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

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 V = VAB = VA –VB V = VCD = VC –VD VA VB VC VD A B C D C1 C2 Q1 Q2 V=V Ceq Qtotal

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

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

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

Example: Parallel or Series Series: Isolated Islands (Constant Q) What is the potential difference across each capacitor? C1=10mF C2=20mF C3=30mF 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)

Example: Series or Parallel? Neither: Circuit Compilation Needed! 10F 5F 10V In the circuit shown, what is the charge on the 10F capacitor? 10 F 10F 10V 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

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 = volume = Ad General expression for any region with vacuum (or air)

Example 10mF (C1) Initial charge on 10mF = (10mF)(120V)= 1200mC 10mF 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)= 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

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