: u = K ( 1 2 ɛ oc o V 2) = Ku o Figure 9: This tables shows the dielectric constants for various materials. Figure 10: This tables shows the dielectric constant and dielectric strength of some insulating materials. 10

11 Ex. 34: A parallel-plate capacitor has capacitance C o = 8.00 pf when there is air between the plates. The separation between the plates is 1.50 mm. (a) What is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates is not to exceed V/m? (b) A dielectric with K = 2.70 is inserted between the plates of the capacitor, completely filling the volume between the planes. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed V/m? 11

12 A Capacitor with a Fixed Charge Q Figure 11: This figure shows the voltage across a capacitor with a fixed charge Q. The voltage across the capacitor is reduced when a dielectric is introduced between the plates. V = V o K 12

13 4.1 Induced Charge and Polarization A Capacitor with a Fixed Charge Q Once again we look at a capacitor with a fixed charge Q before-and-after a dielectric is introduced. The polarization of a dielectric in an electric field E gives rise to thin layers of bound charges on the surfaces of the dielectric creating a total charge density of +σ i and σ i where σ i is the induced surface charge density on the dielectric. This has an effect of reducing the electric field E across the capacitor. E = E o K (when Q is constant) (13) Figure 12: This figure shows the initial and final electric field E o and E before-and-after a dielectric is introduced into a capacitor with fixed total charge Q. E o = σ ɛ o E = σ σ i ɛ o 13

14 Using these equations along with Eq. 13 above, we can calculate the induced surface charge density σ i. σ i = σ ( 1 1 ) K (induced surface charge density) (14) The produce Kɛ o is called the permittivity of the dielectric, and it s denoted by ɛ: ɛ = K ɛ o (definition of permittivity) E = σ ɛ (15) 4.2 A capacitor with and without a dielectric C = KC o = Kɛ o A d = ɛ A d (16) 4.3 Energy storage with and without a dielectric The energy density stored in a capacitor containing a dielectric is: u = 1 2 Kɛ o E 2 = 1 2 ɛe2 (17)..., and the energy density is reduced by the factor K as well: u = u o K 14

15 Ex. 66: A parallel-plate capacitor is made from two plates 12.0 cm on each side and 4.50 mm apart. Half of the space between these plates contains only air, but the other half is filled with Plexiglas of dielectric constant An 18.0-V battery is connected across the plates. (a) What is the capacitance of this combination? (Hint: Can you think of this capacitor as equivalent to two capacitors in parallel?) (b) How much energy is stored in the capacitor? (c) If we remove the Plexiglas but change nothing else, how much energy will be stored in the capacitor? Figure 13: This figure is for problem