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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 Capacitance and Dielectrics
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Goals for Chapter 24 To understand capacitors & calculate capacitance Symbol: Unit: FARADs (Coulombs/Volt)
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Introduction How is a ROUND capacitor like a FLAT pair of plates? UNROLL it!
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Goals for Chapter 24 To analyze networks of capacitors To calculate energy stored in a capacitor To examine dielectrics & how they affect capacitance
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Introduction How do camera flash units store energy? Capacitors are devices that store electric potential energy. Energy of capacitor is stored in E field.
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Introduction
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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.
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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”
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Capacitors and capacitance The definition of capacitance is C = Q/V ab –Q = “charge stored” (in Coulombs) –Q is held symmetrically on each plate (same +Q as –Q) –A voltage difference V ab is the electrical “pressure” that pushes & keeps charge there +Q-Q
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Capacitors and capacitance The definition of capacitance is C = Q/V ab. –V ab = pressure that pushes and keeps charge there –C increases as Q increases more capacity! –C decreases as V ab increases more pressure required to hold the charge there, so less effective in storing it temporarily!
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Capacitors and capacitance The definition of capacitance is C = Q/V ab. –Units of Capacitance: [Coulombs/Volt] = FARADs –But [Volt] = [Joules/Coulomb] –Farads = Coulombs 2 /Joule –How much charge can you store per joule of work to keep them there?
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Parallel-plate capacitor TWO parallel conducting plates Separated by distance that is small compared to their dimensions.
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Evolution in Capacitors Film dielectrics “Wet” electrolytic capacitors Ceramic Capacitors Polymer Capacitors http://www.capacitorlab.com/capacitor-types-polymer/
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Parallel-plate capacitor The capacitance of a parallel-plate capacitor is C = 0 A/d
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Parallel-plate capacitor The capacitance of a parallel-plate capacitor is C = 0 A/d. Note! C is engineered! You control Area & distance by design! C increases with Area C decreases with separation
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Parallel-plate capacitor The capacitance of a parallel-plate capacitor is C = 0 A/d. – 0 has units of [Farads-meters/area] = [Farads/meter] Recall “ 0 ” from Coulomb’s Law: Force = (1/4 0 ) q 1 q 2 /r 2 –Units of 0 were Coulomb 2 /Newton-meter 2 So [Farads/meter] = [ Coulomb 2 /Newton-meter 2 ] [Farads] = [ Coulomb 2 /Newton-meter] [Farads] = [Coulomb 2 /Joule]
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Parallel-plate capacitor Example 24.2: Plates 2.00 m 2 in area; 5.00 mm apart; 10 kV applied Potential Difference. C=? Q on each plate = ? E field between plates = ?
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Parallel-plate capacitor Example 24.2: Plates 2.00 m 2 in area; 5.00 mm apart; 10 kV applied Potential Difference. C= 0 A/d = 8.85 x 10 -12 F/m *2.00 m 2 /.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 = 0 = (Q/Area)/ 0 = 3.54 x 10 -5 C/2.00 m 2 / 0
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A spherical capacitor Example 24.3 – Two concentric spherical shells separated by vacuum.. What is C?
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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 V ab = V a – V b = “Voltage of a relative to b”
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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 V ab = V a – V b = kQ/r a – kQ/r b = kQ (1 /r a – 1/r b ) V ab = Q/4 0 (1 /r a – 1/r b ) C = Q/V = 4 0 /(1 /r a – 1/r b ) = 4 0 (r a r b )/ ( r b – r a )
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A spherical capacitor Example 24.3 – Two concentric spherical shells separated by vacuum. What is C? C = Q/V = 4 0 /(1 /r a – 1/r b ) = 4 0 (r a r b )/ ( r b – r a ) Check! –C increases as Area increases? YES! –C decreases as separation increases? YES!!
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A cylindrical capacitor Example 24.4 – Linear charge density - on outer cylinder of radius r b, + on inner cylinder of radius r a. What is C?
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A cylindrical capacitor Example 24.4 – Linear charge density - on outer cylinder of radius r b, + on inner cylinder of radius r a. What is C? C = Q/V ab ; find Q and find V ab ! Q = L V = [ ] ln (r 0 /r) –r 0 = distance where V was defined to be zero! –Say r 0 = r b here, so V ab = ( ln (r b /r a ) Does this make sense?? YES!
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A cylindrical capacitor Example 24.4 – Linear charge density - on outer cylinder of radius r b, + on inner cylinder of radius r a. What is C? C = Q/Vab = L/[( ln (r b /r a )] C = L/ ln (r b /r a ) Check: –Units = Farads/Meter x Meters = Farads –C increases as L increases, –and C increaes as r b closer to r a !
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Capacitors in series Capacitors are in series if connected one after the other
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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!)
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Capacitors in series Capacitors are in series if connected one after the other Equivalent capacitance of a series combination: 1/C eq = 1/C 1 + 1/C 2 + 1/C 3 + …
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Capacitors in series Capacitors are in series if connected one after the other Equivalent capacitance of a series combination: 1/C eq = 1/C 1 + 1/C 2 + 1/C 3 + … NOTE C eq is always LESS than the smallest capacitor in series! Say C1 = 10 mF, C2 = 20 mF, C3 = 30 mF 1/C eq = 1/10 + 1/20 + 1/30 = 6/60 + 3/60 + 2/60 = 11/60 C eq = 60/11 = 5.45 mF (< C1 !!) WHY?
