Chapter 24 Capacitance

hapter 4 apacitance onceptual Problems If the voltage across a parallel-plate capacitor is ouble, its capacitance (a) oubles (b) rops by half (c) remains the same. Determine the oncept The capacitance of a parallel-plate capacitor is a function of the surface area of its plates, the separation of these plates, an the electrical properties of the matter between them. The capacitance is, therefore, inepenent of the voltage across the capacitor. (c) is correct. If the charge on an isolate spherical conuctor is ouble, its selfcapacitance (a) oubles (b) rops by half (c) remains the same. Determine the oncept The capacitance of an isolate spherical capacitor is given by 4π R, where R is its raius. The capacitance is, therefore, inepenent of the charge of the capacitor. (c) is correct. True or false: The electrostatic energy ensity is uniformly istribute in the region between the conuctors of a cylinrical capacitor. Determine the oncept False. The electrostatic energy ensity is not uniformly istribute because the magnitue of the electric fiel strength is not uniformly istribute, 4 If the istance between the plates of a charge an isolate parallelplate capacitor is ouble, what is the ratio of the final store energy to the initial store energy? Determine the oncept The energy store in the electric fiel of a parallel-plate capacitor is relate to the potential ifference across the capacitor by. If is constant, is irectly proportional to an oubling oubles. Hence the ratio of the initial store energy to the final store energy is. 5 [SSM] parallel-plate capacitor is connecte to a battery. The space between the two plates is empty. If the separation between the capacitor plates is triple while the capacitor remains connecte to the battery, what is the ratio of the final store energy to the initial store energy? Determine the oncept The energy store in a capacitor is given by an the capacitance of a parallel-plate capacitor by. We can 87

88 hapter 4 combine these relationships, using the efinition of capacitance an the conition that the potential ifference across the capacitor is constant, to express as a function of. Express the energy store in the capacitor: se the efinition of capacitance to express the charge of the capacitor: Express the capacitance of a parallel-plate capacitor in terms of the separation of its plates: () where is the area of one plate. Substituting for an in uation () yiels: Because, tripling the separation of the plates will reuce the energy store in the capacitor to one-thir its previous value. Hence the ratio of the final store energy to the initial store energy is /. 6 If the capacitor of Problem 5 is isconnecte from the battery before the separation between the plates is triple, what is the ratio of the final store energy to the initial store energy? Picture the Problem Let represent the initial potential ifference between the plates, the energy store in the capacitor initially, the initial separation of the plates, an,, an these physical quantities when the plate separation has been triple. We can use to relate the energy store in the capacitor to the potential ifference across it an E to relate the potential ifference to the separation of the plates. Express the energy store in the capacitor before the tripling of the separation of the plates:

apacitance 89 Express the energy store in the capacitor after the tripling of the separation of the plates: ' ' because the charge on the plates oes not change. Express the ratio of to an simplify to obtain: ' ' ' The potential ifferences across the capacitor plates before an after the plate separation, in terms of the electric fiel E between the plates, are given by: Substituting for an to obtain: For : E an ' E' because E epens solely on the charge on the plates an, as observe above, the charge oes not change uring the separation process. ' E' E ' ' The ratio of the final store energy to the initial store energy is. 7 True or false: (a) (b) The uivalent capacitance of two capacitors in parallel is always greater than the larger of the two capacitance values. The uivalent capacitance of two capacitors in series is always less than the least of the two capacitance values if the charges on the two plates that are connecte by an otherwise isolate conuctor sum to zero. (a) True. The uivalent capacitance of two capacitors in parallel is the sum of the iniviual capacitances. (b) True. The uivalent capacitance of two capacitors in series is the reciprocal of the sum of the reciprocals of the iniviual capacitances. 8 Two uncharge capacitors have capacitances an, respectively, an are connecte in series. This series combination is then connecte across the terminals a battery. Which of the following is true?

9 hapter 4 (a) The capacitor has twice the charge of the other capacitor. (b) The voltage across each capacitor is the same. (c) The energy store by each capacitor is the same. () The uivalent capacitance is. (e) The uivalent capacitance is /. (a) False. apacitors connecte in series carry the same charge. (b) False. The voltage across a capacitor whose capacitance is is / an the voltage across the secon capacitor is /( ). (c) False. The energy store in a capacitor is given by. () False. This woul be the uivalent capacitance if they were connecte in parallel. (e) True. Taking the reciprocal of the sum of the reciprocals of an yiels /. 9 [SSM] ielectric is inserte between the plates of a parallel-plate capacitor, completely filling the region between the plates. ir initially fille the region between the two plates. The capacitor was connecte to a battery uring the entire process. True or false: (a) (b) (c) () The capacitance value of the capacitor increases as the ielectric is inserte between the plates. The charge on the capacitor plates ecreases as the ielectric is inserte between the plates. The electric fiel between the plates oes not change as the ielectric is inserte between the plates. The energy storage of the capacitor ecreases as the ielectric is inserte between the plates. Determine the oncept The capacitance of the capacitor is given by, the charge on the capacitor is given by, an the energy store in the capacitor is given by. (a) True. s the ielectric material is inserte, increases from (air) to its value for the given ielectric material. (b) False. Because, an increases, must increase.

apacitance 9 (c) True. E /, where is the plate separation. () False. The energy storage of a capacitor is inepenent of the presence of ielectric an is given by. apacitors an B (Figure 4-) have ientical plate areas an gap separations. The space between the plates of each capacitor is half-fille with a ielectric as shown. Which has the larger capacitance, capacitor or capacitor B? Explain your answer. Picture the Problem We can treat configuration as two capacitors in parallel an configuration B as two capacitors in series. Fining the uivalent capacitance of each configuration an examining their ratio will allow us to ecie whether or B has the greater capacitance. In both cases, we ll let be the capacitance of the ielectric-fille capacitor an be the capacitance of the air capacitor. In configuration we have: Express an : Substitute for an an simplify to obtain: an ( ) In configuration B we have: B b Express an : an

