University Physics (PHY 2326)
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1 Chapter 23 University Physics (PHY 2326) Lecture 5 Electrostatics Electrical energy potential difference and electric potential potential energy of charged conductors Capacitance and capacitors 3/26/2015 1
2 24 The definition of capacitance Capacitor: two conductors (separated by an insulator) usually oppositely charged a +Q b -Q The capacitance, C, of a capacitor is defined as a ratio of the magnitude of a charge on either conductor to the magnitude of the potential difference between the conductors Q C V 3/26/2015 2
3 Sources of potential: Battery --sustained -- 3/26/2015 3
4 Capacitors charge separation on capacitor Apply to capacitor E capcitor = - Q e 0 A (constant) = ( Q e 0 A )d 3/26/2015 where C = e 0 A d 4
5 The parallel-plate capacitor The capacitance of a device depends on the geometric arrangement of the conductors C e 0 A d d +Q -Q A A where A is the area of one of the plates, d is the separation, e 0 is a constant called the permittivity of free space, k e 1 4e 0 e 0 = C 2 /N m 2 3/26/2015 5
6 Problem: parallel-plate capacitor A parallel plate capacitor has plates 2.00 m 2 in area, separated by a distance of 5.00 mm. A potential difference of 10,000 V is applied across the capacitor. Determine the capacitance the charge on each plate 3/26/2015 6
7 A parallel plate capacitor has plates 2.00 m 2 in area, separated by a distance of 5.00 mm. A potential difference of 10,000 V is applied across the capacitor. Determine the capacitance the charge on each plate Solution: Given: V=10,000 V A = 2.00 m 2 d = 5.00 mm Find: C=? Q=? Since we are dealing with the parallel-plate capacitor, the capacitance can be found as A C e C N m d F 3.54 nf m Once the capacitance is known, the charge can be found from the definition of a capacitance via charge and potential difference: 9 5 Q CV F 10000V C m 3/26/2015 7
8 Spherical capacitors 3/26/2015 8
9 Cylindrical capacitors 3/26/2015 9
10 Energy stored in a capacitor The potential energy stored in a capacitor is U = Q 2 /2C = 1/2 CV 2 = 1/2 QV. The capacitor energy is stored in the electric field between the plates. The energy density is u = 1/2 e 0 E 2. The Z machine shown below can produce up to W using capacitors in parallel! 3/26/
11 Dielectrics A dielectric is a nonconducting material. Most capacitors have dielectric between their plates. (See Figure at upper right.) The dielectric constant of the material is K = C/C 0 > 1. Dielectric increases the capacitance and the energy density by a factor K. 3/26/
12 Molecular model of induced charge - II Figure below shows polarization of the dielectric and how the induced charges reduce the magnitude of the resultant electric field. 3/26/
13 Dielectric breakdown If the electric field is strong enough, dielectric breakdown occurs and the dielectric becomes a conductor. The dielectric strength is the maximum electric field the material can withstand before breakdown occurs. Table 24.2 shows the dielectric strength of some insulators. 3/26/
14 Circuit geometry --generic circuit elements --imagine battery connection parallel series 3/26/
15 Capacitors: series equivalent Q 1 = Q 2 = Q equivalent Q 1 C 1 + Q 2 C 2 = Q equivalent 1 C C 2 = C equivalent 1 C equivalent 3/26/
16 Capacitors: parallel equivalent Q 1 + Q 2 = Q equivalent C 1 + C 2 = C equivalent 3/26/
17 Capacitors series parallel V V 1 +V 2 = V eq V 1 =V 2 Q Q 1 =Q 2 Q 1 +Q 2 = Q eq 1 C eq = 1 C C 2 + C eq = C 1 + C 2 + 3/26/
18 Problem --Find the charge on (Q) and potential difference (V) across each capacitor. --What is the total energy stored in the system? 12 V 5 mf 2 mf 8 mf 2 mf 3/26/
19 Problem (ans) C equ = C 2 + C 8 (parallel) = 2mF +8mF 12 V 5 mf 2 mf 8 mf =10mF 2 mf 1 = C equ C 5 C 10 C 2 (series) 12 V 5 mf 2 mf 10 µf = 1 5mF mF + 1 2mF = 8 10mF Þ C equ = 5mF 4 12 V 5/4 µf 3/26/
20 Problem (ans) 12 V 5/4 µf Q 5 = Q 10 = Q 2 = Q tot =15mC (in series) 12 V 5 mf 10 µf 2 mf 12 V 5 mf 2 mf 8 mf 2 mf 3/26/
21 Dielectrics change the potential difference The potential between to parallel plates of a capacitor changes when the material between the plates changes. It does not matter if the plates are rolled into a tube or if they are flat. K = C C 0 V = V 0 K E = E 0 K 3/26/
