Physics 2113 Jonathan Dowling Physics 2113 Lecture: 16 MON 23 FEB Capacitance I
Capacitors and Capacitance Capacitor: any two conductors, one with charge +Q, other with charge Q Potential DIFFERENCE between conductors = V +Q Q Q = CV where C = capacitance For fixed Q the larger the capacitance the larger the stored voltage. Units of capacitance: Farad (F) = Coulomb/Volt Uses: storing and releasing electric charge/energy. Most electronic capacitors: micro-farads (µf), pico-farads (pf) 10 12 F New technology: compact 1 F capacitors
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!)
Capacitance depends only on GEOMETRICAL factors and on the MATERIAL that separates the two conductors. It does not depend on the voltage across it or the charge on it at all. (a) Same (b) Same
Electrolytic (1940-70) Electrolytic (new) Paper (1940-70) Capacitors Variable air, mica Tantalum (1980 on) Ceramic (1930 on) Mica (1930-50)
Parallel Plate Capacitor We want capacitance: C = Q/V E field between conducting plates: E = σ Q = ε 0 ε 0 A Area of each plate = A Separation = d charge/area = σ = Q/A Relate E to potential difference V: V d!! = E dx 0 d 0 = σ ε 0 dx = Qd ε 0 A What is the capacitance C? Q V ε A d 0 C = = d C 2 m 2 Units : Nm 2 = C2 = CC m Nm J +Q σ = C V Q
Capacitance and Your iphone! Q V ε A d 0 C = =
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 = ε 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!!) C 2 m 2 Units : Nm 2 = C2 = CC m Nm J Lesson: difficult to get large values of capacitance without special tricks! = C V F [ ] = Farad
Isolated Parallel Plate Capacitor: ICPP Q V A parallel plate capacitor of capacitance C is charged using a battery. Charge = Q, potential voltage difference = V. Battery is then disconnected. If the plate separation is INCREASED, does the capacitance C: (a) Increase? Q is fixed! d increases! (b) Remain the same? C decreases (= ε (c) Decrease? 0 A/d) V=Q/C; V increases. If the plate separation is INCREASED, does the capacitance C: (a) Increase? (b) Remain the same? (c) Decrease? ε A d 0 C = = +Q Q
Parallel Plate Capacitor & Battery: ICPP 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? Q V ε A d 0 C = = E V is fixed constant by battery! C decreases (=ε 0 A/d) Q=CV; Q decreases E = σ/ε 0 = Q/ε 0 A decreases Battery does work on capacitor to maintain constant V! = σ Q = ε 0 ε 0 A +Q Q
Spherical Capacitor V What is the electric field inside the capacitor? (Gauss Law) E = Q 4πε r Relate E to potential difference between the plates: b! = E dr a 0 2! b b a Radius of outer plate = b Radius of inner plate = a Concentric spherical shells: Charge +Q on inner shell, Q on outer shell kq kq = dr = 1 1 2 r r = kq a b a
Spherical Capacitor What is the capacitance? C = Q/V = = Q πε 4 0 Q 1 a 4 0 = πε ( b ab a) 1 b 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, C = 4πε 0a
Cylindrical Capacitor V What is the electric field in between the plates? Gauss Law! E = Q 2πε 0 rl Relate E to potential difference between the plates: b!! = E dr a b = a C = Q/V = 2πε L 0 ln b a Q Q ln dr = 2πε 0rL 2πε 0 r L b a Radius of outer plate = b Radius of inner plate = a Length of capacitor = L +Q on inner rod, Q on outer shell = Q 2πε 0 L ln b a cylindrical Gaussian surface of radius r
Summary Any two charged conductors form a capacitor. Capacitance : C= Q/V Simple Capacitors: Parallel plates: Spherical: Cylindrical: C = ε 0 A/d C = 4πε 0 ab/(b-a) C = 2πε 0 L/ln(b/a)]
Parallel plates: Spherical: Cylindrical: C = C = ε 0 A/d 1 1 1 4πε 0 a 1 b C = 2πε 0 L/ln(b/a)] Hooked to battery V is constant. Q=VC (a) d increases -> C decreases -> Q decreases (b) a inc -> separation d=b-a dec. -> C inc. -> Q increases (c) b increases -> separation d=b-a inc.-> C dec. -> Q decreases
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 C parallel = C 1 + C 2 PAR-V (Parallel V the Same) PARALLEL: V is same for all capacitors Total charge = sum of Q V = V AB = V A V B A C Q 1 V A C 1 V B Q 2 C 2 V C V D Q total C eq B D V = V CD = V C V D ΔV=V
Capacitors in Series: Q=Constant Q 1 = Q 2 = Q = Constant V AC = V AB + V BC Isolated Wire: Q=Q 1 =Q 2 =Constant SERI-Q (Series Q the Same) Q Ceq = Q C + Q 1 C 2 A Q 1 Q 2 B C 1 C 2 C 1 C series = 1 C 1 + 1 C 2 Q = Q 1 = Q 2 SERIES: Q is same for all capacitors Total potential difference = sum of V C eq
PARALLEL: V is same for all capacitors Total charge = sum of Q SERIES: Q is same for all capacitors Total potential difference = sum of V (a) Parallel: Voltage is same V on each but charge is q/2 on each since q/2+q/2=q. (b) Series: Charge is same q on each but voltage is V/2 on each since V/2+V/2=V.
Compiling in parallel and in series In parallel : C par = C 1 + C 2 V par = V 1 = V 2 Q par = Q 1 + Q 2 Q 1 Q 2 C 1 C 2 Q eq C eq In series : Q 1 Q 2 1/C ser = 1/C 1 + 1/C 2 V ser = V 1 + V 2 Q ser = 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 = 120V = Constant Q 1 = (10 µf)(120v) = 1200 µc Q 2 = (20 µf)(120v) = 2400 µc Q 3 = (30 µf)(120v) = 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 120V C par = C 1 + C 2 + C 3 = 10 + 20 + 30 ( )µf = 60µF
Example: Parallel or Series Series: Isolated Islands (Constant Q) What is the potential difference across each capacitor? Q = C ser V Q is same for all capacitors Combined C ser is given by: C 1 =10µF C 2 =20µF C 3 =30µF 1 C ser = 1 (10µF) + 1 (20µF) + 1 (30µF) 120V C eq = 5.46 µf (solve above equation) Q = C eq V = (5.46 µf)(120v) = 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: 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 Compilation Needed! 10µF In the circuit shown, what is the charge on the 10µF capacitor? 5µF 5µF 10V The two 5µF capacitors are in parallel Replace by 10µF Then, we have two 10µF capacitors in series So, there is 5V across the 10 µf capacitor of interest Hence, Q = (10µF )(5V) = 50µC 10µF 10 µf 10V