9 hapter 4 Substitute for an an simplify to obtain: B ( ) Divie B by B 4 ( ) ( ) 4 Because ( ) < for > : > B [SSM] (a) Two ientical capacitors are connecte in parallel. This combination is then connecte across the terminals of a battery. How oes the total energy store in the parallel combination of the two capacitors compare to the total energy store if just one of the capacitors were connecte across the terminals of the same battery? (b) Two ientical capacitors that have been ischarge are connecte in series. This combination is then connecte across the terminals of a battery. How oes the total energy store in the series combination of the two capacitors compare to the total energy store if just one of the capacitors were connecte across the terminals of the same battery? Picture the Problem The energy store in a capacitor whose capacitance is an across which there is a potential ifference is given by. Let represent the capacitance of the each of the two ientical capacitors. (a) The energy store in the parallel system is given by: parallel When the capacitors are connecte in parallel, their uivalent capacitance is: parallel

apacitance 9 Substituting for yiels: an simplifying parallel ( ) () If just one capacitor is connecte to the same battery the store energy is: Diviing uation () by uation () an simplifying yiels: () or capacitor parallel capacitor parallel capacitor (b) The energy store in the series system is given by: series When the capacitors are connecte in series, their uivalent capacitance is: series Substituting for yiels: an simplifying ( ) () series 4 Diviing uation () by uation series 4 () an simplifying yiels: or capacitor series capacitor Two ientical capacitors that have been ischarge are connecte in series across the terminals of a - battery. When only one of the capacitors is connecte across the terminals of this battery, the energy store is. What is the total energy store in the two capacitors when the series combination is connecte to the battery? (a) 4, (b), (c), () /, (e) /4 Picture the Problem We can use the expression to express the ratio of the energy store in the single capacitor an in the ientical-capacitors-in-series combination. Express the energy store in the capacitors when they are connecte to the - battery:

94 hapter 4 Express the uivalent capacitance of the two ientical capacitors connecte in series: Substitute for ( ) to obtain: 4 Express the energy store in one capacitor when it is connecte to the - battery: Express the ratio of to : 4 () is correct. Estimation an pproximation [SSM] Disconnect the coaxial cable from a television or other evice an estimate the iameter of the inner conuctor an the iameter of the shiel. ssume a plausible value (see Table 4 ) for the ielectric constant of the ielectric separating the two conuctors an estimate the capacitance per unit length of the cable. Picture the Problem The outer iameter of a "typical" coaxial cable is about 5 mm, while the inner iameter is about mm. From Table 4- we see that a reasonable range of values for is -5. We can use the expression for the capacitance of a cylinrical capacitor to estimate the capacitance per unit length of a coaxial cable. The capacitance of a cylinrical ielectric-fille capacitor is given by: π L R ln R where L is the length of the capacitor, R is the raius of the inner conuctor, an R is the raius of the secon (outer) conuctor. Divie both sies by L to obtain an expression for the capacitance per unit length of the cable: L π R ln R R k ln R

apacitance 95 If : L 9 ( N m / ) 8.988.5mm ln.5mm.nf/m If 5: L 9 ( N m / ) 8.988 5.5mm ln.5mm. nf/m reasonable range of values for /L, corresponing to 5, is:.nf/m L. nf/m 4 You are part of an engineering research team that is esigning a pulse nitrogen laser. To create the high-energy ensities neee to operate such a laser, the electrical ischarge from a high-voltage capacitor is use. Typically, the energy ruirement per pulse (i.e., per ischarge) is J. Estimate the capacitance ruire if the ischarge is to creates a spark across a gap of about. cm. ssume that the ielectric breakown of nitrogen is the same as the value for normal air. Picture the Problem The energy store in a capacitor is given by. Relate the energy store in a capacitor to its capacitance an the potential ifference across it: The potential ifference across the spark gap is relate to the with of the gap an the electric fiel E in the gap: E Substitute for in the expression for to obtain: E Substitute numerical values an evaluate : ( ) J 6 ( /m) (.cm) μf

96 hapter 4 5 [SSM] Estimate the capacitance of the Leyen jar shown in the Figure 4-4. The figure of a man is one-tenth the height of an average man. Picture the Problem Moeling the Leyen jar as a parallel-plate capacitor, we can use the uation for the capacitance of a ielectric-fille parallel-plate capacitor that relates its capacitance to the area of its plates an their separation (the thickness of the glass) to estimate the capacitance of the jar. See Table 4- for the ielectric constants of various materials. The capacitance of a ielectric-fille parallel-plate capacitor is given by: Let the plate area be the sum of the area of the lateral surface of the jar an its base: Substitute for an simplify to obtain: where is the ielectric constant. lateral base π Rh πr area where h is the height of the jar an R is its insie raius. π R ( πrh πr ) ( h R) If the glass of the Leyen jar is Bakelite of thickness. mm an the raius an height of the jar are 4. cm an 4 cm, respectively, then: π N m (.9)( 4. cm) 8.854 [ ( 4 cm) 4. cm] 4. mm. nf apacitance 6 n isolate conucting sphere that has a. cm raius has an electric potential of. k (the potential far from the sphere is zero). (a) How much charge is on the sphere? (b) What is the self-capacitance of the sphere? (c) By how much oes the self-capacitance change if the sphere s electric potential is increase to 6. k? Picture the Problem The charge on the spherical conuctor is relate to its raius an potential accoring to k/r an we can use the efinition of capacitance to fin the self-capacitance of the sphere.

