Chapter 26 Capacitance Probem Set #5 ue: Ch 26 2, 3, 5, 7, 9, 5, 22, 26, 29, 6, 63, 64 The ieas of energy storage in fies can be carrie a step further by unerstaning the concept of "Capacitance." Lecture Outine. The Definition of Capacitance 2. Capacitors in Circuits 3. nergy Storage in Capacitors an ectric Fies 4. Dieectrics in Capacitors. The Definition of Capacitance R Consier a sphere with a tota charge,, an a raius, R. From previous probems we know that the potentia at the surface is, = k R. Putting more charge on the sphere stores more energy, but the ratio of energy or potentia to charge epens ony on R, not on or. That is, = R k. It's true for a charge objects that the ratio of potentia to votage epens ony on the shape, so this ratio is efine as the capacitance. C The Defintion of Capacitance The units of capacitance are Couomb ot Fara F. xampe : Cacuate the capacitance of two equa but oppositey charge pates of area,, an separation,. Negect any ege effects. The potentia ifference between the pates is, = r r s. The fie between the pates is just the sum of the fies ue to the iniviua pates (see Ch 23 exampe 9), r = r r = σ k ˆ σ k ˆ = σ k ˆ = q 2ε o 2ε o ε o ε o Using r s = z ˆ = k ˆ. k, the votage on the capacitor can be written as, q k ˆ z ˆ q k = ε o ε o z = q q z = ε o ε o. ppying the efinition of capacitance, C = q q ε o C = ε o. Note that the capacitance ony epens on the shape. q q z 26
xampe 2: Fin the capacitance of two concentric cyinrica conuctors of raii a an b with a ength,. Show that the resut is consistent with exampe. ssume the cyiners have equa an opposite charges,. Then the potentia ifference between them is, = r r s where r = 2kλ ˆ r = 2k r r ˆ r from exampe 7 of chapter 24. Using r s = r ˆ r, the votage on the capacitor can be written as, = 2k r ˆ 2k b r r ˆ r = r r = 2k a n b a = 2k n b a. Using the efinition of capacitance, C = 2k n b C = 2k n b. a a gain the resut epens ony on geometry. When a an b get very arge the concentric cyiners ook ike parae pates. The istance between the pates is = b a. In terms of a an, b a = a = a a. Now the capacitance can be written, C = 2k n( a ). In the imit a an is sma, the Tayor expansion of the ogarithm can be use, n δ In this imit, C = 2k a ( ) δ δ 2 = 4πε o a 2 ( ) n a = ε o 2πa ( ) a ( 2a ) a. = ε o as expecte. b a b a 2. Capacitors in Circuits Bring some capacitors C C 2 CN Capacitors in Series: By the Law of Conservation of nergy the sum of the votages on the capacitors must equa the appie votage. = 2 L N Using the efinition of capacitance, = 2 L N C s C C 2 C N The Law of Conservation of Charge requires a the charges to be equa, = = 2 =L= N. = L = L = C s C C 2 C N C s C C 2 C N C s Ci 262
C s = C i The ition of Capacitors in Series Capacitors in Parae: By the Law of Conservation of Charge the sum of the charges on the capacitors must equa the suppie charge. C C 2 = 2 L N Using the efinition of capacitance, C p = C C 2 2 LC N N The Law of Conservation of nergy requires a the votages to be equa, = = 2 =L= N. C p = C C 2 LC N C p = C C 2 LC N C p = C i CN C p = C i The ition of Capacitors in Parae xampe 3: Fin (a)the equivaent capacitance, (b)the charge on each capacitor an (c)the potentia ifference for each capacitor in the circuit shown. Given =.50, C =4.00µF, C 2 =8.00µF, an C 3 =6.00µF. (a)c an C 2 are in parae so C p = C C 2 = 2.0µF. Now we can imagine C p in series with C 3 giving a tota capacitance of = C = C p C 3 = 4.00µF C s C p C 3 C p C 3 (b)&(c)using the efinition of capacitance we can fin the tota charge = C = (4µF)(.5) = 6.00µC. This must equa the charge on C 3 by the Law of Conservation of Charge, 3 =6.00µC. Now the votage on C 3 must be C 3 = 3 = 6µC =. ccoring to the Law of C 3 6µF Conservation of nergy that eaves 0.5 on C an C 2. Now the charge on C is, = C = (4µF)(0.5) = 2.00µC an the charge on C 2 is, 2 = C 2 2 = (8µF)(0.5) = 4.00µC. Note that 3 = 2. In summary, (µc) C(µF) () 2.00 4.00 0.500 4.00 8.00 0.500 6.00 6.00.00 C C 2 C 3 263
