1farad = 1F = 1 C V. The permittivity constant can be expressed in terms of this new unit as: C2 = F (5.3)

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1 hapter 5 apacitance and Dielectrics 5.1 The Important Stuff apacitance Electrical energy can be stored by putting opposite charges ±q on a pair of isolated conductors. Being conductors, the respective surfaces of these two objects are all at the same potential so that it makes sense to speak of a potential difference V between the two conductors, though one should really write V for this. (Also, we will usually just talk about the charge q of the conductor pair though we really mean ±q.) Such a device is called a capacitor. The general case is shown in Fig. 5.1(a). A particular geometry known as the parallel plate capacitor is shown in Fig. 5.1(b). It so happens that if we don t change the configuration of the two conductors, the charge q is proportional to the potential difference V. The proportionality constant is called the capacitance of the device. Thus: q = V (5.1) The SI unit of capacitance is then 1 V, a combination which is called the farad1. Thus: 1farad = 1F = 1 V (5.2) The permittivity constant can be expressed in terms of this new unit as: ɛ 0 = alculating apacitance 2 = F (5.3) N m 2 m For various simple geometries for the pair of conductors we can find expressions for the capacitance. Parallel-Plate apacitor 1 Named in honor of the...uh... Austrian physicist Jim Farad ( ) who did some electrical experiments in... um... Berlin. That s it, Berlin. 71

2 72 HAPTER 5. APAITANE AND DIELETRIS +q -q +q -q (a) (b) Figure 5.1: (a) Two isolated conductors carrying charges ±q: A capacitor! (b) A more common configuration of conductors for a capacitor: Two isolated parallel conducting sheets of area A, separated by (small) distance d. The most common geometry we encounter is one where the two conductors are parallel plates (as in Fig. 5.1(b), with the stipulation that the dimensions of the plates are large compared to their separation to minimize the fringing effect. For a parallel-plate capacitor with plates of area A separated by distance d, the capacitance is given by = ɛ 0A (5.4) d ylindrical apacitor In this geometry there are two coaxial cylinders where the radius of the inner conductor is a and the inner radius of the outer conductor is b. The length of the cylinders is L; we stipulate that L is large compared to b. For this geometry the capacitance is given by L = 2πɛ 0 ln(b/a) (5.5) Spherical apacitor In this geometry there are two concentric spheres where the radius of the inner sphere is a and the inner radius of the outer sphere is b. For this geometry the capacitance is given by: ab = 4πɛ 0 (5.6) b a apacitors in Parallel and in Series Parallel ombination: Fig. 5.2 shows a configuration where three capacitors are combined in parallel across the terminals of a battery. The battery gives a constant potential

3 5.1. THE IMPORTANT STUFF 73 + _ V Figure 5.2: Three capacitors are combined in parallel across a potential difference V (produced by a battery) _ V 3 Figure 5.3: Three capacitors are combined in series across a potential difference V (produced by a battery). difference V across the plates of each of the capacitors. The charges q 1, q 2 and q 3 which collect on the plates of the respective capacitors are not the same, but will be found from q 1 = c 1 V q 2 = 2 V q 3 = 3 V. The total charge on the plates, q = q 1 + q 2 + q 3 is related to the potential difference V by q = equiv V, where equiv is the equivalent capacitance of the combination. In general, the equivalent capacitance for a set of capacitors which are in parallel is given by equiv = i i Parallel (5.7) Series ombination: Fig. 5.3 shows a configuration where three capacitors are combined in series across the terminals of a battery. Here the charges which collect on the respective capacitor plates are the same (q) but the potential differences across the capacitors are different. These potential differences can be found from V 1 = q 1 V 2 = q 2 V 3 = q 3 where the individual potential differences add up to give the total: V 1 + V 2 + V 3 = V. In general, the effective capacitance for a set of capacitors which are in series is 1 equiv = i 1 i Series (5.8)