apacitance 97 (a) Relate the potential of the spherical conuctor to the charge on it an to its raius: Substitute numerical values an evaluate : k r r k (.cm)(. k) 8.988.n 9 N m.5 n (b) se the efinition of capacitance to relate the self-capacitance of the sphere to its charge an potential:.5 n.k.pf (c) It oesn t. The self-capacitance of a sphere is a function of its raius. 7 The charge on one plate of a capacitor is. μ an the charge on the other plate is. μ. The potential ifference between the plates is 4. What is the capacitance of the capacitor? Picture the Problem We can use its efinition to fin the capacitance of this capacitor. se the efinition of capacitance to obtain:.μ 4 75. nf 8 Two isolate conucting spheres of ual raius R have charges an, respectively. Their centers are separate by a istance that is large compare to their raius. Estimate the capacitance of this unusual capacitor. Picture the Problem Let the separation of the spheres be an their raii be R. Outsie the two spheres the electric fiel is approximately the fiel ue to point charges of an, each locate at the centers of spheres, separate by istance. We can erive an expression for the potential at the surface of each sphere an then use the potential ifference between the spheres an the efinition of capacitance an to estimate the capacitance of the two-sphere system. The capacitance of the two-sphere system is given by: The potential at any point outsie the two spheres is: Δ where Δ is the potential ifference between the spheres. k ( ) k( ) r r

98 hapter 4 For a point on the surface of the sphere with charge : where r an r are the istances from the given point to the centers of the spheres. r R an r δ where δ < R Substitute to obtain: k( ) k( ) R δ For δ << : an k R k R k k The potential ifference between the spheres is: Substitute for Δ in the expression for to obtain: Δ k k k k R R k R k R π R R π R For >> R: π R The Storage of Electrical Energy 9 [SSM] (a) The potential ifference between the plates of a.-μf capacitor is. How much energy is store in the capacitor? (b) How much aitional energy is ruire to increase the potential ifference between the plates from to? Picture the Problem Of the three uivalent expressions for the energy store in a charge capacitor, the one that relates to an is.

apacitance 99 (a) Express the energy store in the capacitor as a function of an : Substitute numerical values an evaluate : (. F)( ) 5.mJ μ (b) Express the aitional energy ruire as the ifference between the energy store in the capacitor at an the energy store at : Δ ( ) ( ) (. μf)( ) 45. mj 5. mj The charges on the plates of a -μf capacitor are ±4. μ. (a) How much energy is store in the capacitor? (b) If charge is transferre until the charges on the plates are ual to ±. μ, how much store energy remains? Picture the Problem Of the three uivalent expressions for the energy store in a charge capacitor, the one that relates to an is. (a) Express the energy store in the capacitor as a function of an : Substitute numerical values an evaluate : ( 4. μ) μf.8μj (b) Express the energy remaining ( ) when half the charge is remove: (. μ). μj μf (a) Fin the energy store in a.-nf capacitor when to the charges on the plates are ±5. μ. (b) How much aitional energy is store if charges are increase from ±5. μ to ±. μ? Picture the Problem Of the three uivalent expressions for the energy store in a charge capacitor, the one that relates to an is. (a) Express the energy store in the capacitor as a function of an :

hapter 4 Substitute numerical values an ( 5.μ) evaluate : (b) Express the aitional energy ruire as the ifference between the energy store in the capacitor when its charge is 5 μ an when its charge is μ: Δ ( 5. μ). nf (. μ) ( 5. μ) (. μ). nf.65 mj.5 mj.65 mj.65mj.88 mj What is the maximum electric energy ensity in a region containing ry air at stanar conitions? Picture the Problem The energy per unit volume in an electric fiel varies with the square of the electric fiel accoring tou E. ner stanar conitions, ielectric breakown occurs at approximately E. M/m. Express the energy per unit volume in an electric fiel: u E Substitute numerical values an evaluate u: u 8.854 4J/m N m M. m [SSM] n air-gap parallel-plate capacitor that has a plate area of. m an a separation of. mm is charge to. (a) What is the electric fiel between the plates? (b) What is the electric energy ensity between the plates? (c) Fin the total energy by multiplying your answer from Part (b) by the volume between the plates. () Determine the capacitance of this arrangement. (e) alculate the total energy from, an compare your answer with your result from Part (c). Picture the Problem Knowing the potential ifference between the plates, we can use E / to fin the electric fiel between them. The energy per unit volume is given by u E an we can fin the capacitance of the parallelplate capacitor using. (a) Express the electric fiel between the plates in terms of their separation an the potential ifference between them: E. mm k/m

apacitance (b) Express the energy per unit volume in an electric fiel: u E Substitute numerical values an evaluate u: u 8.854 44.7 mj/m N m 44.mJ/m ( k/m) (c) The total energy is given by: u u ( 44.7 mj/m )(.m )(. mm) 88.5μJ () The capacitance of a parallelplate capacitor is given by: Substitute numerical values an 8.854 (. m ) evaluate : N m. mm 7.7 nf 7.7 nf (e) The total energy is given by: Substitute numerical values an evaluate : ( 7.7nF)( ) 88.5μJ, in agreement with ( c). 4 soli metal sphere has raius of. cm an a concentric metal spherical shell has an insie raius of.5 cm. The soli sphere has a charge 5. n. (a) Estimate the energy store in the electric fiel in the region between the spheres. Hint: You can treat the spheres essentially as parallel flat slabs separate by.5 cm. (b) Estimate the capacitance of this two-sphere system. (c) Estimate the total energy store in the electric fiel from / an compare it to your answer in Part (a). Picture the Problem The total energy store in the electric fiel is the prouct of the energy ensity in the space between the spheres an the volume of this space. (a) The total energy store in the electric fiel is given by: u where u is the energy ensity an is the volume between the spheres.

