Physics 2102 Lecture 8Capacitors in parallel and in seriesExample 1Example 2Example 3Energy Stored in a CapacitorEnergy Stored in Electric FieldExampleDielectric ConstantSlide 10ExampleSummaryPhysics 2102 Physics 2102 Lecture 8Lecture 8Capacitors IICapacitors IIPhysics 2102Jonathan DowlingCapacitors in parallel and in seriesCapacitors in parallel and in series•In series : –1/Ceq = 1/C1 + 1/C2–Veq=V1 +V2–Qeq=Q1=Q2C1C2Q1Q2C1C2Q1Q2•In parallel : –Ceq = C1 + C2–Veq=V1=V2–Qeq=Q1+Q2CeqQeqExample 1Example 1What is the charge on each capacitor?10 F30 F20 F120V• Q = CV; V = 120 V• Q1 = (10 F)(120V) = 1200 C • Q2 = (20 F)(120V) = 2400 C• Q3 = (30 F)(120V) = 3600 CNote that:•Total charge (7200 C) is shared between the 3 capacitors in the ratio C1:C2:C3 — i.e. 1:2:3Example 2Example 2What is the potential difference across each capacitor?10 F30 F20 F120V• Q = CV; Q is same for all capacitors• Combined C is given by:)30(1)20(1)10(11FFFCeqμμμ++=• Ceq = 5.46 F• Q = CV = (5.46 F)(120V) = 655 C• V1= Q/C1 = (655 C)/(10 F) = 65.5 V• V2= Q/C2 = (655 C)/(20 F) = 32.75 V• V3= Q/C3 = (655 C)/(30 F) = 21.8 VNote: 120V is shared in the ratio of INVERSE capacitances i.e.1:(1/2):(1/3) (largest C gets smallest V)Example 3Example 3In the circuit shown, what is the charge on the 10F capacitor?10 F10 F10V10 F5 F5 F10V•The two 5F capacitors are in parallel•Replace by 10F •Then, we have two 10F capacitors in series•So, there is 5V across the 10F capacitor of interest•Hence, Q = (10F )(5V) = 50CEnergy Stored in a CapacitorEnergy Stored in a Capacitor•Start out with uncharged capacitor•Transfer small amount of charge dq from one plate to the other until charge on each plate has magnitude Q•How much work was needed?dq∫=QVdqU0∫==QCQdqCq02222CV=Energy Stored in Electric FieldEnergy Stored in Electric Field•Energy stored in capacitor:U = Q2/(2C) = CV2/2 •View the energy as stored in ELECTRIC FIELD•For example, parallel plate capacitor: Energy DENSITY = energy/volume = u ===CAdQU22=⎟⎠⎞⎜⎝⎛AddAQ022ε2022 AQε2220200EAQ εεε=⎟⎟⎠⎞⎜⎜⎝⎛=volume = AdGeneral expression for any region with vacuum (or air)Example Example •10F capacitor is initially charged to 120V. 20F capacitor is initially uncharged.•Switch is closed, equilibrium is reached.•How much energy is dissipated in the process?10F (C1)20F (C2)Initial energy stored = (1/2)C1Vinitial2 = (0.5)(10F)(120)2 = 72mJ Final energy stored = (1/2)(C1 + C2)Vfinal2 = (0.5)(30F)(40)2 = 24mJ Energy lost (dissipated) = 48mJInitial charge on 10F = (10F)(120V)= 1200CAfter switch is closed, let charges = Q1 and Q2. Charge is conserved: Q1 + Q2 = 1200CAlso, Vfinal is same: 2211CQCQ=221QQ =• Q1 = 400C• Q2 = 800C• Vfinal= Q1/C1 = 40 VDielectric ConstantDielectric Constant•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 are 10–200+Q-–QDIELECTRICC = A/dExample Example •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?dielectric slabExampleExample•Initial values: capacitance = C; charge = Q; potential difference = V; electric field = E;•Battery remains connected•V is FIXED; Vnew = V (same)•Cnew = C (increases)•Qnew = (C)V = Q (increases).•Since Vnew = V, Enew = E (same)dielectric slabEnergy stored? u=0E2/2 => u=0E2/2 = E2/2SummarySummary• Any two charged conductors form a capacitor.• Capacitance : C= Q/V• Simple Capacitors:Parallel plates: C = 0 A/dSpherical : C = 4 0 ab/(b-a)Cylindrical: C = 2 0 L/ln(b/a)• Capacitors in series: same charge, not necessarily equal potential; equivalent capacitance 1/Ceq=1/C1+1/C2+…• Capacitors in parallel: same potential; not necessarily same charge; equivalent capacitance Ceq=C1+C2+…• Energy in a capacitor: U=Q2/2C=CV2/2; energy density u=0E2/2• Capacitor with a dielectric: capacitance increases
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