Physics 2102Physics 2102Lecture: 11 FRI 06 FEBLecture: 11 FRI 06 FEBCapacitance ICapacitance IPhysics 2102Jonathan Dowling25.1–3Tutoring Lab Now OpenTuesdays• Lab Location: 102 Nicholson (acrossthe hall from class)• Lab Hours:MTWT: 12:00N–5:00PMF: 12:N–3:00PMCapacitors and CapacitanceCapacitors and CapacitanceCapacitor: any two conductors,one with charge +Q, otherwith charge –Q+Q–QUses: storing and releasingelectric charge/energy.Most electronic capacitors:micro-Farads (µF),pico-Farads (pF) — 10–12 FNew technology:compact 1 F capacitorsPotential DIFFERENCE betweenconductors = VQ = CCV where C = capacitanceUnits of capacitance: Farad (F) = Coulomb/VoltCapacitanceCapacitance• Capacitance depends onlyon GEOMETRICAL factorsand on the MATERIALthat separates the twoconductors• e.g. Area of conductors,separation, whether thespace in between is filledwith air, plastic, etc.+Q–Q(We first focus on capacitorswhere gap is filled by AIR!)Electrolytic (1940-70)Electrolytic (new)Paper (1940-70)Tantalum (1980 on)Ceramic (1930 on)Mica (1930-50)Variableair, micaCapacitorsCapacitorsParallel Plate CapacitorParallel Plate Capacitor-QE field between the plates: (Gauss’ Law)Relate E to potential difference V:What is the capacitance C!?Area of eachplate = ASeparation = dcharge/area = σ= Q/A!"=dxdEV0!!AQddxAQd000!!"==dAVQC0!==AQE00!!"==We want capacitance: C!=!Q/V!! Units :C2Nm2m2m" # $ % & ' =C2Nm" # $ % & ' =CCJ" # $ % & ' =CV" # $ % & ' (+QCapacitance and Your Capacitance and Your iPhoneiPhone!!dAVQC0!==Parallel Plate Capacitor Parallel Plate Capacitor —— Example Example• A huge parallel plate capacitorconsists of two square metalplates of side 50 cm, separatedby an air gap of 1 mm• What is the capacitance?Lesson: difficult to get large valuesof capacitance without specialtricks!C = ε0A/d= (8.85 x 10–12 F/m)(0.25 m2)/(0.001 m)= 2.21 x 10–9 F(Very Small!!)! Units :C2Nm2m2m" # $ % & ' =C2Nm" # $ % & ' =CCJ" # $ % & ' =CV" # $ % & ' ( F[ ]= FaradIsolatedIsolated Parallel Plate Capacitor Parallel Plate Capacitor• A parallel plate capacitor of capacitanceC is charged using a battery.• Charge = Q, potential difference = V.• Battery is then disconnected.• If the plate separation is INCREASED,does Potential Difference V:(a) Increase?(b) Remain the same?(c) Decrease?+Q–Q• Q is fixed!• C decreases (=ε0A/d)• V=Q/C; V increases.Parallel Plate Capacitor & BatteryParallel Plate Capacitor & Battery• A parallel plate capacitor ofcapacitance C is charged using abattery.• Charge = Q, potential difference = V.• Plate separation is INCREASED whilebattery remains connected.+Q–Q• V is fixed by battery!• C decreases (=e0A/d)• Q=CV; Q decreases• E = Q/ε0A decreasesDoes the Electric Field Inside:(a) Increase?(b) Remain the Same?(c) Decrease?Spherical CapacitorSpherical CapacitorWhat is the electric field insidethe capacitor? (Gauss’ Law)Radius of outerplate = bRadius of innerplate = aConcentric spherical shells:Charge +Q on inner shell,–Q on outer shellRelate E to potential differencebetween the plates:204 rQE!"=!"=bardEV!!babarkQdrrkQ!"#$%&'==(2!"#$%&'=bakQ11Spherical CapacitorSpherical CapacitorWhat is the capacitance?C = Q/V =Radius of outerplate = bRadius of innerplate = aConcentric spherical shells:Charge +Q on inner shell,–Q on outer shellIsolated sphere: let b >> a,!"#$%&'=baQQ1140())(40abab!="#aC04!"=Cylindrical CapacitorCylindrical CapacitorWhat is the electric field inbetween the plates? Gauss’ Law!Relate E to potential differencebetween the plates:Radius of outerplate = bRadius of innerplate = acylindricalGaussiansurface ofradius rLength of capacitor = L+Q on inner rod, –Q on outer shellrLQE02!"=!"=bardEV!!babaLrQdrrLQ!"#$%&=='002ln2()()! =Q2"#0Llnba$ % & ' ( ) ! C = Q /V =2"#0Llnba$ % & ' ( )SummarySummary• Any two charged conductors form a capacitor.•Capacitance : C= Q/V•Simple Capacitors:Parallel plates: C = ε0 A/dSpherical: C = 4π e0 ab/(b-a)Cylindrical: C = 2π ε0
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