Physics 2102 Gabriela Gonzlez Physics 2102 Capacitors Capacitors

Physics 2102 Gabriela González Physics 2102 Capacitors

Capacitors and Capacitance Capacitor: any two conductors, one with charge +Q, other with charge -Q -Q Potential DIFFERENCE between +Q Uses: storing and releasing conductors = V Q = CV -- C = capacitance Units of capacitance: Farad (F) = Coulomb/Volt electric charge/energy. Most electronic capacitors: micro-Farads (m. F), pico-Farads (p. F) -- 10 -12 F New technology: compact 1 F capacitors

Capacitance • Capacitance depends only on GEOMETRICAL factors and on the MATERIAL that separates the two +Q conductors -Q • e. g. Area of conductors, separation, whether the space in between is filled (We first focus on capacitors with air, plastic, etc. where gap is filled by AIR!)

Electrolytic (1940 -70) Electrolytic (new) Paper (1940 -70) Capacitors Variable air, mica Tantalum (1980 on) Ceramic (1930 on) Mica (1930 -50

Capacitors and Capacitance Capacitor: any two conductors, one with charge +Q, other with charge -Q +Q -Q Potential DIFFERENCE between Uses: storing and releasing conductors = V Q = CV C = capacitance Units of capacitance: Farad (F) = Coulomb/Volt electric charge/energy. Most electronic capacitors: micro-Farads (m. F), pico-Farads (p. F) -- 10 -12 F New technology: compact 1 F capacitors

Parallel Plate Capacitor We want capacitance: C=Q/V E field between the plates: (Gauss’ Law) Area of each plate = A Separation = d charge/area = s= Q/A Relate E to potential difference V: What is the capacitance C? +Q -Q

Parallel Plate Capacitor -- example • A huge parallel plate capacitor consists of two square metal plates of side 50 cm, separated by an air gap of 1 mm • What is the capacitance? • C = 0 A/d = (8. 85 x 10 -12 F/m)(0. 25 m 2)/(0. 001 m) = 2. 21 x 10 -9 F (small!!) Lesson: difficult to get large values of capacitance without special tricks!

Isolated Parallel Plate Capacitor • A parallel plate capacitor of capacitance C is charged using a battery. +Q • Charge = Q, potential difference = V. • Battery is then disconnected. • If the plate separation is INCREASED, does potential difference V: (a) Increase? (b) Remain the same? • Q is fixed! (c) Decrease? • C decreases (= 0 A/d) • Q=CV; V increases. -Q

Parallel Plate Capacitor & Battery • A parallel plate capacitor of capacitance C is charged using a battery. • Charge = Q, potential difference = V. • Plate separation is INCREASED while battery remains connected. Does the electric field inside: (a) Increase? (b) Remain the same? (c) Decrease? • V is fixed by battery! • C decreases (= 0 A/d) • Q=CV; Q decreases • E = Q/ 0 A decreases +Q -Q

Spherical Capacitor What is the electric field inside the capacitor? (Gauss’ Law) Relate E to potential difference between the plates: Radius of outer plate = b Radius of inner plate = a Concentric spherical shells: Charge +Q on inner shell, -Q on outer shell

Spherical Capacitor What is the capacitance? C = Q/V = Radius of outer plate = b Radius of inner plate = a Concentric spherical shells: Charge +Q on inner shell, -Q on outer shell Isolated sphere: let b >> a,

Cylindrical Capacitor What is the electric field in between the plates? Relate E to potential difference between the plates: Radius of outer plate = b Radius of inner plate = a Length of capacitor = L +Q on inner rod, -Q on outer shell cylindrical surface of radius r

Cylindrical Capacitor What is the capacitance C? C = Q/V = Radius of outer plate = b Radius of inner plate = a Length of capacitor = L Charge +Q on inner rod, -Q on outer shell Example: co-axial cable.

