Lecture 5 Capacitance Chp. 26CapacitanceParallel Plate CapacitorElectric Field of Parallel Plate CapacitorShow Demo Model, calculate its capacitance , and show how to charge it up with a battery.Demo ContinuedDielectricsSlide 8Permanent dipolesPowerPoint PresentationFind the capacitance of a ordinary piece of coaxial cable (TV cable)Model of coaxial cable for calculation of capacitanceSpherical capacitor or sphereCapacitance of two concentric spherical shellsElectric Potential Energy of CapacitorGraphical interpretation of integrationWhere is the energy stored in a capacitor?How much energy is stored in the Earth’s atmospheric electric field? (Order of magnitude estimate)Parallel Combination of CapacitorsSeries Combination of CapacitorsSample problemSample problem (continued)Warm up set 5What is the electric field in a sphere of uniform distribution of positive charge. (nucleus of protons)Lecture 5 Capacitance Chp. 26•Cartoon - Capacitance definition and examples.•Opening Demo - Discharge a capacitor•Warm-up problem•Physlet •Topics•Demos• Circular parallel plate capacitior•Cylindrical capacitor•Concentric spherical capacitor•Leyden jar capacitor•Dielectric Slab sliding into demo •Show how to calibrate electroscopeCapacitance•Definition of capacitance A capacitor is a useful device in electrical circuits that allows us to store charge and electrical energy in a controllable way. The simplest to understand consists of two parallel conducting plates of area A separated by a narrow air gap d. If charge +Q is placed on one plate, and -Q on the other, the potential difference between them is V, and then the capacitance is defined as C=Q/V. The SI unit is C/V, which is called the Farad, named after the famous and creative scientist Michael Faraday from the early 1800’s.•Applications–Radio tuner circuit uses variable capacitor–Blocks DC voltages in ac circuits–Act as switches in computer circuits–Triggers the flash bulb in a camera–Converts AC to DC in a filter circuitParallel Plate CapacitorElectric Field of Parallel Plate Capacitor€ E =σε0€ q = ε0EAGauss Law€ E =qε0A€ σ =qAAqSESV0ε==Area A+++++++++++--------------------E+ q- qSSAqVqCAqS00εε===Coulomb/Volt = FaradCircular parallel plate capacitorr rsr = 10 cmA = r2 = (.1)2A = .03 m 2S = 1 mm = .001 mSAC0ε=FC10103−×=pFC 300=001.03.)10(11−=C}FaradVoltCoulombp = pico = 10-12Show Demo Model, calculate its capacitance , and show how to charge it up with a battery.Demo ContinuedDemonstrate1. As S increases, voltage increases.2. As S increases, capacitance decreases.3. As S increases, E0 and q are constant.Dielectrics•A dielectric is any material that is not a conductor, but polarizes well. Even though they don’t conduct they are electrically active.–Examples. Stressed plastic or piezo-electric crystal will produce a spark.–When you put a dielectric in a uniform electric field (like in between the plates of a capacitor), a dipole moment is induced on the molecules throughout the volume. This produces a volume polarization that is just the sum of the effects of all the dipole moments. If we put it in between the plates of a capacitor, the surface charge densities due to the dipoles act to reduce the electric field in the capacitor.Dielectrics•The amount that the field is reduced defines the dielectric constant  from the formula E = E0 / , where E is the new field and E0 is the old field without he dielectric.•Since the electric field is reduced and hence the voltage difference is reduced (since E = Vd), the capacitance is increased.–C = Q / V = Q / (V0 / ) =  C0 is typically between 2 – 6 with water equal to 80–Show demo dielectric slab sliding in between plates. Watch how capacitance and voltage change. Also show aluminum slab.Permanent dipolesInduced dipoles++_ _E0 = the applied fieldE’ = the field due to induced dipolesE = E0 - E’VqC =SEV0=00εσ=EAq=σAqE00ε=AqSV0ε=SAC0ε=κ0EE =κSEV0=κ0VV =VqC =0VqCκ=0CC κ=SFind the capacitance of a ordinary piece of coaxial cable (TV cable)metal braid with - qouter insulator• signal wire radius a with + qInsulator (dielectric )radius ba = 0.5 mmb = 2.0 mm  2For a long wire we found that€ Er=2kλrwhere r is radial to the wire. •r€ Va−Vb= − E. dsba∫= −2kλdrrba∫= −2kλ ln rE. ds = Edscos180 = −Eds = Edrab€ V = 2kλ lnbaVa is higher than VbLQ=λ€ k =14πε0 air€ V =Q2πε0Llnba€ C =QV=Q2πε0LQlnba€ C =2πε0Llnba€ CL=2πε0lnba38.11064ln1061111 −−×=×=LCmPFLC43=mPFLC86=0 (air) = 2ds = - dr because path of integration is radially inward€ Va−Vb= −2kλ lnaborModel of coaxial cable for calculation of capacitanceSignal wireOuter metal braidSpherical capacitor or sphereRecall our favorite example for E and V is spherical symmetryQRThe potential of a charged sphere is V = (kQ)/R with V = 0 at r =  .The capacitance is RkRRkQQVQC04πε====Where is the other plate (conducting shell)?It’s at infinity where it belongs, since that’s where the electric lines of flux terminate.k = 1010 and R in meters we have€ C =R1010=10−10R(m) =10−12R(cm)PFcmRC )(=Earth: C = (6x108 cm)PF = 600 FMarble: 1 PFBasketball: 15 PFYou: 30 PFDemo: Leyden jar capacitorDemo: Show how you measured capacitance of electroscopeCapacitance of two concentric spherical shells+q- q€ Va−Vb= − E. dsba∫= − Edrba∫E. ds = Edscos180 = −Eds = Edrds = - drIntegration pathE€ Va−Vb= − Edrba∫= −ba∫kq /r2dr = −kqdrr2ba∫V = kq1rba= kq(1a−1b) = kq(b − a)ab€ C = q /V =abk(b − a)= 4πε0abb − aabElectric Potential Energy of Capacitor•As we begin charging a capacitor, there is initially no potential difference between the plates. As we remove charge from one plate and put it on the other, there is almost no energy cost. As it charges up, this changes.+ -+q -qAt some point during the charging, we have a charge q on the positive plate.The potential difference between the plates is V = q/C. As we transfer an amount dq of positive charge from the negative plate to the positive one, its potential energy increases by an amount dU..dqCqVdqdU ==The total potential energy increase isCQCqdqCqUQ22220===∫AlsoCQCVQVU22212121===using C = Q/ VGraphical interpretation of integration∫=QVdqU0where V = q/CQdqq/cqVV = q/c∑=Δ=NiiiqqCU11= Area under the triangleArea under the triangle is the value of the integraldqCqQ∫0Area of