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Physics 2113 Jonathan Dowling Physics 2113 Lecture: 17 WED 25 FEB Capacitance II 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! V = VAB = VA –VB Q1 A • VAB = VCD = V • Qtotal = Q1 + Q2 • CeqV = C1V + C2V • Ceq = C1 + C2 C C1 VA Q2 VB C2 VC VD B D V = VCD = VC –VD • Equivalent parallel capacitance = sum of capacitances Cparallel = C1 + C2 PAR-V (Parallel: V the Same) Qtotal V=V Ceq Capacitors in Series: Q=Constant • Q1 = Q2 = Q = Constant • VAC = VAB + VBC Isolated Wire: Q=Q1=Q2=Constant Q1 SERI-Q: Series Q the Same Q Q Q = + Ceq C1 C2 1 Cseries B A 1 1 = + C1 C2 SERIES: • Q is same for all capacitors • Total potential difference = sum of V Q2 C1 C C2 Q = Q 1 = Q2 Ceq Capacitors in Parallel and in Series • In parallel : Cpar = C1 + C2 Vpar = V1 = V2 Qpar = Q1 + Q2 Q1 C1 Qeq Q2 C2 Ceq • In series : 1/Cser = 1/C1 + 1/C2 Vser = V1 + V2 Qser= Q1 = Q2 Q1 Q2 C1 C2 Example: Parallel or Series? Parallel: Circuit Splits Cleanly in Two (Constant V) What is the charge on each capacitor? • Qi = CiV • V = 120V on ALL Capacitors (PAR-V) • Q1 = (10 F)(120V) = 1200 C • Q2 = (20 F)(120V) = 2400 C • Q3 = (30 F)(120V) = 3600 C Note that: • Total charge (7200 C) is shared between the 3 capacitors in the ratio C1:C2:C3 — i.e. 1:2:3 C1=10 F C2=20 F C3=30 F 120V Cpar = C1 + C2 + C3 = (10 + 20 + 30)mF = 60mF Example: Parallel or Series Series: Isolated Islands (Constant Q) What is the potential difference across each capacitor? • Q = CserV • Q is same for all capacitors (SERI-Q) • Combined Cser is given by: 1 1 1 1 = + + Cser (10mF) (20mF) (30mF) C1=10mF C2=20mF C3=30mF 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 Simplification Needed! In the circuit shown, what is the charge on the 10 F capacitor? 5 F • 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 by symmetry • Hence, Q = (10 F )(5V) = 50 C 10 F 5 F 10V 10 F 10 F 10V 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 Q Q q Q CV U = ò Vdq = ò dq = = C 2C 2 0 0 2 2 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 = 2 2 æ ö Q e Q e E Q Q 0 0 = ç ÷ u= = = = 2CAd 2æç e 0 A ö÷ Ad 2e 0 A2 2 çè e 0 A ÷ø 2 2 2 2 è d ø volume = Ad General expression for any region with vacuum (or air) Dielectric Constant • If the space between capacitor plates is filled by a dielectric, the capacitance DIELECTRIC INCREASES by a factor • This is a useful, working definition for dielectric constant. • Typical values of are 10–200 but it is always greater than 1! +Q –Q C = 0 A/d The and the constant o are both called dielectric constants. The has no units (dimensionless). Trick: Just substitute o for o in all the previous formulas! Atomic View Emol Molecules set up counter E field Emol that somewhat cancels out capacitor field Ecap. This avoids sparking (dielectric breakdown) by keeping field inside dielectric small. Ecap Hence the bigger the dielectric constant the more charge you can store on the capacitor. Example: Battery Connected — Voltage V is Constant but Charge Q Changes • Capacitor has charge Q, voltage V • Battery remains connected while dielectric slab is inserted. • Do the following increase, dielectric decrease or stay the same: slab – Potential difference? – Capacitance? – Charge? – Electric field? Example: Battery Connected — Voltage V is Constant but Charge Q Changes • 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). dielectric slab • Since Vnew = V, Enew = V/d=E (same) Energy stored? u=0E2/2 => u= 0E2/2 = E2/2 increases Example: Battery Disconnected — Voltage V Changes but Charge Q is Constant • Capacitor has charge Q, voltage V • Battery remains is disconnected then dielectric slab is inserted. • Do the following increase, decrease or stay the same: – Potential difference? – Capacitance? – Charge? – Electric field? dielectric slab Example: Battery Disconnected — Voltage V Changes but Charge Q is Constant • Initial values: capacitance = C; charge = Q; potential difference = V; electric field = E; • Battery remains disconnected • Q is FIXED; Qnew = Q (same) • Cnew = C (increases) • Vnew = Q/Cnew = Q/( C) (decreases). dielectric slab • Since Vnew < V, Enew = Vnew/d = E/ (decreases) Energy stored? u = e E 2 / 2 0 ® ke 0 E 2 new / 2 = ke 0 ( E / k ) / 2 = e 0 E 2 / (k 2 ) decreases 2 Summary • Any two charged conductors form a capacitor. • Capacitance : C= Q/V • Simple Capacitors: Parallel plates: C = 0 A/d Spherical : 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 C’=κC