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Capacitors in parallel Capacitors are connected in parallel between a and b if potential difference V ab is the same for all the capacitors.
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Capacitors in parallel Capacitors are connected in parallel between a and b if potential difference V ab is the same for all the capacitors. The equivalent capacitance of a parallel combination is the sum of the individual capacitances: C eq = C 1 + C 2 + C 3 + ….
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Capacitors in parallel Capacitors are connected in parallel between a and b if potential difference V ab is the same for all the capacitors. The equivalent capacitance of a parallel combination is the sum of the individual capacitances: C eq = C 1 + C 2 + C 3 + …. Note! C eq is always MORE than largest capacitor in parallel! Say C1 = 10 mF, C2 = 20 mF, C3 = 30 mF C eq = 10 + 20 + 30 = 60 mF > C3 WHY??
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Calculations of capacitance Example 24.6, a capacitor network: Find C eq?
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Calculations of capacitance Example 24.6, a capacitor network: Find C eq?
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Calculations of capacitance Example 24.6, a capacitor network: Find C eq?
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Energy stored in a capacitor The potential energy stored in a capacitor is U = Q 2 /2C U increases with more stored charge; U is less for fixed charge on a larger capacitor U = Q 2 /2C but C = Q/V…. so U = 1/2 QV. U = Q 2 /2C and Q = CV … so U = 1/2 CV 2
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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 0 E 2.
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Pull capacitors apart when connected to a battery? Start with a capacitor C = 0 A/d, connected to voltage source V, storing charge Q +V - -Q +Q d
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Pull capacitors apart when connected to a battery? Pull plates apart - this takes work against field. +V - -Q +Q
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Pull capacitors apart when connected to a battery? One way to see this…..Capacitance decreases C = Q/V but V stays the same (fixed by battery) so Q +V - -q* +q* d*
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Pull capacitors apart when connected to a battery? +V - -q* +q* d* +q -q
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Pull capacitors apart when connected to a battery? +V - -q* +q* d* -q E
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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 decreases and E decreases Energy = ½ QV decreases +V - - q* + q* d* -q E
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Pull capacitors apart when NOT connected?? Start with a capacitor C = 0 A/d, charged by voltage source V, storing charge Q +V - -Q +Q d
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Pull capacitors apart when NOT connected?? Start with a capacitor C = 0 A/d, charged by voltage source V, storing charge Q, then disconnected -Q +Q d
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Pull capacitors apart when connected to a battery? Pull plates apart - this takes work against field. But charge can’t go anywhere! -Q +Q
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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 +Q d*
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Pull capacitors apart when connected to a battery? Another way to see this…..Q same, so same, so E the same! -q* +q* d* E
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Pull capacitors apart when connected to a battery? Another way to see this…..Q same, so same, so E the same! -q* +q* d* E constant V* Energy density increases! (more volume in the capacitor)
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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 0 E 2. The Z machine shown below can produce up to 2.9 10 14 W using capacitors in parallel!The Z machine
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Example 24.7 C1 = 8.0 F, charged to 120 V. Switch is open. What is Q on C1? Energy stored?
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Example 24.7 What is Q on C1? C = Q/V so Q = CV = 960 8.0 C What is energy stored? U = ½ CV 2 or ½ QV = 0.58 J
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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?
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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! Q 0 spreads out over both plates = Q 1 + Q 2 Q 0 decreases to Q 1 Q 0 spreads here to Q 2
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Example 24.7 Q 0 = Q 1 + Q 2 (and Q = CV!) Q 0 = C 1 V 1 + C 2 V 2 But in parallel, V 1 = V 2 Q 0 = (C 1 + C 2 )V 1 so V 1 = V 2 = 960 C/(C 1 +C 2 ) = 80V Q 1 = C 1 V 1 = 640 C and Q 2 = C 2 V 2 = 320 C Q 0 = 960 C decreases to Q 1 = 640 C Q 0 spreads out to Q 2 = 320 C
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Dielectrics Dielectric is nonconducting material. Most capacitors have dielectric between plates. Dielectric increases the capacitance and the energy density by a factor K.Dielectric increases the capacitance and the energy density by a factor K. Dielectric constant of material is K = C/C 0 > 1. Dielectric reduces E field between the plates. Dielectric reduces VOLTAGE between plates w/ fixed Q on each.
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Table 24.1—Some dielectric constants Dielectrics increase energy density
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Examples with and without a dielectric Refer to Problem-Solving Strategy 24.2. Follow Example 24.10 to see the effect of the dielectric. Follow Example 24.11 to see how the dielectric affects energy storage. Use Figure 24.16 below.
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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.
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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.
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Molecular model of induced charge - II Figure 24.20 below shows polarization of the dielectric and how the induced charges reduce the magnitude of the resultant electric field.
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Gauss’s law in dielectrics Follow the text discussion of Gauss’s law in dielectrics, using Figure 24.22 at the right. Follow Example 24.12 for a spherical capacitor
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