Capacitors Physics 102: Lecture 04

Recall from last lecture….. Electric Fields, Electric Potential

Comparison: Electric Potential Energy vs. Electric Potential  V AB : the difference in electric potential between points B and A  U AB : the change in electric potential energy of a charge q when moved from A to B  U AB = q  V AB q AB

Electric Potential: Summary E field lines point from higher to lower potential For positive charges, going from higher to lower potential is “downhill” For negative charges, going from lower to higher potential is “downhill” For a battery, the + terminal is at a higher potential than the – terminal Positive charges tend to go “downhill”, from + to - Negative charges go in the opposite direction, from - to +  U AB = q  V AB

Important Special Case Uniform Electric Field Two large parallel conducting plates of area A +Q on one plate - Q on other plate Then E is uniform between the two plates: E=4  kQ/A zero everywhere else This result is independent of plate separation This is call a parallel plate capacitor

Parallel Plate Capacitor: Potential Difference A B E=E 0 d V =V A – V B = +E 0 d A B d E=2 E 0 V =V A – V B = +2E 0 d Potential difference is proportional to charge: Double Q  Double V Charge Q on platesCharge 2Q on plates 10 E 0 =4  kQ/A

Capacitance: The ability to store separated charge C  Q/V Any pair conductors separated by a small distance. (e.g. two metal plates) Capacitor stores separated charge –Positive Q on one conductor, negative Q on other –Net charge is zero E d Q=CV U =(½) Q V Stores Energy Units: 1 C/Volt = 1 Farad (F)

Why Separate Charge? Camera Flash Defibrillator—see ex in text AC -> DC Tuners –Radio –Cell phones

Capacitance Practice How much charge is on a 0.9 F capacitor which has a potential difference of 200 Volts? 14 How much energy is stored in this capacitor?

Capacitance Practice How much charge is on a 0.9 F capacitor which has a potential difference of 200 Volts? 14 Q = CV How much energy is stored in this capacitor? U = ½ Q V = (0.9)(200) = 180 Coulombs = ½ (180) (200) = 18,000 Joules!

Capacitance of Parallel Plate Capacitor V = Ed E=4  kQ/A (Between two large plates) So: V = 4  kQd/A Recall: C  Q/V So: C = A/(4  kd) Define:  0 =1  k)=8.85x C 2 /Nm 2 A d A E+ - V 16 C =  0 A/d Parallel plate capacitor

Parallel Plate Capacitor C =  0 A/d Calculate the capacitance of a parallel plate capacitor made from two large square metal sheets 1.3 m on a side, separated by 0.1 m. A d A 18

Parallel Plate Capacitor C =  0 A/d Calculate the capacitance of a parallel plate capacitor made from two large square metal sheets 1.3 m on a side, separated by 0.1 m. A d A 18

Dielectric Placing a dielectric between the plates increases the capacitance. C =  C 0 Capacitance with dielectric Dielectric constant (  > 1) Capacitance without dielectric 1919

A parallel plate capacitor given a charge q. The plates are then pulled a small distance further apart. What happens to the charge q on each plate of the capacitor? ACT: Parallel Plates 22 1) Increases2) Constant3) Decreases Remember charge is real/physical. There is no place for the charges to go. +q-q d pull

A parallel plate capacitor given a charge q. The plates are then pulled a small distance further apart. Which of the following apply to the situation after the plates have been moved? 1)The capacitance increases True False 2)The electric field increases True False 3)The voltage between the plates increases True False 4)The energy stored in the capacitor increases True False Preflight 4.1 +q-q d pull C =  0 A/dC decreases! E= Q/(  0 A)Constant V= Ed :24 66% 59% 52% 35%

A parallel plate capacitor given a charge q. The plates are then pulled a small distance further apart. Which of the following apply to the situation after the plates have been moved? ACT/Preflight 4.1 U= ½ QV Q constant, V increased :25 The energy stored in the capacitor A) increasesB) constantC) decreases Plates are attracted to each other, you must pull them apart, so the potential energy of the plates increases. +q-q d pull

Two identical parallel plate capacitors are shown in end- view in A) of the figure. Each has a capacitance of C. A) B) If the two are joined as in (B) of the figure, forming a single capacitor, what is the final capacitance? 1) 2C 2) C 3) C/2 “because C=EA/d. none of these variables are changing so C must be the same when the 2 are joined to form a single capacitor” “2 capacitors become 1 which means that the total capacitance got halved ” “More area, means more capacitance.” 63% 7% 30% 26 ACT/Preflight 4.2

