Physics 2102 Lecture: 08 THU 11 FEB Capacitance I Physics 2102 Jonathan Dowling 25.1–4
Capacitors and Capacitance Capacitor: any two conductors, one with charge +Q, other with charge –Q +Q –Q Uses: storing and releasing electric charge/energy. Most electronic capacitors: micro-Farads ( F), pico-Farads (pF) — 10 –12 F New technology: compact 1 F capacitors Uses: storing and releasing electric charge/energy. Most electronic capacitors: micro-Farads ( F), pico-Farads (pF) — 10 –12 F New technology: compact 1 F capacitors Potential DIFFERENCE between conductors = V C Q = CV where C = capacitance Units of capacitance: Farad (F) = Coulomb/Volt
Capacitance Capacitance depends only on GEOMETRICAL factors and on the MATERIAL that separates the two conductors e.g. Area of conductors, separation, whether the space in between is filled with air, plastic, etc. +Q –Q (We first focus on capacitors where gap is filled by AIR!)
Electrolytic ( ) Electrolytic (new) Paper ( ) Tantalum (1980 on) Ceramic (1930 on) Mica ( ) Variable air, mica Capacitors
Parallel Plate Capacitor -Q E field between the plates: (Gauss’ Law) Relate E to potential difference V: What is the capacitance C ? Area of each plate = A Separation = d charge/area = = Q/A We want capacitance: C = Q/V +Q
Capacitance and Your iPhone!
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? Lesson: difficult to get large values of capacitance without special tricks! Lesson: difficult to get large values of capacitance without special tricks! C = 0 A/d = (8.85 x 10 –12 F/m)(0.25 m 2 )/(0.001 m) = 2.21 x 10 –9 F (Very Small!!)
Isolated Parallel Plate Capacitor A parallel plate capacitor of capacitance C is charged using a battery. 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? (c) Decrease? +Q –Q Q is fixed! C decreases (= 0 A/d) V=Q/C; V increases.
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. +Q –Q V is fixed by battery! C decreases (= 0 A/d) Q=CV; Q decreases E = Q/ 0 A decreases Does the Electric Field Inside: (a) Increase? (b) Remain the Same? (c) Decrease?
Spherical Capacitor What is the electric field inside the capacitor? (Gauss’ Law) Radius of outer plate = b Radius of inner plate = a Concentric spherical shells: Charge +Q on inner shell, –Q on outer shell Relate E to potential difference between the plates:
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? Gauss’ Law! Relate E to potential difference between the plates: Radius of outer plate = b Radius of inner plate = a cylindrical Gaussian surface of radius r Length of capacitor = L +Q on inner rod, –Q on outer shell
Summary Any two charged conductors form a capacitor. Capacitance : C= Q/V Simple Capacitors: Parallel plates: C = 0 A/d Spherical: C = 4 e 0 ab/(b-a) Cylindrical: C = 2 0 L/ln(b/a)]
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 AB = V CD = V Q total = Q 1 + Q 2 C eq V = C 1 V + C 2 V C eq = C 1 + C 2 Equivalent parallel capacitance = sum of capacitances AB C D C1C1 C2C2 Q1Q1 Q2Q2 C eq Q total V = V AB = V A –V B V = V CD = V C –V D VAVA VBVB VCVC VDVD V=V PAR-V (Parallel V the Same)
Capacitors in Series: Q=Constant Q 1 = Q 2 = Q = Constant V AC = V AB + V BC A BC C1C1 C2C2 Q1Q1 Q2Q2 C eq Q = Q 1 = Q 2 SERIES: Q is same for all capacitors Total potential difference = sum of V Isolated Wire: Q=Q 1 =Q 2 =Constant SERI-Q (Series Q the Same)
Capacitors in parallel and in series In series : 1/C ser = 1/C 1 + 1/C 2 V ser = V 1 + V 2 Q ser = Q 1 = Q 2 C1C1 C2C2 Q1Q1 Q2Q2 C1C1 C2C2 Q1Q1 Q2Q2 In parallel : C par = C 1 + C 2 V par = V 1 = V 2 Q par = Q 1 + Q 2 C eq Q eq
Example: Parallel or Series? What is the charge on each capacitor? C 1 =10 F C 3 =30 F C 2 =20 F 120V Q i = C i V V = 120V = Constant Q 1 = (10 F)(120V) = 1200 C Q 2 = (20 F)(120V) = 2400 C Q 3 = (30 F)(120V) = 3600 C Note that: Total charge (7200 C) is shared between the 3 capacitors in the ratio C 1 :C 2 :C 3 — i.e. 1:2:3 Parallel: Circuit Splits Cleanly in Two (Constant V)
Example: Parallel or Series What is the potential difference across each capacitor? C 1 =10 F C 3 =30 F C 2 = 20 F 120V Q = C ser V Q is same for all capacitors Combined C ser is given by: C eq = 5.46 F (solve above equation) Q = C eq V = (5.46 F)(120V) = 655 C V 1 = Q/C 1 = (655 C)/(10 F) = 65.5 V V 2 = Q/C 2 = (655 C)/(20 F) = V V 3 = Q/C 3 = (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) Series: Isolated Islands (Constant Q)
Example: Series or Parallel? In the circuit shown, what is the charge on the 10 F capacitor? 10 F 10V 10 F 5F5F 5F5F 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 Neither: Circuit Compilation Needed!