4 74 HAPTER 5. APAITANE AND DIELETRIS Energy Stored in a apacitor When we consider the work required to charge up a capacitor by moving a charge q from on plate to another we arrive at the potential energy U of the charges, which we can view as the energy stored in the electric field between the plates of the capacitor. This energy is: U = q2 2 = 1 2 V 2 (5.9) If we associate the energy in Eq. 5.9 with the region where there is any electric field, the interior of the capacitor (the field is effectively zero outside) then we arrive at an energy per unit volume for the electric field, i.e. an energy density, u. It is: u = 1 2 ɛ 0E 2 (5.10) This result also holds for any electric field, regardless of its source apacitors and Dielectrics If we fill the region between the plates of a capacitor with an insulating material the capacitance will be increased by some numerical factor κ: = κ air. (5.11) The number κ (which is unitless) is called the dielectric constant of the insulating material. 5.2 Worked Examples apacitance 1. Show that the two sets of units given for ɛ 0 in Eq. 5.3 are in fact the same. Start with the new units for ɛ 0, F. From Eq. 5.2 we substitute 1F = 1 so that m V 1 F m = 1 /V m = 1 V m Now use the definition of the volt from Eq. 4.4: 1V = 1J/ = 1N m/ to get 1 F = 1 2 m N m = 1 m So we arrive at the original units of ɛ 0 given in Eq N m 2 2. The capacitor shown in Fig. 5.4 has capacitance 25µF and is initially un-

5 5.2. WORKED EXAMPLES 75 S + _ Figure 5.4: Battery and capacitor for Example 2. r d Figure 5.5: apacitor described in Example 3. charged. The battery provides a potential difference of 120V. After switch S is closed, how much charge will pass through it? When the switch is closed, then the charge q which collects on the capacitor plates is given by q = V. Plugging in the given values for the capacitance and the potential difference V, we find: q = V = ( F)(120V) = = 3.0m This is the amount of charge which has been exchanged between the top and bottom plates of the capacitor. So 3.0m of charge has passed through the switch alculating apacitance 3. A parallel plate capacitor has circular plates of 8.2cm radius and 1.3mm separation. (a) alculate the capacitance. (b) What charge will appear on the plates if a potential difference of 120V is applied? (a) The capacitor is illustrated in Fig The area of the plates is A = πr 2 so that with

6 76 HAPTER 5. APAITANE AND DIELETRIS r = 8.2cm and d = 1.3mm and using Eq. 5.4 we get: = ɛ 0A d = ɛ 0πr 2 d = ( F m )π( m) 2 ( m) = F = 140pF (b) When a potential of 120V is applied to the plates of the capacitor the charge which appears on the plates is q = V = ( F)(120V) = = 17n 4. You have two flat metal plates, each of area 1.00m 2, with which to construct a parallel-plate capacitor. If the capacitance of the device is to be 1.00 F, what must be the separation between the plates? ould this capacitor actually be constructed? In Eq. 5.4 (formula for for a parallel-plate capacitor) we have and A. We can solve for the separation d: = ɛ 0A = d = ɛ 0A d Plug in the numbers: d = ( F m )(1.00m2 ) (1.00 F) = m This is an extremely tiny length if we are thinking about making an actual device, because the typical size of an atom is on the order of m. Our separation d is ten times smaller than that, so the atoms in the plates would not be truly separated! So a suitable capacitor could not be constructed. 5. A 2.0 µf spherical capacitor is composed of two metal spheres, one having a radius twice as large as the other. If the region between the spheres is a vacuum, determine the volume of this region. The capacitance of a ( air filled ) spherical capacitor is ab = 4πɛ 0 (b a). where a and b are the radii of the concentric spherical plates. Here we are given that b = 2a, so we then have: = 4πɛ 0 2a 2 a = 8πɛ 0a