3. nergy Storage in Capacitors an ectric Fies q q q Suppose we are trying to put a tota charge on a capacitor. How much energy wi it take? ssume at some point the charge on the capacitor is q an the potentia ifference is, using the eectric potentia energy we can fin the energy neee to a a sma amount of charge, q U = q U = q. Using the efinition of capacitance U = q q C. U = 2 To fin the tota energy to charge the capacitor from q=0 to q= integrate, 2 C = 2 C 2 U U = q q 0 U = 0 C 2 Store nergy in Capacitors 2 C = 2 C 2. xampe 4: parae pate capacitor of area,, an pate separation,, remains connecte to a battery as the pates are pue apart unti they are separate by 2. Fin the change in store energy. The initia store energy is U o = 2 C o o 2 where C o is the initia capacitance, C o = ε o, an o is the initia potentia ifference ue to the battery. The fina potentia energy is U = 2 C o 2 where C is the fina capacitance, C = ε o 2. The battery keeps the potentia ifference constant. The change in store energy is, U U U o = 2 C o 2 2 C o o 2 = 2 ε o 2 o 2 2 ε o How can the energy rop when work is one to pu the pates apart? o 2 = 4 ε o o 2 2 We can attribute this energy to the fie, instea of to the capacitor. Using the capacitance of parae pates an the reationship between the fie an the potentia, U = 2 C 2 = 2 ε o ( ) 2 = 2 ε o 2 = 2 ε o 2 vo. In terms of the energy ensity, u U vo = 2 ε o 2 Store nergy in ectric Fies This expression turns out to be true for a fies. 264
4. Dieectrics in Capacitors Dieectrics are insuators. ectrons are not free to fow from one moecue to another. The atoms in a ieectric can have ipoe moments. In a typica chunk of ieectric materia these ipoes are ranomy aigne an therefor prouce no net fie as shown. σo σo o When a ieectric is pace between the pates of a capacitor with a surface charge ensity σ o the resuting eectric fie, o, tens to aign the ipoes with the fie. This resuts in a net charge ensity σ i inuce on the surfaces of the ieectric which in turns creates an inuce eectric fie, i, in the opposite irection to the appie fie. The tota fie insie the ieectric is reuce to, = o i The ieectric constant is efine as the ratio of the appie fie to the tota fie, κ o. Substituting for an soving for the inuce fie, κ o o i = κ i ( ) o. Note that κ= is a perfect insuator such as a vacuum an κ= is a perfect conuctor. σi i σi How oes the introuction of a ieectric affect the capacitance of a capacitor? Reca the cacuation of the capacitance of parae pates starts with the cacuation of the potentia ifference, = r r r s = o κ r s = r κ o r s = κ o. The potentia ifference wi be smaer by a factor of κ. ppying the efinition of capacitance, C = = κ = κc o. The capacitance is arger by a κ o o factor of κ. C = κc o Capacitors with Dieectrics 265
xampe 5: parae pate capacitor with 00cm 2 area an 2.00mm pate separation is connecte to a 0.0 battery. Fin the capacitance, charge, eectric fie an store energy before an after it is isconnecte from the battery an pace in oi of ieectric constant 5.00. Using the capacitance of parae pates, C o = ε o = ( ) 0.000 8.85x0 2.00222 = 44.3pF. Using the efinition of capacitance, C o o o = C o = 443pC. The fie can be foun from the votage, o = o x = o = 0.0 0.00200 = 5000 m. ( ) 2 The energy in a capacitor is, U o = 2 o = 443x0 2 2 C o 2 44.3x0 2 = 2.2x0 9 J. The new capacitance with the ieectric is, C = κc o = 5.00 44.3 = 22pF. The charge remains the same because the battery is isconnecte, = o = 443pC. The new fie is smaer by a factor of κ, = o κ = 5000 5.00 = 000 m. The new energy is, U = 2 2 C = ( 443x0 2 ) 2 2 22x0 2 = 0.443x0 9 J. Where oes the energy go? xampe 6: Repeat exampe 5 assuming the battery remains connecte. The new capacitance with the ieectric is sti, C = κc o = 5.00 44.3 = 22pF. This time the votage remains the same but the charge changes, = C = (22) (0.0) = 220pC. Since the votage is the same, the fie must be the same, = o = 5000 m. 2 The new energy is, U = 2 C = 2 Where oes the energy come from? Chapter 26 Summary The Definition of Capacitance C The Capacitance of Parae Pates C = ε o The ition of Capacitors in Series C s = The ition of Capacitors in Parae C p = Store nergy in Capacitors U = 2 2 ( 220x0 2 ) 2 22x0 2 =.x0 9 J. Ci C i C = 2 C 2 Store nergy ectric Fies u U vo = 2 ε o 2 Capacitors with Dieectrics C = κc o 266