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

Goals for Chapter 24 To understand capacitors & calculate capacitance Symbol: Unit: FARADs (Coulombs/Volt)

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 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/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

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!

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?

Parallel-plate capacitor TWO parallel conducting plates Separated by distance that is small compared to their dimensions.

Evolution in Capacitors Film dielectrics “Wet” electrolytic capacitors Ceramic Capacitors Polymer Capacitors

Parallel-plate capacitor The capacitance of a parallel-plate capacitor is C =  0 A/d

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

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]

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 = ?

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 F/m *2.00 m 2 /.005 m = 3.54 x F Q on each plate from C = Q/V so Q = CV = 3.54 x C E field between plates from V = Ed so E = V/d = 2.00 x V/m Or… E =  0 = (Q/Area)/  0 = 3.54 x C/2.00 m 2 /  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 V ab = V a – V b = “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 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 )

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!!

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?

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!

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 !

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/C eq = 1/C 1 + 1/C 2 + 1/C 3 + …

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?

Capacitors in parallel Capacitors are connected in parallel between a and b if potential difference V ab is the same for all the capacitors.

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 + ….

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 = = 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 = 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

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.

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

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 = Q/V but V stays the same (fixed by battery) so Q +V - -q* +q* d*

Pull capacitors apart when connected to a battery? +V - -q* +q* d* +q -q

Pull capacitors apart when connected to a battery? +V - -q* +q* d* -q E

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

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

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

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 +Q d*

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

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)

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  W using capacitors in parallel!The Z machine

Example 24.7 C1 = 8.0  F, 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 =  C What is energy stored? U = ½ CV 2 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! Q 0 spreads out over both plates = Q 1 + Q 2 Q 0 decreases to Q 1 Q 0 spreads here to Q 2

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

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.

Table 24.1—Some dielectric constants Dielectrics increase energy density

Examples with and without a dielectric Refer to Problem-Solving Strategy Follow Example to see the effect of the dielectric. Follow Example to see how the dielectric affects energy storage. Use Figure below.

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 (right) and (below) show the effect of an applied electric field on polar and nonpolar molecules.

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.

Gauss’s law in dielectrics Follow the text discussion of Gauss’s law in dielectrics, using Figure at the right. Follow Example for a spherical capacitor