22 Field lines as dielectrics change Moving from part (a) to part (b) of Figure shows the change induced by the dielectric. In vacuum, energy density is u = 1 2 e 0 E 2 In dielectric e = Ke 0 C = Ke 0 A d = e A d u = 1 2 ee 2 3/26/
23 Examples to consider, capacitors with and without dielectrics If capacitor is disconnected from circuit, inserting a dielectric changes decreases electric field, potential and increases capacitance, but the amount of charge on the capacitor is unchanged. If the capacitor is hooked up to a power supply with constant voltage, the voltage must remain the same, but capacitance and charge increase 3/26/
24 Q24.8 You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the charges on the plates remain constant. What effect does adding the dielectric have on the potential difference between the capacitor plates? A. The potential difference increases. B. The potential difference remains the same. C. The potential difference decreases. D. not enough information given to decide 3/26/
25 A24.8 You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the charges on the plates remain constant. What effect does adding the dielectric have on the potential difference between the capacitor plates? A. The potential difference increases. B. The potential difference remains the same. C. The potential difference decreases. D. not enough information given to decide 3/26/
26 Q24.9 You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the charges on the plates remain constant. What effect does adding the dielectric have on the energy stored in the capacitor? A. The stored energy increases. B. The stored energy remains the same. C. The stored energy decreases. D. not enough information given to decide 3/26/
27 A24.9 You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the charges on the plates remain constant. What effect does adding the dielectric have on the energy stored in the capacitor? A. The stored energy increases. B. The stored energy remains the same. C. The stored energy decreases. D. not enough information given to decide 3/26/
28 Q24.10 You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the potential difference between the plates remains constant. What effect does adding the dielectric have on the amount of charge on each of the capacitor plates? A. The amount of charge increases. B. The amount of charge remains the same. C. The amount of charge decreases. D. not enough information given to decide 3/26/
29 A24.10 You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the potential difference between the plates remains constant. What effect does adding the dielectric have on the amount of charge on each of the capacitor plates? A. The amount of charge increases. B. The amount of charge remains the same. C. The amount of charge decreases. D. not enough information given to decide 3/26/
30 Q24.11 You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the potential difference between the plates remains constant. What effect does adding the dielectric have on the energy stored in the capacitor? A. The stored energy increases. B. The stored energy remains the same. C. The stored energy decreases. D. not enough information given to decide 3/26/
31 A24.11 You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the potential difference between the plates remains constant. What effect does adding the dielectric have on the energy stored in the capacitor? A. The stored energy increases. B. The stored energy remains the same. C. The stored energy decreases. D. not enough information given to decide 3/26/
32 1. A capacitor is basically two parallel conducting plates with insulating material in between. The capacitor doesn t have to look like metal plates. 2. When a capacitor is connected to an external potential, charges flow onto the plates and create a potential difference between the plates. 3. Capacitors in circuits 3/26/2015 symbols analysis follow from conservation of energy conservation of charge Capacitor for use in high-performance audio systems
