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Chapter 24 Capacitance and Dielectrics PowerPoint® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Lectures by Wayne Anderson Copyright © 2012 Pearson Education Inc.modified Scott Hildreth Chabot College 2016 Goals for Chapter 24 • To understand capacitors & calculate capacitance Goals for Chapter 24 • To understand capacitors & calculate capacitance • Unit: FARADs (Coulombs/Volt) • Symbol: Introduction • How is a ROUND capacitor like a FLAT pair of plates? • UNROLL it! Goals for Chapter 24 • To analyze networks of capacitors • To calculate energy stored in a capacitor • To examine dielectrics & how they affect capacitance Introduction • How do camera flash units store energy? • Capacitors are devices that store electric potential energy. • Energy of capacitor is stored in E field. Introduction Capacitors and capacitance • Any two conductors separated by an “insulator” form a capacitor. • Insulator will allow E field between the conductors • Insulator will not allow charge to flow from one conductor through itself to the other. Capacitors and capacitance • Capacitors are like parking lots for charge… • They are built with physical dimensions in mind, but how many cars are in them depends upon the “pressure” to park there! Capacitors and capacitance • The more charge you can hold, the larger the capacitor! “capacity” • Charge a capacitor by pushing it there with a potential voltage “pressure” Capacitors and capacitance • The definition of capacitance is C = Q/Vab – Q = “charge stored” (in Coulombs) +Q – Q is held symmetrically on each plate (same +Q as –Q) -Q a b – A voltage difference Vab is the electrical “pressure” that pushes & keeps charge there Capacitors and capacitance • C = Q/Vab so Q = CVab • Q increases as C is made larger, for a given pressure – more capacity means more charge stored! • Q increases as Vab increases – For any particular capacitor, more pressure Vab allows it to hold more charge! • But no voltage applied means no charge stored… Capacitors and capacitance • The definition of capacitance is C = Q/Vab. – Units of Capacitance: [Coulombs/Volt] = FARADs – How much charge can you store per joule of work to keep them there? – But [Volt] = [Joules/Coulomb] – Farads = Coulombs2/Joule Parallel-plate capacitor • TWO parallel conducting plates • Separated by distance that is small compared to their dimensions. Batteries vs. Capacitors • Batteries store MORE energy, for longer times, but take more time to deliver it (or be recharged) “Permanent Charge Storage” • Capacitors store less energy, but deliver it instantly, and recharge quickly. “Temporary Charge Storage” Evolution in Capacitors • Film dielectrics • “Wet” electrolytic capacitors • Ceramic Capacitors • Polymer Capacitors • Super Capacitors! • Graphene! Parallel-plate capacitor • The capacitance of a parallel-plate capacitor is C = 0A/d Parallel-plate capacitor • The capacitance of a parallel-plate capacitor is C = 0A/d. • Note! C is engineered! You control Area & distance by design! • C increases with Area • C decreases with separation Parallel-plate capacitor • The capacitance of a parallel-plate capacitor is C = 0A/d. – So….0 = C x d/A = [Farads x meters/meters2] – 0 has units of [Farads/meter] • Recall “0” from Coulomb’s Law: Force = (1/4p 0) q1q2/r2 – Units of 0 were Coulomb2/Newton-meter2 • So [Farads/meter] = [Coulomb2/Newton-meter2] • [Farads] = [Coulomb2/Newton-meter] • [Farads] = [Coulomb2/Joule] Parallel-plate capacitor Example 24.2: Plates 2.00 m2 in area; 5.00 mm apart; 10 kV applied Potential Difference. • C=? • Q on each plate = ? • E field between plates = ? Parallel-plate capacitor Example 24.2: Plates 2.00 m2 in area; 5.00 mm apart; 10 kV applied Potential Difference. • C= 0 A/d = 8.85 x 10-12 F/m *2.00 m2/.005 m = 3.54 x 10-9 F • Q on each plate from C = Q/V so Q = CV = 3.54 x 10-5 C • E field between plates from V = Ed so E = V/d = 2.00 x 10-6 V/m • Or… E = s/0 = (Q/Area)/0 = 3.54 x 10-5 C/2.00 m2 / 0 A spherical capacitor • Example 24.3 – Two concentric spherical shells separated by vacuum.. What is C? A spherical