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Capacitance Capacitor- a device for storing charge and energy, can be discharged rapidly to release energy. Applications •Camera flash •automobile starting system •capacitors in electronic devices •computer memories (store information) •Laser flash lamp laser fusion Energy stored in a capacitor 2 x106 J released in 1-2 x10-6 s. High Power ~ 1012 W Parallel plate capacitor - + - - - + + - + + FIELD LINES IN BLACK (VECTORS) POTENTIAL CONTOURS IN RED (NO ARROWS, BECAUSE NOT A VECTOR) +q conductors + + + + + + + + + + + + + + + + + + + + + + - - - - - - - - - - - - - - - - - - - - - - - - - -q potential source Eg. battery Parallel Plate Capacitor Work is done to separate charges Capacitor Stores Electrical Energy ΔV Circuit Diagram Wire = conductor Voltage source + ΔV - conductor +q capacitor C -q Measuring voltage C V V Voltmeter Ideal voltmeter Draws no charge Perfect insulator Parallel Plate Capacitor Metal plates with area A, separated by a gap, d containing an insulating material. (eg. Air) A +q ΔV d -q Capacitance – describes the ability to store separated charge q C= !V Units C/V, farad (F) A 100 microfarad capacitor is charged to 100 V. How much charge does it store? q C= "V q = C( "V ) = 100 x10!6 (100) = 10!2 C This is a lot of charge. Recall that 10-2 C on a sphere of 1m radius generates a potential of q kq r=1m V = e = 9 x107V r How does the capacitor store this charge without high potentials? How do you get a large capacitance? Gaussian surface- a cylinder with sides perpendicular to the plane. E is constant at ends. Flux through sides is zero. q " E = 2 AE = !o E + + + + + + q E= 2 A! o + + + + + + + " E= 2! o + + E A where q ! = A Parallel plate capacitor two “infinite” planes of charge area A separated by distance d where d<< A, carry charge +q, -q A +q/A=+σ ++++++++++++ ----------------The charges are at the inner surface of the capacitor d -q/A=-σ Field inside the capacitor plates By superposition of charges due to sheet of charge Eout =0 E+ ! Ein= "o A q/A=σ ++++++++++++ ----------------- EEout =0 ! ! Ein=E+ +E- = 2 = 2" o " 0 d What is the capacitance of a parallel plate capacitor with plates of area A separated by distance d? (in air) +q A ΔV d -q Recall from Gauss’ law rearrange thus ! q #V E= = = " o A" o d A! o q "V = E field increases with charge density d A! o C = d to increase C Increase A Decrease d How do you stabilize the separated charge i.e. make ΔV as small as possible for a given q (high capacitance) +q A + + + + + + + + + + + + + + + + + + + + + + - - - - - - - - - - - - - - - - - - - - - - - - - -q area A as Large as possible distance d as SMALL as possible d Thin film capacitors Metal film separated by thin insulators metal insulator Metal Making the area large and the insulating gap small increases C A parallel plate capacitor with 2 plates each with area 1.0 m2 separated by a distance of 1.0 mm holds +q,-q, q=10-6C (a) Find the capacitance. (b) Find the E field in the capacitor (c) Find ΔV across the plates A=1 m2 q ΔV d=1mm -q C= A! o d E= q A! o ΔV= Ed 1(8.9 x10!12 ) = = 8.9 x10!9 F = 8.9nF 0.001 1x10!6 5 = = 1.1 x 10 V /m !12 (1)(8.9 x10 ) = 1.1x105 (1x10 !3 ) = 1.1x102V A parallel plate capacitor (infinite) with a constant charge q has its separation increased by a factor of 2. How does this change E and ΔV? q q Eo do -q C ΔVo E= Eo ΔV= 2ΔVo 2do C E= -q q is constant A! o !V = dE = 2!Vo A parallel plate capacitor (infinite) with a constant voltage