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Physics 2102 Jonathan Dowling Physics 2102 Lecture 8 Capacitors II Capacitors in parallel and in series • In parallel : – Ceq = C1 + C2 – Veq=V1=V2 – Qeq=Q1+Q2 • In series : – 1/Ceq = 1/C1 + 1/C2 – Veq=V1 +V2 – Qeq=Q1=Q2 Q1 C1 Q2 C2 Qeq Q1 Q2 C1 C2 Ceq Example 1 What is the charge on each capacitor? • Q = CV; V = 120 V • Q1 = (10 mF)(120V) = 1200 mC • Q2 = (20 mF)(120V) = 2400 mC • Q3 = (30 mF)(120V) = 3600 mC Note that: • Total charge (7200 mC) is shared between the 3 capacitors in the ratio C1:C2:C3 — i.e. 1:2:3 10 mF 20 mF 30 mF 120V Example 2 What is the potential difference across each capacitor? • Q = CV; Q is same for all capacitors 10 mF • Combined C is given by: 1 1 1 1 Ceq (10mF ) (20mF ) (30mF ) 20 mF 120V 30 mF • Ceq = 5.46 mF • Q = CV = (5.46 mF)(120V) = 655 mC • V1= Q/C1 = (655 mC)/(10 mF) = 65.5 V Note: 120V is shared in the • V2= Q/C2 = (655 mC)/(20 mF) = 32.75 V ratio of INVERSE capacitances • V3= Q/C3 = (655 mC)/(30 mF) = 21.8 V i.e.1:(1/2):(1/3) (largest C gets smallest V) Example 3 10 mF In the circuit shown, what is the charge on the 10mF capacitor? 5 mF • The two 5mF capacitors are in parallel • Replace by 10mF • Then, we have two 10mF capacitors in series • So, there is 5V across the 10mF capacitor of interest • Hence, Q = (10mF )(5V) = 50mC 5 mF 10V 10 mF 10 mF 10V Energy 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? Q Q dq 2 2 q Q CV U Vdq dq C 2C 2 0 0 Energy 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 = 2 Q Q 0 Q 0E 2 Q U 2 2CAd 2 0 A Ad 2 0 A 2 A 2 0 2 2 2 d volume = Ad General expression for any region with vacuum (or air) Example • 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) Initial charge on 10mF = (10mF)(120V)= 1200mC 20mF (C2) After switch is closed, let charges = Q1 and Q2. Charge is conserved: Q1 + Q2 = 1200mC • Q1 = 400mC Q2 Also, Vfinal is same: Q1 Q2 • Q2 = 800mC Q1 C1 C 2 2 • Vfinal= Q1/C1 = 40 V 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 Dielectric Constant DIELECTRIC +Q -–Q C = A/d • 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 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 slab Example • 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 slab Energy stored? u=0E2/2 => u=0E2/2 = E2/2 Summary • Any two charged conductors form a capacitor. • Capacitance : C= Q/V • Simple Capacitors: Parallel plates: C = 0 A/d Spherical : C = 4p 0 ab/(b-a) Cylindrical: C = 2p 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’=C