capacitor • Example 24.3 – Two concentric spherical shells separated by vacuum.. What is C? • C = Q/V so we need V! • Get V from E field! • V = kq/r in general for spherical charge distribution • Vab = Va – Vb = “Voltage of a relative to b” A spherical capacitor • Example 24.3 – Two concentric spherical shells separated by vacuum.. What is C? • C = Q/V so we need V! • Get V from E field! • V = kq/r in general for spherical charge distribution • Vab = Va – Vb = kQ/ra – kQ/rb = kQ (1/ra – 1/rb) • Vab = Q/4p0 (1/ra – 1/rb) • C = Q/V = 4p0/(1/ra – 1/rb) = 4p0 (ra rb)/ (rb – ra) A spherical capacitor • Example 24.3 – Two concentric spherical shells separated by vacuum. What is C? • C = Q/V = 4p0/(1/ra – 1/rb) = 4p0 (ra rb)/ (rb – ra) • Check! – C increases as Area increases? • YES! – C decreases as separation increases? • YES!! A cylindrical capacitor • Example 24.4 – Linear charge density -l on outer cylinder of radius rb, + l on inner cylinder of radius ra. What is C? A cylindrical capacitor • Example 24.4 – Linear charge density - l on outer cylinder of radius rb, +l on inner cylinder of radius ra. What is C? • C = Q/Vab; find Q and find Vab! • Q=lL • V = [l/2p0 ] ln (r0/r) – r0 = distance where V was defined to be zero! – Say r0 = rb here, so Vab = (l/2p0) ln (rb/ra) • Does this make sense?? YES! A cylindrical capacitor • Example 24.4 – Linear charge density - l on outer cylinder of radius rb, +l on inner cylinder of radius ra. What is C? • C = Q/Vab = l L/[(l/2p0) ln (rb/ra)] • C = 2p0 L/ ln (rb/ra) • Check: – Units = Farads/Meter x Meters = Farads – C increases as L increases, – and C increaes as rb closer to ra! Capacitors in series • Capacitors are in series if connected one after the other Capacitors in series • Capacitors are in series if connected one after the other • NOTE: • Charges are the same on all plates in the series (even with different capacitances!) Capacitors in series • Capacitors are in series if connected one after the other • Equivalent capacitance of a series combination: 1/Ceq = 1/C1 + 1/C2 + 1/C3 + … Capacitors in series • Capacitors are in series if connected one after the other • Equivalent capacitance of a series combination: 1/Ceq = 1/C1 + 1/C2 + 1/C3 + … • NOTE • Ceq is always LESS than the smallest capacitor in series! • Say C1 = 10 mF, C2 = 20 mF, C3 = 30 mF • 1/Ceq = 1/10 + 1/20 + 1/30 = 6/60 + 3/60 + 2/60 = 11/60 • Ceq = 60/11 = 5.45 mF (< C1 !!) • WHY? Capacitors in parallel • Capacitors are connected in parallel between a and b if potential difference Vab is the same for all the capacitors. Capacitors in parallel • Capacitors are connected in parallel between a and b if potential difference Vab is the same for all the capacitors. • The equivalent capacitance of a parallel combination is the sum of the individual capacitances: Ceq = C1 + C2 + C3 + … . Capacitors in parallel • Capacitors are connected in parallel between a and b if potential difference Vab is the same for all the capacitors. • The equivalent capacitance of a parallel combination is the sum of the individual capacitances: Ceq = C1 + C2 + C3 + … . • Note! • Ceq is always MORE than largest capacitor in parallel! • Say C1 = 10 mF, C2 = 20 mF, C3 = 30 mF • Ceq = 10 + 20 + 30 = 60 mF > C3 • WHY?? Calculations of capacitance • Example 24.6, a capacitor network: • Find C eq? Calculations of capacitance • Example 24.6, a capacitor network: • Find C eq? Calculations of capacitance • Example 24.6, a capacitor network: • Find C eq? Energy stored in a capacitor • The potential energy stored in a capacitor is • U = Q2/2C (in JOULES) • U increases with more stored charge; • U is less for fixed charge on a larger capacitor • U = Q2/2C but C = Q/V…. so • U = 1/2 QV. • U = Q2/2C and Q = CV … so • U = 1/2 CV2 Energy stored in a capacitor • Potential energy stored in capacitor: U = Q2/2C = 1/2 CV2 = 1/2 QV • Capacitor energy stored in electric field between the plates. Energy stored in a capacitor • Capacitor energy stored in electric field between the plates. • Energy Density is u = 1/2 0E2 [Joules/m3] How can we visualize energy in the field? What happens when you pull plates