7 5.2. WORKED EXAMPLES 77 + _ 1 V 3 2 Figure 5.6: onfiguration of capacitors for Example 6. We are given the value of so we can solve for a: a = 8πɛ 0 = ( F) 8π( F m ) = m (!) so that b = 2a = m. Then the volume of the enclosed region between the two plates is: V enc = 4 3 πb3 4 3 πa3 = 4 3 π((2a)3 a 3 ) = 4 3 π(7a3 ) = m apacitors in Parallel and in Series 6. In Fig. 5.6, find the equivalent capacitance of the combination. Assume that 1 = 10.0µF, 2 = 5.00µF, and 3 = 4.00µF The configuration given in the figure is that of a series combination of two capacitors ( 1 and 2 ) combined in parallel with a single capacitor ( 3 ). We can use the reduction formulae Eq. 5.8 and Eq. 5.7 to give a single equivalent capacitance. First combine the series capacitors with Eq The equivalent capacitance is: 1 equiv = µF µF = 0.300µF 1 = equiv = 3.33µF After this reduction, the configuration is as shown in Fig. 5.7(a). Now we have two capacitors in parallel. By Eq. 5.7 the equivalent capacitance is just the sum of the two values: equiv = 3.33µF µF = 7.33µF The final equivalent capacitance is shown in Fig. 5.7(b). The equivalent capacitance of the combination is 7.33µF. 7. How many 1.00µF capacitors must be connected in parallel to store a charge of 1.00 with a potential of 110V across the capacitors?

8 + _ V 3.33 mf 4.00 mf + _ V mf 78 HAPTER 5. APAITANE AND DIELETRIS (a) (b) Figure 5.7: (a) Series capacitors in previous figure have been combined as a single equivalent capacitor. (b) Parallel combination in (a) has been combined to give a single equivalent capacitor. n capacitors + _ V... Figure 5.8: n capacitors in parallel, for Example 7. In this problem we imagine a configuration like that shown in Fig. 5.8, where we have n capacitors with = 1.00µF connected in parallel across a potential difference of V = 110V. Since parallel capacitors simply add to give the equivalent capacitance (see Eq. 5.7) we have equiv = n, and the potential difference across the combination is related to the total charge q tot on the plates by q tot = equiv V = nv. We then use this to solve for n: n = q tot V = (1.00) ( F)(110V) = So one would need to hook up n = 9090 capacitors (!) to store the 1.00 of charge. 8. Each of the uncharged capacitors in Fig. 5.9 has a capacitance of 25.0µF. A potential difference of 4200 V is established when the switch is closed. How many coulombs of charge then pass through the meter A? The (total) charge which passes through the (current) meter A is the total charge which collects on the plates of the three capacitors. We note that for each capacitor the potential difference across the plates (after the switch is closed) is 4200V. So the charge on each capacitor is q = V = ( F)(4200V) = and the total charge is q Total = 3(0.105) =

9 5.2. WORKED EXAMPLES 79 A 4200 V Figure 5.9: onfiguration of capacitors for Example mf 3.0 mf a 20 mf b 6.0 mf Figure 5.10: ombination of capacitors for Example 9. So this is the amount of charge which passes through meter A. We could also note that the equivalent capacitance of the three parallel capacitors is equiv = 3(25.0µF) = 75.0µF and with 4200 V across the leads of the equivalent capacitance the total charge which collects on the plates is q Total = equiv V = ( F)(4200V) = Four capacitors are connected as shown in Fig (a) Find the equivalent capacitance between points a and b. (b) alculate the charge on each capacitor if V ab = 15V. (a) To get the equivalent capacitance of the set of capacitors between a and b: First note that the 15µF and 3.0µF capacitors are in series so they combine as: 1 = 1 equiv 15µF µF = 0.40µF 1 = equiv = 2.5µF After this reduction, the configuration is as shown in Fig. 5.11(a). The reduced circuit now