Summary • Any two charged conductors form a capacitor. • Capacitance : C= Q/V • Simple Capacitors: Parallel plates: C = 0 A/d Spherical : C = 4 p 0 ab/(b-a) Cylindrical: C = 2 p 0 L/ln(b/a)

Capacitors in Parallel • A wire is a conductor, so it is an equipotential. • 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 • Equivalent parallel capacitance = sum of capacitances PARALLEL: • V is same for all capacitors • Total charge in Ceq = sum of charges A C Q 1 C 1 Q 2 C 2 Qtotal B D Ceq

Capacitors in series • Q 1 = Q 2 = Q (WHY? ? ) • VAC = VAB + VBC Q 1 Q 2 B A C 1 C 2 Q SERIES: • Q is same for all capacitors • Total potential difference in Ceq = sum of V C Ceq

Capacitors in parallel and in series • In parallel : – Ceq = C 1 + C 2 – Veq=V 1=V 2 – Qeq=Q 1+Q 2 C 1 Qeq C 2 Q 2 Ceq • In series : – 1/Ceq = 1/C 1 + 1/C 2 – Veq=V 1 +V 2 – Qeq=Q 1=Q 2 Q 1 Q 2 C 1 C 2

Example 1 What is the charge on each capacitor? • Q = CV; V = 120 V • Q 1 = (10 m. F)(120 V) = 1200 m. C • Q 2 = (20 m. F)(120 V) = 2400 m. C • Q 3 = (30 m. F)(120 V) = 3600 m. C Note that: • Total charge (7200 m. C) is shared between the 3 capacitors in the ratio C 1: C 2: C 3 -- i. e. 1: 2: 3 10 m. F 20 m. F 30 m. F 120 V

Example 2 What is the potential difference across each capacitor? • Q = CV; Q is same for all capacitors • Combined C is given by: 10 m. F 20 m. F 120 V 30 m. F • Ceq = 5. 46 m. F • Q = CV = (5. 46 m. F)(120 V) = 655 m. C • V 1= Q/C 1 = (655 m. C)/(10 m. F) = 65. 5 V Note: 120 V is shared in the • V 2= Q/C 2 = (655 m. C)/(20 m. F) = 32. 75 V ratio of INVERSE capacitances • V 3= Q/C 3 = (655 m. C)/(30 m. F) = 21. 8 V i. e. 1: (1/2): (1/3) (largest C gets smallest V)

Example 3 10 m. F In the circuit shown, what is the charge on the 10 F capacitor? 5 m. 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 5 V across the 10 F capacitor of interest • Hence, Q = (10 F )(5 V) = 50 C 5 m. F 10 V 10 m. F 10 V

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? dq

Energy Stored in Electric Field • 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)

Example • 10 F capacitor is initially charged to 120 V. 20 F capacitor is initially uncharged. • Switch is closed, equilibrium is reached. • How much energy is dissipated in the process? Initial charge on 10 m. F = (10 m. F)(120 V)= 1200 m. C 10 F (C 1) 20 F (C 2) After switch is closed, let charges = Q 1 and Q 2. Charge is conserved: Q 1 + Q 2 = 1200 m. C • Q 1 = 400 m. C Also, Vfinal is same: • Q 2 = 800 m. C • Vfinal= Q 1/C 1 = 40 V Initial energy stored = (1/2)C 1 Vinitial 2 = (0. 5)(10 m. F)(120)2 = 72 m. J Final energy stored = (1/2)(C 1 + C 2)Vfinal 2 = (0. 5)(30 m. F)(40)2 = 24 m. J Energy lost (dissipated) = 48 m. J

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

Summary • Capacitors in series and in parallel: • in series: charge is the same, potential adds, equivalent capacitance is given by 1/C=1/C 1+1/C 2 • in parallel: charge adds, potential is the same, equivalent capaciatnce is given by C=C 1+C 2. • Energy in a capacitor: U=Q 2/2 C=CV 2/2; energy density u= 0 E 2/2 • Capacitor with a dielectric: capacitance increases C’= C