difference ΔV has its separation increased by a factor of 2. How does this change E and q? ΔVo d qo Eo o qo q= 2 ΔVo -qo C Eo E= 2 2do C ΔV=dE q E= A! o ΔV= ΔVo Capacitor combinations Capacitors connected in series and parallel Electrical circuit elements + Voltage source Circuit diagram resistor R capacitor C V Conductor one capacitor ΔV + C - q C= !V +q -q Vc=ΔV Two Capacitors in Parallel q q ΔV C1 q1 q2 C2 = ΔV Ceq Ceq q q = !V equivalent capacitance What single capacitor has the same properties as two capacitors in parallel? Two Parallel Capacitors equivalent capacitance q C1 ΔV q1 q2 C2 !V = !V1 = !V2 q = q1 + q2 = C1!V + C2 !V Ceq C1!V + C2 !V q = = !V !V Ceq = C1 + C2 Ceq q = !V Parallel Capacitors For N capacitors in parallel Ceq = C1 + C2 + ........CN Ceq is the sum of capacitances Like a larger capacitor, larger area Find the equivalent capacitance ΔV 5µF 3µF 10µF A. 15 uF Ceq =C1 +B. C217+ uF C3 Ceq = 5 + C. 3 +1810uF= 18 µ F D. 20 uF Find the equivalent capacitance ΔV 5µF 3µF 10µF Ceq =C1 + C2 + C3 Ceq = 5 + 3 + 10 = 18 µ F Two Capacitors in Series What is the equivalent capacitance? q C1 ΔV C2 q q +q ΔV 1 -q +q ΔV 2 -q Ceq q = !V the charge on both capacitors in series is q Two Capacitors in Series q C1 ΔV C2 q q q = !V +q -q Ceq +q -q q=q1=q2 q q !V = !V1 + !V2 = + C1 C2 1 !V 1 q q = = ( + ) Ceq q q C1 C2 1 1 1 = + Ceq C1 C2 For N capacitors in series 1 1 1 1 = + + ......... Ceq C1 C2 CN Capacitors in series Ceq is smaller than the smallest capacitance. You store less charge on series capacitors than you would on either one of them alone with the same voltage! Physical Argument Take a parallel plate capacitor and place a thin metal plate with the same area in the middle of the gap. +++++++++++++ ++++++++++++ ΔV/2 ΔV ----------- ΔV/2 ----------+++++++++++++ ----------- C the component capacitances are larger than the total C1=C2= 2C Ceq = C The equivalent capacitance is less than the component capacitances Ceq <C1 or C2 C1 C2 Find the equivalent capacitance 3µF 5µF ΔV 10µF 1 1 1 1 1 1 1 = + + = + + = 0.633 Ceq C1 C2 C3 5 3 10 Ceq = 1.58 µ F 34. Find the equivalent capacitance. X Cseries=6µF X 1 Cseries Ceq= 4.00+2.00+6.00=12.00 µF 1 1 4 1 = + = = 24 8 24 6 34. Find the charge on each capacitor. C1 C2 X X C3 C4 q = C !V q1=C1ΔV=4x10-6(36)=1.44x10-4 C q2 =C1ΔV=2x10-6(36)=0.72x10-4 C q3=q4=CseriesΔV=6x10-6(36)=2.16x10-4C Cseries =6µF 34.Find the voltage drop across each capacitor. ΔV3 ΔV1 ΔV2 ΔV4 q !V = C ΔV1=ΔV2=36V q 2.16 x10!4 "V3 = = = 9.0V !6 C3 24 x10 q 2.16 x10!4 "V4 = = = 27V !6 C4 8 x10 series capacitors The larger C has the smaller voltage drop 42. Find the equivalent capacitance between a and b. X 1 1 1 1 = + = Cser 7 5 35 /12 X Cser = 2.9 µ F 4.0 µF a 2.9 µF 6.0 µF b Ceq=4.0+2.9+6.0 =12.9 µF Two identical capacitors. charge one capacitor at 10 V , disconnect, connect the charged capacitor to the uncharged capacitor. What is the voltage drop across the each capacitor? One way to do this problem q is constant but divided between the two capacitors ΔV0 =10V q/2 +q ΔV -q C q/2 -q/2 -q/2 C the charge on each capacitor is reduced by 2 fold thus the voltage across each capacitor is reduced by 2fold !V = !Vo 2 charge one capacitor at 10 V , disconnect, connect the charged capacitor to the uncharged capacitor. What is the voltage drop across the each capacitor? Another way to do this problem q is constant but is placed on an equivalent capacitor ΔVo =10V Ceq =2C +q -q C C The voltage is reduced by 2 fold !V = !Vo q q = = Ceq 2C 2 ΔV