SummarySlide 2Slide 3Slide 4CapacitanceSlide 6Capacitance, contParallel-Plate CapacitorSlide 9Parallel-plate CapacitorSlide 11Slide 12Slide 13(a) C decreases (b) Q stays the same (c) E stays the same (d) DV increasesDielectricDielectric Constant, K A measure of how effective it is in reducing an electric field across the platesSlide 17(a) C increases (b) Q increases (c) E stays the same (d) DV remains the sameElectric Field EnergySlide 20Energy Stored in a CapacitorA Summary of the various formulas for charge Q, potential difference V, capacitance and energy WSummarySummaryElectric potential: PE per unit charge, V=PE/qPotential difference: work done to move the charge from one point to another, Vab=Vb-Va=PE/q=Wab/q or Wab=qVabFor uniform electric field: Vab=-EdFor electric field due to a point charge: V=kQ/r, (V=0 at r=∞)VrElectric Potential due to a Point ChargeVr+V=V=kkQ/Q/r r (V=0 at r=∞)rrThe sketch shows cross sections of equipotential surfaces between two charged conductors shown in solid black. Points on the equipotential surfaces near the conductors are labeled A, B, C, ..., H.1. What is the magnitude of the potential difference between points A and H? (a) 100 V (b) 200 V (c) 400 V (d) 600 V (e) 700 V 2. What is the direction of the electric field at point E? (a) toward B (b) toward G (c) toward H (d) toward C (e) toward F3. How much work is required to move a +6.0 µC point charge from B to F to D to A?(a) -1.2x10–3 J (b) +1.2x10–3 J (c) +3.6x10–3 J (d) –3.6x10–3 J (e) zero joulesXXXThe sketch below shows cross section of equipotential surfaces between two charged conductors that are shown below in solid black. Various points on the equipotential surfaces near the conductors are labeled A,B,C…,I. At which of the labeled points will the electric field have the greatest magnitude? (a) G (c) A (e) D (b) I (d) H At which of the labeled points will an electron have the greatest potential energy? (a) A (c) G (e) I (b) D (d) H What is the potential difference between points B and E? (a) 10 V (c) 40 V (e) 60 V (b) 30 V (d) 50 V XXXCapacitanceCapacitanceCapacitor:Capacitor:A Device to Store Energy in the Form of an Electric FieldCapacitanceA capacitor is a device used in a variety of electric circuitsThe capacitance, C, of a capacitor is defined as the ratio of the magnitude of the charge on either conductor (plate) to the magnitude of the potential difference between the conductors (plates)Capacitance, cont Units: Farad (F)1 F = 1 C / VA Farad is very largeOften will see µF or pFVQCParallel-Plate CapacitorThe capacitance of a device depends on the geometric arrangement of the conductorsFor a parallel-plate capacitor whose plates are separated by air:dACo++++++++--------d+-Q -Q Capacitance C=Q/VUnit: farad, F. 1 F=1 coulomb/voltParallel-plate CapacitorParallel-plate CapacitorQ=AE/4πk=εoAE, where A is area of the plate, and εo is the permittivity of free space, 8.85x10-12 C2/N-m2Or E=charge density)/εo V=EdC=Q/V= εoAE/Ed=εoA/dProperties of Capacitance C: Constant for a given capacitor, independent of Q or VC depends on the geometry of the capacitor and the material between the plates.Question: The energy content of a charged capacitor resides in its (a) plates (b) potential difference(c) charge (d) electric fieldAnswer: dQuestion: The plates of a parallel-plate capacitor of capacitance C are brought together to one-third their original separation. The capacitance is now (a) C/9(b) C/3(c) 3C(d) 9CAnswer: cQUICK QUIZ 16.5You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires connected to the plates from touching each other. When you pull the plates farther apart, do the following quantities increase, decrease, or stay the same? (a) C; (b) Q; (c) E between the plates; (d) V.QUICK QUIZ 16.5 ANSWER(a) C decreases(b) Q stays the same(c) E stays the same (d) V increasesDielectricDielectric Insulating materialsCapacitance increased by a factor Dielectric Constant, KDielectric Constant, KA measure of how effective it is in reducing an electric field across the platesQ -QVoCo=Q/VoQ -Q+++V=Vo/KC=Q/V=KQ/Vo=KCoQUICK QUIZ 16.6A fully charged parallel-plate capacitor remains connected to a battery while you slide a dielectric between the plates. Do the following quantities increase, decrease, or stay the same? (a) C; (b) Q; (c) E between the plates; (d) V.QUICK QUIZ 16.6 ANSWER(a) C increases(b) Q increases(c) E stays the same (d) V remains the sameElectric Field Energy Work must be done to separate positive and negative charges against the Coulomb forces attracting them together. This work is stored as potential energy or electric field energy in a capacitor.V+-eV++++----eVinitialVfinalThe average potential difference during the charge transfer is:<V>=(Vfinal+Vinitial)/2 =(Vf+0)/2=Vf/2The total charge transferred is QThe work done is W=U=Q<V>=(1/2) QV =(1/2) CV2=(1/2) Q2/C(C=Q/V)Q-QEnergy Stored in a CapacitorEnergy stored = ½ Q ΔVFrom the definition of capacitance, this can be rewritten in different formsC2QVC21VQ21Energy22A Summary of the various formulas for charge Q, A Summary of the various formulas for charge Q, potential difference V, capacitance and energy Wpotential difference V, capacitance and energy WKnown QuantitiesUnknownQuantityC, V C, Q Q, V W, C W, V W, Q Q= CV(2WC)1/22W/V V= Q/C(2W/C)1/22W/Q C= Q/V2W/V2Q2/2W W=CV2/2
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