hapter 4 The energy ensity of the fiel is: u E where E is the fiel between the spheres. The volume between the spheres is approximately: ( r ) 4π r r Substitute for u an to obtain: E r ( r r ) π () The magnitue of the electric fiel between the concentric spheres is the sum of the electric fiels ue to each charge istribution: Because the two surfaces are so close together, the electric fiel between them is approximately the sum of the fiels ue to two plane charge istributions: E E E σ σ σ E Substitute for σ to obtain: E 4π r Substitute for E in uation () an simplify: π r ( r r ) Substitute numerical values an evaluate : r 8π 4 r π r r (.n)(.5cm.cm) 5 9 8π 8.854 N m (.cm) 56. J.6μJ

apacitance (b) The capacitance of the twosphere system is given by: Δ where Δ is the potential ifference between the two spheres. The electric potentials at the surfaces an of the spheres are: 4π r 4π r Substitute for Δ an simplify to r r 4π obtain: r r 4π r 4π r Substitute numerical values an evaluate : 4π 8.854 N m (.cm)(.5cm).7 nf.5cm.cm.nf (c) se to fin the total energy store in the electric fiel between the spheres: ( 5. n).5 μj.7 nf a result that agrees to within 5% with the exact result obtaine in (a). 5 parallel-plate capacitor has plates of area 5 cm an is connecte across the terminals of a battery. fter some time has passe, the capacitor is isconnecte from the battery. When the plates are then move.4 cm farther apart, the charge on each plate remains constant but the potential ifference between the plates increases by. (a) What is the magnitue of the charge on each plate? (b) Do you expect the energy store in the capacitor to increase, ecrease, or remain constant as the plates are move this way? Explain your answer. (c) Support your answer to Part (b), by etermining the change in store energy in the capacitor ue to the movement of the plates. Picture the Problem (a) We can relate the charge on the positive plate of the capacitor to the charge ensity of the plate σ using its efinition. The charge ensity, in turn, is relate to the electric fiel between the plates accoring to σ E an the electric fiel can be foun from E Δ/Δ. We can use Δ Δ in Part (c) to fin the increase in the energy store ue to the movement of the plates.

4 hapter 4 (a) Express the charge on the positive plate of the capacitor in terms of the plate s charge ensity σ an surface area : σ Relate σ to the electric fiel E between the plates of the capacitor: σ E Express E in terms of the change in as the plates are separate a istance Δ: E Δ Δ Substitute for σ an E to obtain: E Δ Δ Substitute numerical values an evaluate : ( 8.854 /N m )( 5cm ).n n.4 cm (b) Because work has to be one to pull the plates farther apart, you woul expect the energy store in the capacitor to increase. (c) Express the change in the electrostatic energy in terms of the change in the potential ifference: Substitute numerical values an evaluate Δ: Δ Δ (.n)( ).55 J Δ μ ombinations of apacitors 6 (a) How many.-μf capacitors connecte in parallel woul it take to store a total charge of. m if the potential ifference of across each capacitor is.? Diagram the parallel combination. (b) What woul be the potential ifference across this parallel combination? (c) If the capacitors in Part (a) are ischarge, connecte in series, an then energize until the potential ifference across each is ual to., fin the charge on each capacitor an the potential ifference across the connection.

apacitance 5 Picture the Problem We can apply the properties of capacitors connecte in parallel to etermine the number of.-μf capacitors connecte in parallel it woul take to store a total charge of. m with a potential ifference of. across each capacitor. Knowing that the capacitors are connecte in parallel (Parts (a) an (b)) we etermine the potential ifference across the combination. In Part (c) we can use our knowlege of how potential ifferences a in a series circuit to fin the potential ifference across the combination an the efinition of capacitance to fin the charge on each capacitor. 99 (a) Express the number of capacitors n in terms of the charge q on each an the total charge : Relate the charge q on one capacitor to its capacitance an the potential ifference across it: Substitute for q to obtain: n q q n Substitute numerical values an n. m evaluate n: (. μf)(. ) (b) Because the capacitors are connecte in parallel the potential ifference across the combination is the same as the potential ifference across each of them: parallel combination.

6 hapter 4 (c) With the capacitors connecte in series, the potential ifference across the combination will be the sum of the potential ifferences across the capacitors: series combination ( ).. k se the efinition of capacitance to fin the charge on each capacitor: q ( μ F)( ).μ 7.-μF capacitor an a 6.-μF capacitor are ischarge an then connecte in series, an the series combination is then connecte in parallel with an 8.-μF capacitor. Diagram this combination. What is the uivalent capacitance of this combination? Picture the Problem The capacitor array is shown in the iagram. We can fin the uivalent capacitance of this combination by first fining the uivalent capacitance of the.-μf an 6.-μF capacitors in series an then the uivalent capacitance of this capacitor with the 8.-μF capacitor in parallel. Express the uivalent capacitance for the.-μf an 6.-μF capacitors in series: Solve for 6 : Fin the uivalent capacitance of a.-μf capacitor in parallel with an 8.-μF capacitor:. µf 6. µf. μ F 8. µf 6. 6 μf 6. μf 8. μ F 8. μf. μf 8 Three capacitors are connecte in a triangle as shown in Figure 4-5. Fin an expression for the uivalent capacitance between points a an c in terms of the three capacitance values. Picture the Problem Because we re intereste in the uivalent capacitance across terminals a an c, we nee to recognize that capacitors an are in series with each other an in parallel with capacitor.

apacitance 7 Fin the uivalent capacitance of an in series: Fin the uivalent capacitance of an in parallel: 9.-μF capacitor an a.-μf capacitor are connecte in parallel across the terminals of a 6.- battery. (a) What is the uivalent capacitance of this combination? (b) What is the potential ifference across each capacitor? (c) Fin the charge on each capacitor. () Fin the energy store in each capacitor. Picture the Problem Because the capacitors are connecte in parallel we can a their capacitances to fin the uivalent capacitance of the combination. lso, because they are in parallel, they have a common potential ifference across them. We can use the efinition of capacitance to fin the charge on each capacitor. (a) Fin the uivalent capacitance of the two capacitors in parallel:. μ F. μf. μf (b) Because capacitors in parallel have a common potential ifference across them: 6. (c) se the efinition of capacitance to fin the charge on each capacitor: () se to fin the energy store in each capacitor: 6. μ an μ 8 μj an 6 μj (. μf)( 6.) (. μf)( 6.) ( 6. μ)( 6. ) (.μ)( 6. ).-μf capacitor an a.-μf capacitor are connecte in parallel across the terminals of a 6.- battery. (a) What is the uivalent capacitance of this combination? (b) What is the potential ifference across each capacitor?