apart? +Q +V - d -Q Pull capacitors apart when connected to a battery? Start with a capacitor C = 0A/d, connected to voltage source V, storing charge Q +Q +V - d -Q Pull capacitors apart when connected to a battery? Pull plates apart - this takes work against field. +V +Q - -Q Pull capacitors apart when connected to a battery? One way to see this…..Capacitance decreases C = 0A/d so if d C +q* +V - d* -q* Pull capacitors apart when connected to a battery? One way to see this…..Capacitance decreases C = Q/V but V is fixed by battery, so C +q* +V - d* -q* =Q Pull capacitors apart when connected to a battery? +q* +q +V - d* -q -q* Pull capacitors apart when connected to a battery? • Suppose q stayed the same? • Then E would be constant! • So DV = increases because dl increases!! • So DVcap > DV battery!! • And charge HAS to flow back into the battery!!!! +q* +q +V - * d -q -q* Pull capacitors apart when connected to a battery? d increases, A constant => C decreases V fixed, so C = Q/V => Q = CV decreases Q decreases so s decreases and E decreases Energy = ½ QV decreases + q* E +V - d* -q q* Pull capacitors apart when connected to a battery? +q* +V E - d* -q -q* Pull capacitors apart when NOT connected?? Start with a capacitor C = 0A/d, charged by voltage source V, storing charge Q +Q +V - d -Q Pull capacitors apart when NOT connected?? Start with a capacitor C = 0A/d, charged by voltage source V, storing charge Q, then disconnect from battery! +Q d -Q Pull capacitors apart when connected to a battery? Pull plates apart - this takes work against field. But charge can’t go anywhere! +Q -Q Pull capacitors apart when connected to a battery? One way to see this…..Capacitance decreases C = Q/V but Q stays the same so V increase +Q d* -Q Pull capacitors apart when connected to a battery? Another way to see this…..Q same, so s same, so E the same! +q* E d* -q* Pull capacitors apart when connected to a battery? Another way to see this…..Q same, so s same, so E the same! +q* E constant V* d* -q* Energy density increases! (more volume in the capacitor) Energy stored in a capacitor • The potential energy stored in a capacitor is U = Q2/2C = 1/2 CV2 = 1/2 QV. • The capacitor energy is stored in the electric field between the plates. The energy density is u = 1/2 0E2. • The Z machine shown below can produce up to 2.9 1014 W using capacitors in parallel! Example 24.7 • C1 = 8.0 mF, charged to 120 V. Switch is open. • What is Q on C1? Energy stored? Example 24.7 • What is Q on C1? • C = Q/V so Q = CV = 960 8.0 mC • What is energy stored? • U = ½ CV2 or ½ QV = 0.58 J Example 24.7 • Close the switch; after steady state, what is V across each capacitor? What is charge on each? • What is final energy of the system? Example 24.7 • Close the switch; after steady state, what is V across each capacitor? What is charge on each? • Becomes a PARALLEL network; pressure will be unequal initially, and charge flows from C1 to C2 until the PRESSURE (voltages) are the same! • Q0 spreads out over both plates = Q1 + Q2 Q0 decreases to Q1 Q0 spreads here to Q2 Example 24.7 • Q0 = Q1 + Q2 (and Q = CV!) • Q0 = C1V1 + C2V2 • But in parallel, V1 = V2 • Q0 = (C1 + C2)V1 so V1 = V2 = 960 mC/(C1+C2) = 80V • Q1 = C1V1 = 640 mC and Q2 = C2V2 = 320 mC Q0 = 960 mC decreases to Q1 = 640 mC Q0 spreads out to Q2 = 320 mC Dielectrics • Dielectric is nonconducting material. Most capacitors have dielectric between plates. • Dielectric increases the capacitance and the energy density by a factor K. • Dielectric constant of material is K = C/C0 > 1. • Dielectric reduces E field between the plates, so for fixed voltage, more Q • Dielectric reduces VOLTAGE between plates w/ fixed Q on each (Q = CV) Table 24.1—Some dielectric constants Dielectrics increase energy density 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. Molecular model of induced charge - I • Figures 24.17 (right) and 24.18 (below) show the effect of an applied electric field on polar and nonpolar molecules. Molecular model of induced charge - II • Polarization of dielectric shows how induced charges reduce magnitude of resultant electric field.