2Capacitors and Capacitance Capacitor: any two conductors, one with charge +Q, other with charge -Q-QPotential DIFFERENCE between conductors = V+QUses: storing and releasing electric charge/energy.Most electronic capacitors:micro-Farads (mF),pico-Farads (pF) FNew technology:compact 1 F capacitorsQ = CV -- C = capacitanceUnits of capacitance:Farad (F) = Coulomb/Volt
3CapacitanceCapacitance depends only on GEOMETRICAL factors and on the MATERIAL that separates the two conductorse.g. Area of conductors, separation, whether the space in between is filled with air, plastic, etc.+Q-Q(We first focus on capacitorswhere gap is filled by AIR!)
5Capacitors and Capacitance Capacitor: any two conductors, one with charge +Q, other with charge -Q-Q+QPotential DIFFERENCE between conductors = VUses: storing and releasing electric charge/energy.Most electronic capacitors:micro-Farads (mF),pico-Farads (pF) FNew technology:compact 1 F capacitorsQ = CV C = capacitanceUnits of capacitance:Farad (F) = Coulomb/Volt
6Parallel Plate Capacitor We want capacitance: C=Q/VArea of each plate = ASeparation = dcharge/area = s= Q/AE field between the plates: (Gauss’ Law)+Q-QRelate E to potential difference V:What is the capacitance C?
7Parallel Plate Capacitor -- example A huge parallel plate capacitor consists of two square metal plates of side 50 cm, separated by an air gap of 1 mmWhat is the capacitance?C = e0A/d =(8.85 x F/m)(0.25 m2)/(0.001 m)= 2.21 x 10-9 F(small!!)Lesson: difficult to get large valuesof capacitance without specialtricks!
8Isolated Parallel Plate Capacitor A parallel plate capacitor of capacitance C 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-QQ is fixed!C decreases (=e0A/d)Q=CV; V increases.
9Parallel Plate Capacitor & Battery A parallel plate capacitor of capacitance C is charged using a battery.Charge = Q, potential difference = V.Plate separation is INCREASED while battery remains connected.+Q-QDoes the electric field inside:(a) Increase?(b) Remain the same?(c) Decrease?V is fixed by battery!C decreases (=e0A/d)Q=CV; Q decreasesE = Q/ e0A decreases
10Spherical Capacitor What is the electric field inside the capacitor? (Gauss’ Law)Radius of outer plate = bRadius of inner plate = aConcentric spherical shells:Charge +Q on inner shell,-Q on outer shellRelate E to potential differencebetween the plates:
11Spherical Capacitor What is the capacitance? Radius of outer plate = b C = Q/V =Radius of outer plate = bRadius of inner plate = aConcentric spherical shells:Charge +Q on inner shell,-Q on outer shellIsolated sphere: let b >> a,
12Cylindrical Capacitor What is the electric field in between the plates?Radius of outer plate = bRadius of inner plate = aLength of capacitor = L+Q on inner rod, -Q on outer shellRelate E to potential differencebetween the plates:cylindrical surface of radius r
13Cylindrical Capacitor What is the capacitance C?C = Q/V =Radius of outer plate = bRadius of inner plate = aLength of capacitor = LCharge +Q on inner rod,-Q on outer shellExample: co-axial cable.