8 hapter 4 (c) Fin the charge on each capacitor. () Fin the energy store in each capacitor. Picture the Problem We can use the properties of capacitors in series to fin the uivalent capacitance an the charge on each capacitor. We can then apply the efinition of capacitance to fin the potential ifference across each capacitor. (a) Because the capacitors are connecte in series they have ual charges: Express the uivalent capacitance of the two capacitors in series:.μ F.μF Solve for to obtain: (. μf)(. μf). μf. μf 6.67 μf (b) Because the capacitors are in series, they have the same charge. Substitute numerical values to obtain: 4. μ ( 6.67 μf)( 6.) (c) pply the efinition of capacitance to fin the potential ifference across each capacitor: 4. μ. μf an 4. μ. μf 4.. () se to fin the energy store in each capacitor: 8. μj an 4. μj ( 4. μ)( 4. ) ( 4. μ)(. ) Three ientical capacitors are connecte so that their maximum uivalent capacitance, which is 5. μf, is obtaine. (a) Determine how the capacitors are connecte an iagram the combination. (b) There are three aitional ways to connect all three capacitors. Diagram these three ways an etermine the uivalent capacitances for each arrangement.

apacitance 9 Picture the Problem We can use the properties of capacitors connecte in series an in parallel to fin the uivalent capacitances for various connection combinations. (a) If their capacitance is to be a maximum, the capacitors must be connecte in parallel: Fin the capacitance of each capacitor: 5. μf 5. μf (b) () onnect the three capacitors in series: Because the capacitors are in series, their uivalent capacitance is the reciprocal of the sum of their reciprocals: 5. μf 5. μf 5. μf an.67 μf () onnect two in parallel, with the thir in series with that combination: Fin the uivalent capacitance of the two capacitors that are in parallel an then the uivalent capacitance of the network of three capacitors: in, parallel two an ( 5. μ F).μF. μ F 5. μf Solving for yiels:.μf () onnect two in series, with the thir in parallel with that combination:

hapter 4 Fin the uivalent capacitance of the two capacitors connecte in series: or in, series two, in two series 5.μ F 5.μF.5 μf Fin the capacitance uivalent to.5 μf an 5. μf in parallel:.5μ F 5.μF 7.5μF For the circuit shown in Figure 4-6, the capacitors were each ischarge before being connecte to the voltage source. Fin (a) the uivalent capacitance of the combination, (b) the charge store on the positively charge plate of each capacitor, (c) the voltage across each capacitor, an () the energy store in each capacitor. Picture the Problem We can use the properties of capacitors connecte in series an in parallel to fin the uivalent capacitance between the terminals an these properties an the efinition of capacitance to fin the charge on each capacitor. (a) Relate the uivalent capacitance of the two capacitors in series to their iniviual capacitances: Solving for 45 yiels: 4. μ F 5. 4 5 μ 4 5.58μF F Fin the uivalent capacitance of 45 in parallel with the.-μf capacitor: (b) se the efinition of capacitance to fin the charge store on the -μf capacitor: Because the capacitors in series have the same charge: 4.58μF. μf 5.6 μf 5. μf μ.4 m 5 4 5 μ (. F)( ) (.58 F)( ).66 m.6 m (c) Because the.-μf capacitor is connecte irectly across the source, the voltage across it is:

apacitance se the efinition of capacitance to fin 4 an 5 : () se to fin the energy store in each capacitor:.66 m 4. μf 4 4 4 an.66 m 5. μf 5 5 5 4 5 4 4 49.9 mj 5 5. mj an 4 mj 58 4 (.66 m)( 58 ) (.66 m)( 4 ) (.4 m)( ) (a) Show that the uivalent capacitance of two capacitors in series can be written (b) sing only this formula an some algebra, show that must always be less than an, an hence must be less than the smaller of the two values. (c) Show that the uivalent capacitance of three capacitors in series is can be written () sing only this formula an some algebra, show that must always be less than each of, an, an hence must be less than the least of the three values. Picture the Problem We can use the properties of capacitors in series to establish the results calle for in this problem. (a) Express the uivalent capacitance of two capacitors in series: Solve for by taking the reciprocal of both sies of the uation to obtain:

hapter 4 (b) Divie numerator an enominator of this expression by to obtain: Because > < : Divie numerator an enominator of this expression by to obtain: Because > : <, showing that must always be less than both an, an hence must be less than the smaller of the two values. (c) sing our result from part (a) for two of the capacitors, a a thir capacitor in series to obtain: Take the reciprocal of both sies of the uation to obtain: () Rewrite the result of Part (c) as follows: Divie numerator an enominator of this expression by to obtain: Because > < :

apacitance Procee similarly to show that: < an <, showing that must always be less than, an, an hence must be less than the smaller of the three values. 4 For the circuit shown in Figure 4-7 fin (a) the uivalent capacitance between the terminals, (b) the charge store on the positively charge plate of each capacitor, (c) the voltage across each capacitor, an () the total store energy. Picture the Problem Let represent the uivalent capacitance of the parallel combination an the total uivalent capacitance between the terminals. We can use the uations for capacitors in parallel an then in series to fin. Because the charge on is the same as on the.-μf capacitor an we ll know the charge on the.-μf capacitor when we have foun the total charge store by the circuit. We can fin the charges on the.-μf an.5-μf capacitors by first fining the potential ifference across them an then using the efinition of capacitance.,.µf. µf.µf.5 µf, (a) Fin the uivalent capacitance for the parallel combination: The.-μF capacitor is in series with. Their uivalent capacitance is given by: an. μ F.5 μf.5μf. μ F.5μF.494 μ F.4 μf (b) Express the total charge store..5.) by the circuit : (.494 μf )(.494μ