33 Units of capacitance The unit of C is the farad (F), but most capacitors have values of C ranging from picofarads to microfarads (pf to F). 1F 1C V Recall, micro 10-6, nano 10-9, pico If the external potential is disconnected, charges remain on the plates, so capacitors are good for storing charge (and energy). 3/26/
34 Combinations of capacitors It is very often that more than one capacitor is used in an electric circuit We would have to learn how to compute the equivalent capacitance of certain combinations of capacitors C 1 C 2 C 1 C 5 C 3 C 3 C 2 C 4 3/26/
35 a. Parallel combination a Connecting a battery to the parallel combination of capacitors is equivalent to introducing the same potential difference for both capacitors, C 1 +Q 1 C 2 +Q 2 V V V 1 2 V=V ab b Q 1 Q 2 A total charge transferred to the system from the battery is the sum of charges of the two capacitors, Q Q Q 1 2 By definition, Thus, C eq would be 3/26/2015 Q C V Q C V C eq eq eq Q Q Q Q Q Q Q V V V V V V C C C Q Q Q 1 2 V V V
36 Parallel combination: notes Analogous formula is true for any number of capacitors, C C C C (parallel combination) eq It follows that the equivalent capacitance of a parallel combination of capacitors is greater than any of the individual capacitors 3/26/
37 Problem: parallel combination of capacitors A 3 F capacitor and a 6 F capacitor are connected in parallel across an 18 V battery. Determine the equivalent capacitance and total charge deposited. 3/26/
38 A 3 F capacitor and a 6 F capacitor are connected in parallel across an 18 V battery. Determine the equivalent capacitance and total charge deposited. a Given: C 1 +Q 1 C 2 +Q 2 V = 18 V C 1 = 3 F C 2 = 6 F V=V ab b Q 1 Q 2 Find: C eq =? Q=? First determine equivalent capacitance of C 1 and C 2 : C C C F Next, determine the charge 6 4 Q CV 910 F 18V C 3/26/
39 b. Series combination a C 1 +Q 1 Connecting a battery to the serial combination of capacitors is equivalent to introducing the same charge for both capacitors, Q Q Q 1 2 V=V ab b C 2 c Q 1 +Q 2 Q 2 A voltage induced in the system from the battery is the sum of potential differences across the individual capacitors, V V V 1 2 By definition, Thus, C eq would be 3/26/2015 Q C V Q C V V V V V V V V C Q Q Q Q Q Q eq C C C C C C eq eq Q Q Q 1 2 V V V
40 Series combination: notes Analogous formula is true for any number of capacitors, (series combination) C C C C eq It follows that the equivalent capacitance of a series combination of capacitors is always less than any of the individual capacitance in the combination 3/26/
41 Problem: series combination of capacitors A 3 F capacitor and a 6 F capacitor are connected in series across an 18 V battery. Determine the equivalent capacitance. 3/26/
42 A 3 F capacitor and a 6 F capacitor are connected in series across an 18 V battery. Determine the equivalent capacitance and total charge deposited. a Given: V = 18 V C 1 = 3 F C 2 = 6 F V=V ab b C 1 C 2 c +Q 1 Q 1 +Q 2 Q 2 Find: C eq =? Q=? First determine equivalent capacitance of C 1 and C 2 : C eq CC 1 2 C C 1 2 Next, determine the charge 2 F 6 5 Q CV 210 F 18V C 3/26/
43 Capacitor q -q Potential difference=v Any two conductors separated by either an insulator or vacuum for a capacitor The charge of a capacitor is the absolute value of the charge on one of conductors. q constant This constant is called the capacitance and V is geometry dependent. It is the capacity for holding charge at a constant voltage 43
44 Units 1 Farad=1 F= 1 C/V Symbol: Indicates positive potential 44
45 Interesting Fact When a capacitor has reached full charge, q, then it is often useful to think of the capacitor as a battery which supplies EMF to the circuit. 45
46 Simple Circuit +q H -q L Initially, After S is H closed, & L =0 H=+q L=-q S i -i 46
47 Recalling Displacement Current Maxwell thought of the capacitor as a flow device, like a resistor so a displacement current would flow between the plates of the capacitor like this i +q -q i d i This plate induces a negative charge here Which means the positive charge carriers are moving here and thus a positive current moving to the right 47
48 If conductors had area, A Then current density would be J d =i d /A 48
49 Calculating Capacitance Calculate the E-field in terms of charge and geometrical conditions Calculate the voltage by integrating the E-field. You now have V=q*something and since q=cv then 1/something=capacitance 49