14Summary Any two charged conductors form a capacitor. Capacitance : C= Q/VSimple Capacitors: Parallel plates: C = e0 A/d Spherical : C = 4p e0 ab/(b-a) Cylindrical: C = 2p e0 L/ln(b/a)
15Capacitors in Parallel A wire is a conductor, so it is an equipotential.Capacitors in parallel have SAME potential difference but NOT ALWAYS same charge.VAB = VCD = VQtotal = Q1 + Q2CeqV = C1V + C2VCeq = C1 + C2Equivalent parallel capacitance = sum of capacitancesABCDC1C2Q1Q2CeqQtotalPARALLEL:V is same for all capacitorsTotal charge in Ceq = sum of charges
16Capacitors in series A B C C1 C2 Q1 Q2 Q1 = Q2 = Q (WHY??) VAC = VAB + VBCCeqQSERIES:Q is same for all capacitorsTotal potential difference in Ceq = sum of V
17Capacitors in parallel and in series Ceq = C1 + C2Veq=V1=V2Qeq=Q1+Q2C1C2Q1Q2CeqQeqIn series :1/Ceq = 1/C1 + 1/C2Veq=V1 +V2Qeq=Q1=Q2Q1Q2C1C2
18Example 1 What is the charge on each capacitor? 10 mF 120VQ = CV; V = 120 VQ1 = (10 mF)(120V) = 1200 mCQ2 = (20 mF)(120V) = 2400 mCQ3 = (30 mF)(120V) = 3600 mCNote that:Total charge (7200 mC) is shared between the 3 capacitors in the ratio C1:C2:C3 -- i.e. 1:2:3
19Example 2 What is the potential difference across each capacitor? Q = CV; Q is same for all capacitorsCombined C is given by:10 mF30 mF20 mF120VCeq = 5.46 mFQ = CV = (5.46 mF)(120V) = 655 mCV1= Q/C1 = (655 mC)/(10 mF) = 65.5 VV2= Q/C2 = (655 mC)/(20 mF) = VV3= Q/C3 = (655 mC)/(30 mF) = 21.8 VNote: 120V is shared in the ratio of INVERSE capacitances i.e.1:(1/2):(1/3)(largest C gets smallest V)
20Example 3In the circuit shown, what is the charge on the 10F capacitor?10 mF5 mF10VThe two 5F capacitors are in parallelReplace by 10FThen, we have two 10F capacitors in seriesSo, there is 5V across the 10F capacitor of interestHence, Q = (10F )(5V) = 50C10 mF10V
21Energy Stored in a Capacitor Start out with uncharged capacitorTransfer small amount of charge dq from one plate to the other until charge on each plate has magnitude QHow much work was needed?dq
22Energy Stored in Electric Field Energy stored in capacitor:U = Q2/(2C) = CV2/2View the energy as stored in ELECTRIC FIELDFor example, parallel plate capacitor: Energy DENSITY = energy/volume = u =volume = AdGeneral expression for any region with vacuum (or air)
23Example10mF capacitor is initially charged to 120V. 20mF capacitor is initially uncharged.Switch is closed, equilibrium is reached.How much energy is dissipated in the process?10mF (C1)20mF (C2)Initial charge on 10mF = (10mF)(120V)= 1200mCAfter switch is closed, let charges = Q1 and Q2.Charge is conserved: Q1 + Q2 = 1200mCQ1 = 400mCQ2 = 800mCVfinal= Q1/C1 = 40 VAlso, Vfinal is same:Initial energy stored = (1/2)C1Vinitial2 = (0.5)(10mF)(120)2 = 72mJFinal energy stored = (1/2)(C1 + C2)Vfinal2 = (0.5)(30mF)(40)2 = 24mJEnergy lost (dissipated) = 48mJ
24Dielectric Constant C = A/d If the space between capacitor plates is filled by a dielectric, the capacitance INCREASES by a factor This is a useful, working definition for dielectric constant.Typical values of k:+Q- QDIELECTRICC = A/d
25Example Capacitor has charge Q, voltage V Battery remains connected while dielectric slab is inserted.Do the following increase, decrease or stay the same:Potential difference?Capacitance?Charge?Electric field?dielectricslab
26Example (soln)Initial values: capacitance = C; charge = Q; potential difference = V; electric field = E;Battery remains connectedV is FIXED; Vnew = V (same)Cnew = kC (increases)Qnew = (kC)V = kQ (increases).Since Vnew = V, Enew = E (same)dielectricslabEnergy stored? u=e0E2/2 => u=ke0E2/2 = eE2/2
27Summary Capacitors in series and in parallel: in series: charge is the same, potential adds, equivalent capacitance is given by 1/C=1/C1+1/C2in parallel: charge adds, potential is the same, equivalent capaciatnce is given by C=C1+C2.Energy in a capacitor: U=Q2/2C=CV2/2; energy density u=e0E2/2Capacitor with a dielectric: capacitance increases C’=kC