4 hapter 4 The.-μF an.5-μf capacitors, being in parallel, have a common potential ifference. Express this potential ifference in terms of the. across the system an the potential ifference across the.-μf capacitor:.5......494μ..μf.95 sing the efinition of capacitance, fin the charge on the.-μf an.5-μf capacitors: Because the voltage across the parallel combination of the.-μf an.5-μf capacitors is.95, the voltage across the.-μf capacitor is 8.65 an: (c) From (b), the voltages across the.-μf capacitor an the parallel combination of the.-μf an.5-μf capacitors is:. an.5....9μ. an..5.5.48μ...4μ 8.6.5.9 (.μf)(.95) (.5μF)(.95) (.μf)( 8.65). 8.6 () The total store energy is given by: Substitute numerical values an evaluate : (.49 μ F)(.). J μ 5 Five ientical capacitors of capacitance are connecte in a so-calle brige network, as shown in Figure 4-8. (a) What is the uivalent capacitance between points a an b? (b) Fin the uivalent capacitance between points a an b if the capacitor at the center is replace by a capacitor that has a capacitance of. Picture the Problem Note that there are three parallel paths between a an b. We can fin the uivalent capacitance of the capacitors connecte in series in the upper an lower branches an then fin the uivalent capacitance of three

apacitance 5 capacitors in parallel. (a) Fin the uivalent capacitance of the series combination of capacitors in the upper an lower branch: Now we have two capacitors with capacitance in parallel with a capacitor whose capacitance is. Fin their uivalent capacitance: ' (b) If the central capacitance is, then: ' 6 You an your laboratory team have been given a project by your electrical engineering professor. Your team must esign a network of capacitors that has a uivalent capacitance of. μf an breakown voltage of 4. The restriction is that your team must use only.-μf capacitors that have iniviual breakown voltages of. Diagram the combination. Picture the Problem Place four of the capacitors in series. Then the potential across each is when the potential across the combination is 4. The uivalent capacitance of the series combination is.5 μf. If we place four such series combinations in parallel, as shown in the circuit iagram, the total capacitance between the terminals is. μf. 7 Fin the ifferent uivalent capacitances that can be obtaine by using two or three of the following capacitors: a.-μf capacitor, a.-μf capacitor, an a 4.-μF capacitor. Picture the Problem We can connect two capacitors in parallel, all three in parallel, two in series, three in series, two in parallel in series with the thir, an two in series in parallel with the thir.

6 hapter 4 onnect in parallel to obtain:. μ F.μF. μf or. μ F 4.μF 5. μf or. μ F 4.μF 6.μF onnect obtain: all three in parallel to.μf.μf 4.μF 7. μf onnect two in series: (. μf)(. μf) or or. μf. μf (. μf)( 4. μf). μf 4. μf (. μf)( 4. μf). μf 4. μf.667 μf.8 μf.μf onnecting all three in series yiels: (. μf)(. μf)( 4. μf) (. μf)(. μf) (. μf)( 4. μf) (. μf)( 4. μf).57μf onnect two in parallel, an the parallel combination in series with the thir: or or ( 4. μf)(. μf. μf). μf. μf 4. μf.7μf (. μf)( 4. μf. μf). μf. μf 4. μf.857 μf (. μf)( 4. μf. μf). μf. μf 4. μf.4μf

onnect two in series, an the series combination in parallel with the thir: or or (. μf)(. μf) 4.67 μf apacitance 7 4. μf. μf. μf ( 4. μf)(. μf). μf 4. μf. μf.μf (. μf)( 4. μf). μf. μf 4. μf.8 μf 8 What is the uivalent capacitance (in terms of which is the capacitance of one of the capacitors) of the infinite laer of capacitors shown in Figure 4-9? Picture the Problem Let be the capacitance of each capacitor in the laer an let be the uivalent capacitance of the infinite laer less the series capacitor in the first rung. Because the capacitance is finite an non-zero, aing one more stage to the laer will not change the capacitance of the network. The capacitance of the two capacitor combination shown to the right is the uivalent of the infinite laer, so it has capacitance also. The uivalent capacitance of the parallel combination of an The uivalent capacitance of the series combination of an ( ) is, so: Simply this expression to obtain a quaratic uation in : is:

8 hapter 4 Solving for the positive value of yiels: 5. 68 Parallel-Plate apacitors 9 parallel-plate capacitor has a capacitance of. μf an a plate separation of.6 mm. (a) What is the maximum potential ifference between the plates, so that ielectric breakown of the air between the plates oes not occur? (b) How much charge is store at this potential ifference? Picture the Problem The potential ifference across a parallel-plate capacitor, the electric fiel E between its plates, an the separation of the plates are relate accoring to E. We can use this relationship to fin max corresponing to ielectric breakown an the efinition of capacitance to fin the maximum charge on the capacitor. (a) Express the potential ifference across the plates of the capacitor in terms of the electric fiel between the plates E an their separation : E correspons to : max E (.M/m)(.6mm) max max 4.8k (b) sing the efinition of capacitance, fin the charge store at this maximum potential ifference: μ max 9.6m (. F)( 4.8k) 4 n electric fiel of. 4 /m exists between the circular plates of a parallel-plate capacitor that has a plate separation of. mm. (a) What is the potential ifference across the capacitor plates? (b) What plate raius is ruire if the positively charge plate is to have a charge of. μ? Picture the Problem The potential ifference across a parallel-plate capacitor, the electric fiel E between its plates, an the separation of the plates are relate accoring to E. In Part (b) we can use the efinition of capacitance an the expression for the capacitance of a parallel-plate capacitor to fin the ruire plate raius.