50 Parallel plates of area A and distance, D, apart V V V q E da e q EA e f i ED and qd Ae 1 Ae 0 qd D Ae 0 0 E E ds 0 0 q C E C q Ae 0 E ds q Ae 0 ED Area, A Distance=D
51 51 Coaxial Cable Inner conductor of radius a and thin outer conductor radius b +q -q a b L C C a b L a b L a b L q V r dr L q drr E ds E V rl q E q rl E q da E b a f i ln 2 ln 2 ln ln 2 2 ˆ e e e e e e e e
52 52 Spherical Conductor Inner conductor radius A and thin outer conductor of radius B C A B AB AB A B AB A B q A B q V r dr q drr E ds E V r q E q r E q da E B A f i ˆ 4 4 e e e e e e e e
53 Isolated Sphere of radius A V q 4e 0 1 B 1 A Let B V q 4e 0 A C 4e 0 A 53
54 Capacitors in Parallel i E C 1 C2 i 1 i 2 i 3 E q 1 q 2 C 3 q 3 i=i 1 +i 2 +i 3 implies q=q 1 +q 2 +q 3 E i C eq C eq V C V 1 C V 2 C V 3 q C eq C 1 C 2 C 3 54
55 55 Capacitors in Series E C 1 i i q q q C 2 C 3 By the loop rule, E=V 1 +V 2 +V 3 E C eq C C C C C q C q C q C q C q C q C q E C q E eq eq eq
56 Energy Stored in Capacitors V d d Work Work charge Vdq q And V C q Work dq C Or 1 2C q 2 Technically, this is the potential to do work or potential energy, U U=1/2 CV 2 or U=1/2 q 2 /C Work 1 2 CV 2 Recall Spring s Potential Energy U=1/2 kx 2 56
57 Energy Density, u u=energy/volume Assume parallel plates at right Vol=AD U=1/2 CV 2 e 0 A C D 1 e 0 A 2 u V 2AD D 2 1 V 1 u e 0 e 2 0E 2 D 2 2 Area, A Distance=D Volume wherein energy resides
58 Dielectrics Area, A Distance=D Insulator Voltage at which the insulating material allows current flow ( break down ) is called the breakdown voltage 1 cm of dry air has a breakdown voltage of 30 kv (wet air less) 58
59 The capacitance is said to increase because we can put more voltage (or charge) on the capacitor before breakdown. The dielectric strength of vacuum is 1 Dry air is So we can replace, our old capacitance, C air, by a capacitance based on the dielectric strength, k, which is C new =k*c air An example is the white dielectric material in coaxial cable, typically polyethylene (k=2.25) or polyurethane (k=3.4) Dielectric strength is dependent on the frequency of the electric field 59
60 Induced Charge and Polarization in Dielectrics Note that the charges have separated or polarized E Total =E 0 -E i E i E 0 E E 0 E0 k e 0 e 0 i k e e 1 i 1 k
61 Permittivity of the Dielectric eke 0 For real materials, we define a D-field where D=ke 0 E For these same materials, there can be a magnetization based on the magnetic susceptibility, c, : H= c 0 B D da H D ds q i free free enclosed D da enclosed t D 61
62 Capacitor Rule For a move through a capacitor in the direction of current, the change in potential is q/c If the move opposes the current then the change in potential is +q/c. move V a -V b = -q/c V a -V b = +q/c V a i V b 62
63 RC Circuits Initially, S is open so at t=0, i=0 in the resistor, and the charge on the capacitor is 0. Recall that i=dq/dt V B A S R C 63
64 Switch to A Start at S (loop clockwise) and use the loop rule q ir V 0 C q V ir C dq q V R dt C V B A S R C 64
65 65 An Asatz A guess of the solution ) (1 ) ( , : 0 RC t RC p RC t RC t RC t p RC t p e CV t q CV B B CV Be A q q t at CV A or C A V C Be A e C B V C q dt dq R V e RC B dt dq Be A q ansatz My
66 Ramifications of Charge At t=0, q(0)=cv-cv=0 At t=, q=cv (indicating fully charged) What is the current between t=0 and the time when the capacitor is fully charged? i i d d q ( t ) CV e dt dt 1 CV RC e t RC V R e t RC t RC 66
67 Ramifications of Current At t=0, i(0)=v/r (indicates full current) At t=, i=0 which indicates that the current has stopped flowing. Another interpretation is that the capacitor has an EMF =V and thus V B A S R Circuit after a very long time ~V 67
68 Voltage across the resistor and capacitor V Potential across resistor, V R R V ir e R t RC Ve t RC A R S Potential across capacitor, V C V B At t=0, V C =0 and V R= V At t=, V C =V and V R =0 C V V C C q C CV 1 e C V 1 e t RC t RC 68
69 RC Not just a cola RC is called the time constant of the circuit RC has units of time (seconds) and represents the time it takes for the charge in the capacitor to reach 63% of its maximum value When RC=t, then the exponent is -1 or e -1 t=rc 69
