Physics 2113 Jonathan Dowling Physics 2113 Lecture 06

Physics 2113 Jonathan Dowling Physics 2113 Lecture: 06 WED 01 OCT 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! • VAB = VCD = V • Qtotal = Q 1 + Q 2 • Ceq. V = C 1 V + C 2 V • Ceq = C 1 + C 2 V = VAB = VA –VB A Q 1 C 1 VA Q 2 VB C 2 C VD B D V = VCD = VC –VD • Equivalent parallel capacitance = sum of capacitances Qtotal PAR-V (Parallel: V the Same) ΔV=V Ceq

Capacitors in Series: Q=Constant • Q 1 = Q 2 = Q = Constant • VAC = VAB + VBC Isolated Wire: Q=Q 1=Q 2=Constant Q 1 SERI-Q: Series Q the Same Q 2 C B A C 1 C 2 Q = Q 1 = Q 2 SERIES: • Q is same for all capacitors • Total potential difference = sum of V Ceq

Capacitors in Parallel and in Series • In parallel : Cpar = C 1 + C 2 Vpar = V 1 = V 2 Qpar = Q 1 + Q 2 • In series : 1/Cser = 1/C 1 + 1/C 2 Vser = V 1 + V 2 Qser= Q 1 = Q 2 Q 1 C 1 Qeq Q 2 Ceq Q 1 Q 2 C 1 C 2

Example: Parallel or Series? Parallel: Circuit Splits Cleanly in Two (Constant V) What is the charge on each capacitor? • Q i = C i. V • V = 120 V on ALL Capacitors (PAR-V) • Q 1 = (10 μF)(120 V) = 1200 μC • Q 2 = (20 μF)(120 V) = 2400 μC • Q 3 = (30 μF)(120 V) = 3600 μC Note that: • Total charge (7200 μC) is shared between the 3 capacitors in the ratio C 1: C 2: C 3 — i. e. 1: 2: 3 C 1=10 μF C 2=20 μF C 3=30 μF 120 V

Example: Parallel or Series: Isolated Islands (Constant Q) What is the potential difference across each capacitor? • Q = Cser. V • Q is same for all capacitors (SERI-Q) • Combined Cser is given by: C 1=10 m. F C 2=20 m. F C 3=30 m. F 120 V • Ceq = 5. 46 μF (solve above equation) • Q = Ceq. V = (5. 46 μF)(120 V) = 655 μC • V 1= Q/C 1 = (655 μC)/(10 μF) = 65. 5 V • V 2= Q/C 2 = (655 μC)/(20 μF) = 32. 75 V • V 3= Q/C 3 = (655 μC)/(30 μF) = 21. 8 V Note: 120 V 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! In the circuit shown, what is the charge on the 10μF capacitor? 5μF • The two 5μF capacitors are in parallel • Replace by 10μF • Then, we have two 10μF capacitors in series • So, there is 5 V across the 10 μF capacitor of interest by symmetry • Hence, Q = (10μF )(5 V) = 50μC 10μF 5μF 10 V 10 μF 10 V

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 = Q 2/(2 C) = CV 2/2 • 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)

Dielectric Constant • If the space between capacitor plates is DIELECTRIC filled by a dielectric, the capacitance INCREASES by a factor κ • This is a useful, working definition for dielectric constant. • Typical values of κ are 10– 200 +Q –Q C = κε 0 A/d The κ and the constant ε =κε o are both called dielectric constants. The κ has no units (dimensionless).

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 • 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 • Initial values: capacitance = C; charge = Q; potential difference = V; electric field = E; • Battery remains connected • V is FIXED; Vnew = V (same) • Cnew = κC (increases) • Qnew = (κC)V = κQ (increases). dielectric slab • Since Vnew = V, Enew = V/d=E (same) Energy stored? u=ε 0 E 2/2 => u=κ e 0 E 2/2 = ε E 2/2

Summary • Any two charged conductors form a capacitor. • Capacitance : C= Q/V • Simple Capacitors: Parallel plates: C = ε 0 A/d Spherical : C = 4π ε 0 ab/(b-a) Cylindrical: C = 2π ε 0 L/ln(b/a) • Capacitors in series: same charge, not necessarily equal potential; equivalent capacitance 1/Ceq=1/C 1+1/C 2+… • Capacitors in parallel: same potential; not necessarily same charge; equivalent capacitance Ceq=C 1+C 2+… • Energy in a capacitor: U=Q 2/2 C=CV 2/2; energy density u=ε 0 E 2/2 • Capacitor with a dielectric: capacitance increases C’=κC