10 80 HAPTER 5. APAITANE AND DIELETRIS 2.5 mf a 20 mf b a 8.5 mf 20 mf b 6.0 mf (a) (b) Figure 5.11: (a) After reduction of the series pair. (b) After combining two parallel capacitances. has 2.5 µf and 6.0 µf capacitors in parallel which combine as: equiv = 2.5µF + 6.0µF = 8.5µF which gives us the combination shown in 5.11(b). Finally, the 8.5µF and 20µF capacitors in series reduce to: 1 = 1 equiv 8.5µF µF = 0.168µF 1 = equiv = 5.96µF so the equivalent capacitance between points a and b is 5.96µF. (b) Since the equivalent capacitance between a and b is 5.96 µf, the charge which collects on either end of the combination is Q = equiv V ab = ( F)(15V = This is the same as the charge on the far end of the 20µF capacitor (and thus on either plate of that capacitor), so we have the charge on that capacitor: Q 20µF = Now we can find the potential difference across the 20µF capacitor: V 20 µf = Q 20µF = ( ) 20µF ( F) = 4.47V With this value, we can find the potential difference between points a and c (see Fig. 5.12): V ac = 15.0V 4.47V = 10.5V This is now the potential difference across the 6.0µF capacitor, so we can find its charge: Q 6.0µF = ( F)(10.5V) =

11 5.2. WORKED EXAMPLES mf 3.0 mf V ab =15 V a c b 20 mf 6.0 mf Figure 5.12: Point c comes just before the 20 µf capacitor. Find V ac by subtracting V 20 µf from 15 V a S 1 a b (a) S 2 b (b) Figure 5.13: apacitor configuration for Example 10. (a) before switches are closed. (b) After switches are closed, charges redistribute on the plates of 1 and 2. Finally, we note that the potential difference across the 15µF 3.0µF series pair is also 10.5 V. Now, the equivalent capacitance of this pair was 2.50 µf, so that the charge which collects on each end of this combination is Q = equiv V = ( F)(10.5V) = But this is the same as the charge on the outer plates of the two capacitors, and that means that both capacitors have the same charge, namely: Q 15µF = Q 3.0µF = We now have the charges on all four of the capacitors. 10. In Fig. 5.13(a), the capacitances are 1 = 1.0µF and 2 = 3.0µF and both capacitors are charged to a potential difference of V = 100V but with opposite polarity as shown. Switches S 1 and S 2 are now closed. (a) What is now the potential difference between a and b? What are now the charges on capacitors (b) 1 and (c) 2?

12 82 HAPTER 5. APAITANE AND DIELETRIS Let s first find the charges which the capacitors had before the switch was closed. For 1 the magnitude of its charge was Q 1 = 1 V = (1.0µF)(100V) = What we mean here is that the upper plate of 1 had a charge of , because the polarity matters here! So the lower plate of 1 had a charge of For 2, the magnitude of its charge was Q 2 = 2 V = (3.0µF)(100V) = but here we mean that the upper plate of 2 had a charge of because of the polarity indicated in Fig. 5.13(a). So its lower plate had a charge of We note that the total charge on the upper plates is Q 1,upper + Q 1,upper = and the total charge on the lower plates is Now when the switches are closed the charges on the upper plates will redistribute themselves on the upper plates of 1 and 2. Lets call these new charges (on the upper plates) Q 1 and Q 2. We note that since the total charge on the upper plates was negative then it is a net negative charge which shifts around on the upper plates and Q 1 and Q 2 are both negative, as indicated in Fig. 5.13(b). By conservation of charge, the total is still equal to : Q 1 + Q 2 = Though we don t yet know the new potential difference across each capacitor, we do know that it is the same for both. Actually, we know that b must be at the higher potential; we will let the potential change in going from a to b be called V. Now, the potential for each capacitor is found from V = Q/; actually because of the polarities here (the Q s being negative) we need a minus sign, but the fact that the potential differences are the same across both capacitors gives: V = Q 1 1 = Q 2 2 = Q 2 = 2 1 Q 1 = Substituting this result into the previous one gives ( ) 3.0µF Q 1 1.0µF = 3.0Q 1 Q Q 1 = = Q 1 = Having solved for one of the unknowns, we re nearly finished! The change in potential as we go from a to b is then: = V = Q 1 1 = F = 50V (b) The magnitude of the new charge on capacitor 1 is Q 1 =

