Physics 2102 Lecture: 12 MON FEB

Physics 2102 Jonathan Dowling Physics 2102 Lecture: 12 MON FEB Capacitance II 25.4–5

V = VAB = VA –VB C1 Q1 VA VB A B C2 Q2 VC VD D C V = VCD = VC –VD Ceq Qtotal V=V Capacitors in Parallel: V=Constant • An ISOLATED wire is an equipotential surface: V=Constant • Capacitors in parallel have SAME potential difference but NOT ALWAYS same charge! • VAB = VCD = V • Qtotal = Q1 + Q2 • CeqV = C1V + C2V • Ceq = C1 + C2 • Equivalent parallel capacitance = sum of capacitances

Q1 Q2 B C A C1 C2 Capacitors in Series: Q=Constant Isolated Wire: Q=Q1=Q2=Constant • Q1 = Q2 = Q = Constant • VAC = VAB + VBC Q = Q1 = Q2 • SERIES: • Q is same for all capacitors • Total potential difference = sum of V Ceq

C1 Q1 C2 Q2 Qeq Q1 Q2 Ceq C1 C2 Capacitors in parallel and in series • In parallel : • Cpar = C1 + C2 • Vpar= V1 = V2 • Qpar= Q1 + Q2 • In series : 1/Cser = 1/C1 + 1/C2 Vser= V1 + V2 Qser= Q1 = Q2

Parallel: Circuit Splits Cleanly in Two (Constant V) C1=10 F C2=20 F C3=30 F 120V Example: Parallel or Series? • Qi = CiV • V = 120V = Constant • Q1 = (10 F)(120V) = 1200 C • Q2 = (20 F)(120V) = 2400 C • Q3 = (30 F)(120V) = 3600 C What is the charge on each capacitor? • Note that: • Total charge (7200 mC) is shared between the 3 capacitors in the ratio C1:C2:C3— i.e. 1:2:3

Series: Isolated Islands (Constant Q) Example: Parallel or Series C2=20mF C3=30mF C1=10mF What is the potential difference across each capacitor? • Q = CserV • Q is same for all capacitors • Combined Cser is given by: 120V • Ceq = 5.46 F (solve above equation) • Q = CeqV = (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 V Note: 120V is shared in the ratio of INVERSE capacitances i.e. (1):(1/2):(1/3) (largest C gets smallest V)

10 F 10F 10F 10V 5F 5F 10V Example: Series or Parallel? Neither: Circuit Compilation Needed! In the circuit shown, what is the charge on the 10F capacitor? • The two 5F capacitors are in parallel • Replace by 10F • Then, we have two 10F capacitors in series • So, there is 5V across the 10 F capacitor of interest • Hence, Q = (10F )(5V) = 50C

Energy U 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

General expression for any region with vacuum (or air) volume = Ad Energy Stored in Electric Field of Capacitor • 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 =

10mF (C1) 20mF (C2) Example • 10mFcapacitor is initially charged to 120V. • 20mF capacitor is initially uncharged. • Switch is closed, equilibrium is reached. • How much energy is dissipated in the process? Initial charge on 10mF = (10mF)(120V)= 1200mC After switch is closed, let charges = Q1 and Q2. Charge is conserved: Q1 + Q2 = 1200mC • Q1 = 400mC • Q2 = 800mC • Vfinal= Q1/C1 = 40 V Also, Vfinal is same: Initial energy stored = (1/2)C1Vinitial2 = (0.5)(10mF)(120)2 = 72mJ Final energy stored = (1/2)(C1 + C2)Vfinal2= (0.5)(30mF)(40)2 = 24mJ Energy lost (dissipated) = 48mJ

Any two charged conductors form a capacitor. • Capacitance : C= Q/V • Simple Capacitors:Parallel plates: C = e0 A/d Spherical : C = 4pe0 ab/(b-a)Cylindrical: C = 2pe0 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 densityu=e0E2/2 • Capacitor with a dielectric: capacitance increasesC’=kC Summary

Physics 2102 Lecture: 12 MON FEB