apacitance 9 (a) Express the potential ifference across the plates of the capacitor in terms of the electric fiel between the plates E an their separation : E Substitute numerical values an evaluate : 4 (. /m)(.mm) 4. (b) se the efinition of capacitance to relate the capacitance of the capacitor to its charge an the potential ifference across it: The capacitance of a parallel-plate capacitor is given by: π R where R is the raius of the circular plates. Equate these two expressions for : π R R π Substitute numerical values an evaluate R: R 4.4m (.μ)(.mm) π 8.854 N m ( 4.) 4 parallel-plate, air-gap capacitor has a capacitance of.4 μf. The plates are.5 mm apart. (a) What is the area of each plate? (b) What is the potential ifference between the plates if the positively charge plate has a charge of. μ? (c) What is the store energy? () What is the maximum energy this capacitor can store before ielectric breakown of the air between the plates occurs? Picture the Problem We can use the expression for the capacitance of a parallelplate capacitor to fin the area of each plate an the efinition of capacitance to fin the potential ifference when the capacitor is charge to. μ. We can fin the store energy using an the efinition of capacitance an the relationship between the potential ifference across a parallel-plate capacitor an the electric fiel between its plates to fin the charge at which ielectric breakown occurs. Recall that E max, air. M/m.

hapter 4 (a) The capacitance of a parallelplate capacitor is given by: Substitute numerical values an evaluate : (.4μF)(.5 mm) 8.854 7.9m N m 7.96m (b) sing the efinition of capacitance, fin the potential ifference across the capacitor when it is charge to. μ:.μ.9.4 μf (c) Express the store energy as a function of the capacitor s capacitance an the potential ifference across it: Substitute numerical values an evaluate : (.4 μf)(.9) 7 μj 6.7 μj () The maximum energy this capacitor can store before ielectric breakown of the air between the plates occurs is given by: Relate the maximum potential ifference to the maximum electric fiel between the plates: max max E max max Substituting for max yiels: max Emax Substitute numerical values an evaluate max : (.4 F)(.5mm) (. M/m).6 J μ max 4 Design a.-μf parallel-plate capacitor that has air between its plates an that can be charge to a maximum potential ifference of before ielectric breakown occurs. (a) What is the minimum possible separation

apacitance between the plates? (b) What minimum area must each plate of the capacitor have? Picture the Problem The potential ifference across the capacitor plates is relate to their separation an the electric fiel between them accoring to E. We can use this uation with E max. M/m to fin min. In Part (b) we can use the expression for the capacitance of a parallel-plate capacitor to fin the ruire area of the plates. (a) se the relationship between the potential ifference across the plates an the electric fiel between them to fin the minimum separation of the plates: min E max.mm. M/m (b) The capacitance of a parallelplate capacitor is given by: Substitute numerical values an evaluate : (.μf)(.mm) 8.854.76m - /N m ylinrical apacitors 4 In preparation for an experiment that you will o in your introuctory nuclear physics lab, you are shown the insie of a Geiger tube. You measure the raius an the length of the central wire of the Geiger tube to be. mm an. cm, respectively. The outer surface of the tube is a conucing cylinrical shell that has an inner raius of.5 cm. The shell is coaxial with the wire an has the same length (. cm). alculate (a) the capacitance of your tube, assuming that the gas in the tube has a ielectric constant of., an (b) the value of the linear charge ensity on the wire when the potential ifference between the wire an shell is. k? Picture the Problem The capacitance of a cylinrical capacitor is given by π L ln( R R ) where L is its length an R an R the raii of the inner an outer conuctors. (a) The capacitance of the coaxial cylinrical shell is given by: π L R ln R

hapter 4 Substitute numerical values an evaluate : π (.)( 8.854 /N m )(. m) ln.5 cm. mm.546 pf.55 pf (b) se the efinition of capacitance to express the charge per unit length: Substitute numerical values an evaluate λ: λ L L λ (.546 pf)(. k). m 5.5n/m 44 cylinrical capacitor consists of a long wire that has a raius R, a length L an a charge. The wire is enclose by a coaxial outer cylinrical shell that has a inner raius R, length L, an charge. (a) Fin expressions for the electric fiel an energy ensity as a function of the istance R from the axis. (b) How much energy resies in a region between the conuctors that has a raius R, a thickness R, an a volume πrl R? (c) Integrate your expression from Part (b) to fin the total energy store in the capacitor. ompare your result with that obtaine by using the formula ( ) in conjunction with the known expression for the capacitance of a cylinrical capacitor. Picture the Problem The iagram shows a partial cross-sectional view of the inner wire an the outer cylinrical shell. By symmetry, the electric fiel is raial in the space between the wire an the concentric cylinrical shell. We can apply Gauss s law to cylinrical surfaces of raii R < R, R < R < R, an R > R to fin the electric fiel an, hence, the energy ensity in these regions. R R R (a) pply Gauss s law to a cylinrical surface of raius R < R an length L to obtain: E insie ( π RL) R < R an E R<R Because E for R < R : u R<R

apacitance pply Gauss s law to a cylinrical surface of raius R < R < R an length L to obtain: E ( π RL) λl insie R < R< R whereλ is the linear charge ensity. Solve for E R < R< R to obtain: E λ π R R < R< R k RL The energy ensity in the region R < R < R is given by: u R < R< R ER < R< R Substituting for simplifying yiels: E R < R< R an u R < R< R kλ R k R L k RL pply Gauss s law to a cylinrical surface of raius R > R an length L to obtain: E insie ( π RL) R > R an E R>R Because E for R > R : u R>R (b) Express the energy resiing in a cylinrical shell between the conuctors of raius R, thickness R, an volume π RL R: π RLu ( R) k π RL R L k R RL R r (c) Integrate from R R to R R to obtain: R k R L R R k L R ln R