Voltage in Circuits Elements are connected by wires. Any connected region of wire has the same potential. V wire 1 = 0 VV wire 2 = 5 VV wire 3 = 12 VV wire 4 = 15 V V C 1 = 5-0 V= 5 VV C 3 = V= 3 VV C 2 = 12-5 V= 7 V C1C1 C2C2 C3C3 The potential difference across an element is the element’s “voltage.” 28

Capacitors in Parallel Both ends connected together by wire C1C1 C2C2 C eq 15 V 10 V 15 V 10 V 15 V 10 V Add Areas:C eq = C 1 +C 2 remember C=  0 A/d Share Charge: Q eq = Q 1 +Q 2 = V eq 30 Same voltage: V 1 = V 2

Parallel Practice A 4  F capacitor and 6  F capacitor are connected in parallel and charged to 5 volts. Calculate C eq, and the charge on each capacitor. C4C4 C6C6 C eq 5 V 0 V 5 V 0 V 5 V 0 V :34 C eq = C 4 +C 6 Q 4 = C 4 V 4 Q 6 = C 6 V 6 Q eq = C eq V eq = 4  F+6  F = 10  F = (4  F)(5 V) = 20  C = (6  F)(5 V) = 30  C = (10  F)(5 V) = 50  C = Q4+Q6 V = 5 V

Add d: Capacitors in Series Connected end-to-end with NO other exits Same Charge: Q 1 = Q 2 = Q eq C eq C1C1 C2C Q -Q +Q -Q +Q -Q Share Voltage:V 1 +V 2 =V eq 36

Series Practice C eq C4C4 C6C Q -Q +Q -Q +Q -Q A 4  F capacitor and 6  F capacitor are connected in series and charged to 5 volts. Calculate C eq, and the charge on the 4  F capacitor. 5 V 0 V 5 V 0 V Q = CV :38

Comparison: Series vs. Parallel Series Can follow a wire from one element to the other with no branches in between. Parallel Can find a loop of wire containing both elements but no others (may have branches). C1C1 C2C2 C2C2 C1C1

Electromotive Force Battery –Maintains potential difference V –Not constant power –Not constant current –Does NOT produce or supply charges, just “pushes” them

A circuit consists of three initially uncharged capacitors C 1, C 2, and C 3, which are then connected to a battery of emf . The capacitors obtain charges q 1, q 2,q 3, and have voltages across their plates V 1, V 2, and V 3. Which of these are true? C 2 C 3 C 1 E + - -q 2 +q 1 -q 1 +q 3 -q 3 +q 2 V1V1 V2V2 V3V3 1)q 1 = q 2 2)q 2 = q 3 3)V 2 = V 3 4)E = V 1 5)V 1 < V 2 6)C eq > C 1 43 Preflight 4.4

A circuit consists of three initially uncharged capacitors C 1, C 2, and C 3, which are then connected to a battery of emf . The capacitors obtain charges q 1, q 2,q 3, and have voltages across their plates V 1, V 2, and V 3. C 2 C 3 C 1 E + - -q 2 +q 1 -q 1 +q 3 -q 3 +q 2 V1V1 V2V2 V3V3 1) q 1 = q 2 Not necessarily C 1 and C 2 are NOT in series. 2) q 2 = q 3 Yes! C 2 and C 3 are in series. 45 Preflight 4.4

C 2 C 3 C 1 E + - -q 2 +q 1 -q 1 +q 3 -q 3 +q 2 V1V1 V2V2 V3V3 3) V 2 = V 3 Not necessarily, only if C 1 = C 2 4) E = V 1 10V 0V 7V?? Yes! Both ends are connected by wires 48 A circuit consists of three initially uncharged capacitors C 1, C 2, and C 3, which are then connected to a battery of emf . The capacitors obtain charges q 1, q 2,q 3, and have voltages across their plates V 1, V 2, and V 3. Preflight 4.4

C 2 C 3 C 1 E + - -q 2 +q 1 -q 1 +q 3 -q 3 +q 2 V1V1 V2V2 V3V3 5) V 1 < V 2 Nope, V 1 > V 2. (E.g. V 1 = 10-0, V 2 =10-7 6) C eq > C 1 10V 0V 7V?? Yes! C 1 is in parallel with C A circuit consists of three initially uncharged capacitors C 1, C 2, and C 3, which are then connected to a battery of emf . The capacitors obtain charges q 1, q 2,q 3, and have voltages across their plates V 1, V 2, and V 3. Preflight 4.4

Capacitance C = Q/V Parallel Plate: C =  0 A/d Capacitors in parallel: C eq = C 1 +C 2 Capacitors in series: C eq = 1/(1/C 1 +1/C 2 ) Batteries provide fixed potential difference Recap of Today’s Lecture