13 5.2. WORKED EXAMPLES 83 + _ 2 1 V 0 3 S Figure 5.14: onfiguration of capacitors and potential difference with switch for Example 11. (c) Using Q 2 = 3.0Q 1, the magnitude of the new charge on the second capacitor is Q 2 = 3.0 Q 1 = 3.0( ) = When switch S is thrown to the left in Fig. 5.14, the plates of capacitor 1 acquire a potential difference V 0. apacitors 2 and 3 are initially uncharged. The switch is now thrown to the right. What are the final charges q 1, q 2 and q 3 on the capacitors? Initially the only capacitor with a charge is 1, with a charge given by: q 1,init = 1 V 0 (5.12) since the potential across its plates is V 0. Now consider what happens when the switch is thrown to the right and the capacitors have charges q 1, q 2 and q 3. Since 2 and 3 are joined in series, their charges will be equal, so q 2 = q 3 and we only need to find q 2. Also, note that the upper plate of 1 is only connected to the upper plate of 2 so that q 1 and q 2 must add up to give the original charge on 1 : q 1 + q 2 = q 1,init (5.13) Finally, we note that the potential difference across 1 is equal to the potential difference across the 2-3 series combination. The equivalent capacitance of the 2-3 combination is: 1 = = = equiv = 2 3 equiv The potential across 1 is q 1 / 1, and the potential across the series pair is q 2 / equiv. So equating the potential differences gives q 1 = q ( ) = q 2 (5.14) 1 equiv 2 3 And that s all the equations we need; we can now solve for q 1 and q 2. Eq gives q 2 = q 1 (5.15)

14 84 HAPTER 5. APAITANE AND DIELETRIS and then substitute this and also Eq into Eq We get: Factor out q 1 on the left: Now we can isolate q 1 : ( ) 2 3 q 1 = q q 1 = 1 V 0 q 1 = Then go back and use Eq to get q 2 : ( ) 1 ( ) q 1 = 1 V 0 1 ( ) 2 1 ( ) V 0 q 2 = ( ) q 1 = ( ) = V ( ) ( ) V 0 Finally, we recall that q 3 = q 2. This gives us expressions for all three charges in terms of the initial parameters Energy Stored in a apacitor 12. How much energy is stored in one cubic meter of air due to the fair weather electric field of magnitude 150V/m? From Eq we have the energy density of an electric field. (As noted there, the source of the electric field is irrelevant.) We get: u = 1 2 ɛ 0E 2 = 1 2 ( )(150 V 2 N m 2 m )2 = J m 3 So in one cubic meter, J of energy are stored. 13. What capacitance is required to store an energy of 10kW h at a potential difference of 1000 V? First, convert the given energy to some sensible units! ( ) 3600s E = 10kW h = J (1h) s 1h = J

15 + _ 300 V 2.0 mf 4.0 mf + _ 300 V 6.0 mf 5.2. WORKED EXAMPLES 85 (a) (b) Figure 5.15: (a) apacitor configuration for Example 14. (b) Equivalent capacitor. Then use Eq. 5.9 for the energy stored in a capacitor: Plug in the numbers: A capacitance of 72F (big!) is needed. E = 1 2 V 2 = = 2E V 2 = 2( J) (1000V) 2 = 72F 14. Two capacitors, of 2.0 and 4.0 µf capacitance, are connected in parallel across a 300V potential difference. alculate the total energy stored in the capacitors. The capacitors and potential difference are diagrammed in Fig. 5.15(a). For the purpose of finding the total energy in the capacitors we can replace the two parallel capacitors with a single equivalent capacitor of value 6.0 µf (the original two were in parallel, so we sum the values). This is because the charge which collects on the equivalent capacitor is the sum of charges on the plates of the original two capacitors. Then the energy stored is E = 1 2 V 2 = 1 2 ( F)(300V) 2 = 0.27J

16 86 HAPTER 5. APAITANE AND DIELETRIS

17 Appendix A: Useful Numbers onversion Factors Length cm meter km in ft mi 1cm = m = km = in = ft = mi = Mass g kg slug u 1g = kg = slug = u = An object with a weight of 1lb has a mass of kg. onstants: m electron m proton e = = esu 12 2 ɛ 0 = N m 2 k = 1/(4πɛ 0 ) = N m2 2 µ 0 = 4π 10 7 N = N A 2 A 2 = kg N A = kg c = m s = mol 1 87

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