70 Switch to B q( t) The capacitor is fully charged to V or q=cv at t=0 q ir 0 C dq q R dt C dq 1 dt ln q RC Ae t RC and q t RC i( t) k A RC V e t RC A S B If q CV A CV at R t 0, q(0) Ae CV C 0 70
71 Ramifications At t=0, q=cv and i=-v/r At t=, q=0 and i=0 (fully discharging) Where does the charge go? The charge is lost through the resistor 71
72 Three Connection Conventions For Schematic Drawings Connection Between Wires No Connection A B C 72
73 Ground Connectors Equivalently 73
74 Household Wiring hot or black return / neutral or white ground or green Single Phase Rated 20 A (NW-14) Max V 120 VAC Normally, the return should be at 0 V w.r.t. ground In THEORY, but sometimes no! 74
75 The Death of Little Johnny A short develops between the hot lead and the washer case Little Johnny hot neutral Washer X X Uhoh! It leaks! 120V R G R Little Johnny If If R G =, =0, then then Johnny Johnny is is safe dead! R G 75
76 Saving Little Johnny A short develops between the hot lead and the washer case Little Johnny hot neutral Washer Uhoh! It leaks! R G No Path to Johnny! 120V R Little Johnny R G 76
77 Problem In the figure to the left, a potential difference of 20 V is applied across points a and b. a) What is charge on each capacitor if C 1 = 10 F, C 2 =20 F, and C 3 =30 F. b) What is potential difference across points a and d? c) D and b? 77
78 Step 1: Find Equivalent Resistance C1&C2: 10+20=30 C1&C2+C3: (1/30)+(1/30)= 2/30 i.e. 30/2=15 F 78
79 Finding Potentials Q=CV=15*20=300 C So charge on C3 is 300 C The voltage across C3 is 300 C/30 F= 10 V So 10 V across points b & d and therefore, 20-10=10 V across a & d 79
80 Parallel Network If there is 10 V across this network then Q1=10 F*10V= 100 C If there are only 300 C total and 100 C is on this capacitor, then 200 C must be charge of C2 Check: Q2=20 F*10V=200 C 80
81 Problem In the figure, each capacitor C 1 =6.9 F and C 2 =4.6 F. a)compute equivalent capacitance between a and b b)compute the charge on each capacitor if V ab =420 V c)compute V cd when V ab =420 V 81
82 Analysis Look at the right most connection: it is parallel connection between the C2 capacitor and the 3 C1 capacitors 3 C1 s: 6.9/3=2.3 F C2+3C1 s= =6.9 82
83 Repeat Again Now the 6.9 F capacitor is in series with the other C1 capacitors So it is the same circuit as again, so the equivalent is 6.9 F Finally, the total equivalent is 2.3 F 83
84 So 420 V *2.3 F= 966 C For each C1 capacitor on the leftmost network, the voltage across each is 140 V (966/6.9 or 420/3) If there is 140 V across the C2, then 140*4.6=644 C There must be = 322 C in the other branch. 84
85 In the middle network, Each capacitor has V (140/3 or 322 C/6.9) So voltage across c & d is V Then C2 capacitor has 214C and the other C1 capacitors have =107 C 85
86 Problem Two parallel plates have equal and opposite charges. When the space between them is evacuated, the electric field is 3.2 x 10 5 V/m. When the space is filled with a dielectric, the electric field is 2.5 x 10 5 V/m. a) What is the charge density on each surface of the dielectric? b) What is the dielectric constant? 86
87 E Total =E 0 -E i E i E 0 E E i Total 0 E e *( E E i i e 0 Total ) i =8.85e-12*( )*10 5 =4.32x10 6 C/m 2 87
88 Dielectric Constant K=E 0 /E=3.2e5/2.5e5=
89 Problem A 3.4 F capacitor is initially uncharged and then connected in series with a 7.25 kw resistor and an emf source of 180 V which has negligible resistance. a) What is the RC time constant? b) How much time does it take (after connection) for the capacitor to reach 50% of its maximum charge? c) After a long time the EMF source is disconnected from the circuit, how long does it take the current to reach 1% of its maximum value? 89
90 RC Value RC=3F*7.25kW RC=3e-6*7.25e3 RC=21.75 ms 90
91 Time to 50% of max charge Q(t)=C*V*(1-e -t/rc ) Q(t)/CV is the fraction of the maximum charge so let Q(t)/CV =50%.5=1-e -t/rc e -t/rc =.5=1/2 or 2-1 -t/rc=-ln(2) t=rc*ln(2)=21.75*.693 t=15.07 ms 91
92 Since R & C have not changed, RC=21.75 ms I(t)=(V/R)*e -t/rc I(t)/(V/R) is the fraction of maximum current Let I(t)/(V/R) = 1% or =e -t/rc ln(0.01)=-4.605=-t/rc t=21.75*4.605=100 ms 92
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