4 hapter 4 se obtain: an the expression for the capacitance of a cylinrical capacitor to k R ln L R π L R ln R in agreement with the result from Part (b). 45 Three concentric, thin long conucting cylinrical shells have raii of. mm, 5. mm, an 8. mm. The space between the shells is fille with air. The innermost an outermost shells are connecte at one en by a conucting wire. Fin the capacitance per unit length of this configuration. Picture the Problem Note that with the innermost an outermost cyliners connecte together the system correspons to two cylinrical capacitors π L connecte in parallel. We can use to express the capacitance per ln R R ( ) unit length an then calculate an a the capacitances per unit length of each of the cylinrical shell capacitors. Relate the capacitance of a cylinrical capacitor to its length L an inner an outer raii R an R : Divie both sies of the uation by L to express the capacitance per unit length: Express the capacitance per unit length of the cylinrical system: π L ln L L ( R R ) π ln ( R R ) L outer L inner () Substitute numerical values an evaluate the capacitance per unit length of the outer cylinrical shell combination: L outer π ( 8.854 /N m )(.) ln(.8cm.5cm) 8.4 pf/m

apacitance 5 Substitute numerical values an evaluate capacitance per unit length of the inner cylinrical shell combination: L inner π ( 8.854 /N m )(.) ln(.5cm.cm) 6.7 pf/m Substituting numerical results in uation () yiels: L 8.4 pf/m 6.7 pf/m 79 pf/m 46 goniometer is a precise instrument for measuring angles. capacitive goniometer is shown in Figure 4-4a. Each plate of the variable capacitor (Figure 4-4b) consists of a flat metal semicircle that has an inner raius R an an outer raius R. The plates share a common rotation axis, an the with of the air gap separating the plates is. alculate the capacitance as a function of the angle θ an the parameters given. Picture the Problem We can use the expression for the capacitance of a parallel-plate capacitor of variable area an the geometry of the figure to express the capacitance of the goniometer. The capacitance of the parallel-plate capacitor is given by: The area of the plates is: If the top plate rotates through an angle Δθ, then the area is reuce by: π Δ π ( Δ) θ ( R R ) ( R R ) π θ Δθ Δ ( R R ) ( R R ) π θ

6 hapter 4 Substitute for an Δ in the expression for an simplify to obtain: θ ( R R ) ( R R ) ( R R ) ( θ Δ θ ) Δθ 47 capacitive pressure gauge is shown in Figure 4-4. Each plate has an area. The plates are separate by a material that has a ielectric constant, a thickness, an a Young s moulus Y. If a pressure increase of ΔP is applie to the plates, erive an expression for the change in capacitance. Picture the Problem Let be the capacitance of the capacitor when the pressure is P an be the capacitance when the pressure is P ΔP. We ll assume that (a) the change in the thickness of the plates is small, an (b) the total volume of material between the plates is conserve. We can use the expression for the capacitance of a ielectric-fille parallel-plate capacitor an the efinition of Young s moulus to express the change in the capacitance Δ of the given capacitor when the pressure on its plates is increase by ΔP. Express the change in capacitance resulting from the ecrease in separation of the capacitor plates by Δ: Because the volume is constant: Δ ' ' Δ '' or ' ' Δ Substitute for in the expression for Δ an simplify to obtain: Δ Δ Δ ( Δ ) ( Δ ) ( Δ ) From the efinition of Young s moulus: Δ P Y Δ P Y

apacitance 7 Substitute for Δ in the expression for Δ to obtain: Δ P Y P Y Expan obtain: P Y binomially to P Y P Y P Y... Provie P << Y: P Y P Y Substitute in the expression for Δ an simplify to obtain: P P Δ Y Y Spherical apacitors 48 Moel Earth as a conucting sphere. (a) What is its self-capacitance? (b) ssume the magnitue of the electric fiel at Earth s surface is 5 /m. What charge ensity oes this correspon to? Express this value in funamental charge units e per square centimeter Picture the Problem (a) We can use the efinition of capacitance an the expression for the electric potential at the surface of Earth to fin Earth s selfcapacitance. In Part (b) we can use E σ to fin Earth s surface charge ensity. (a) The self-capacitance of Earth is given by: k Because where R is the R raius of Earth: where is the charge on Earth an is the potential at its surface. k R R k

8 hapter 4 Substitute numerical values an evaluate : 67 km 9 N m 8.988.79 mf.787 mf (b) The electric fiel at the surface of Earth is relate to Earth s charge ensity: σ E σ E Substitute numerical values an evaluate σ : σ n m e 8.854 5.8 9 89 e cm N m m m cm.6 49 [SSM] spherical capacitor consists of a thin spherical shell that has a raius R an a thin, concentric spherical shell that has a raius R, where R > R. (a) Show that the capacitance is given by 4π R R /(R R ). (b) Show that when the raii of the shells are nearly ual, the capacitance approximately is given by the expression for the capacitance of a parallel-plate capacitor, /, where is the area of the sphere an R R. Picture the Problem We can use the efinition of capacitance an the expression for the potential ifference between charge concentric spherical shells to show that π R R ( R ). 4 R (a) sing its efinition, relate the capacitance of the concentric spherical shells to their charge an the potential ifference between their surfaces: Express the potential ifference between the conuctors: R R k k R R RR