Physics 102 Lecture 04 Capacitors batteries Phys 102

Physics 102: Lecture 04 Capacitors (& batteries)

Phys 102 so far Basic principles of electricity • Lecture 1 – electric charge & electric force • Lecture 2 – electric field • Lecture 3 – electric potential energy and electric potential Applications of electricity – circuits • • • Lecture 4 – capacitance Lecture 5 – resistance Lecture 6 – Kirchhoff’s rules Lecture 7 – RC circuits Lecture 12 & 13 – AC circuits

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

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

Electric Potential: Summary • E field lines point from higher to lower potential • For positive charges, going from higher to lower potential is “downhill” Positive charges tend to go “downhill”, from + to Negative charges go in the opposite direction, from - to + DUAB = q DVAB + charge E

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 k. Q/A • zero everywhere else • This result is independent of plate separation This is call a parallel plate capacitor +Q A d E –Q A

Parallel Plate Capacitor: Potential Difference Charge Q on plates V =VA – VB = +E 0 d E=E 0 + - + +A B + + d - Charge 2 Q on plates V =VA – VB = +2 E 0 d + E=2 E 0 + + +A + + d B - Potential difference is proportional to charge: Double Q Double V E 0=4 k. Q/A 10

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 Q=CV – Positive Q on one conductor, negative Q on other – Net charge is zero + • Stores Energy U =(½) Q V Units: 1 Coulomb/Volt = 1 Farad (F) + + E - + - d 12

Why Separate Charge? • • Camera Flash Defibrillator—see example in textbook AC → DC Tuners – Radio – Cell phones • Cell membranes

Capacitance of Parallel Plate Capacitor V = Ed E=4 k. Q/A V (Between two large plates) So: V = 4 k. Qd/A Recall: C Q/V So: C = A/(4 kd) Define: 0=1/(4 k)=8. 85 x 10 -12 C 2/Nm 2 C = 0 A/d A +E – A d Parallel plate capacitor 16

Dielectric • Placing a dielectric between the plates increases the capacitance. Dielectric constant (k > 1) C = k C 0 + + + + + -+ -+ -+ d - Capacitance without dielectric Capacitance with dielectric 19

ACT: Parallel Plates -q pull - + - d +q + pull + 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? 1) Increases 2) Constant 3) Decreases Remember charge is real/physical. There is no place for the charges to go. 22

-q Preflight 4. 1 pull - + - d +q + 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 C = ε 0 A/d False 29% True False 44% True False 43% C decreases! 2)The electric field increases E= Q/(ε 0 A) True Constant 3)The voltage between the plates increases V= Ed : 24

ACT/Preflight 4. 1 -q pull - + - d +q + 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? The energy stored in the capacitor A) increases U= ½ QV B) constant C) decreases Q constant, V increased Plates are attracted to each other, you must pull them apart, so the potential energy of the plates increases. : 25

ACT/Preflight 4. 2 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? 71% 1) 2 C 12% 2) C 17% 3) C/2 26

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

Capacitors in Parallel • Both ends connected together by wire • Same voltage: V 1 = V 2 = Veq • Share Charge: Qeq = Q 1+Q 2 • Add Areas: Ceq = C 1+C 2 15 V C 1 C 2 10 V remember C= 0 A/d 15 V Ceq 10 V 30

Parallel Practice A 4 m. F capacitor and 6 m. F capacitor are connected in parallel and charged to 5 volts. Calculate Ceq, and the charge on each capacitor. Ceq = C 4+C 6 = 4 m. F+6 m. F = 10 m. F Q 4 = C 4 V 4 = (4 m. F)(5 V) = 20 m. C Q 6 = C 6 V 6 = (6 m. F)(5 V) = 30 m. C Qeq = Ceq Veq = (10 m. F)(5 V) = 50 m. C = Q 4+Q 6 5 V 5 V C 4 C 6 0 V 0 V V=5 V 5 V Ceq 0 V : 34

Capacitors in Series • Connected end-to-end with NO other exits • Same Charge: Q 1 = Q 2 = Qeq • Share Voltage: V 1+V 2=Veq • Add d: +Q -Q + + - C 1 +Q + Ceq C 2 -Q 36

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

Comparison: Series vs. Parallel Series Parallel • Can follow a wire from one element to the other with no branches in between. C 1 C 2 • Can find a loop of wire containing both elements but no others (may have branches). C 1 C 2

Electromotive Force • Battery + - – Maintains constant potential difference V (electromotive force – emf ε) – Does NOT produce or supply charges, just “pushes” them. Analogy to pump… 40

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 ε. 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? 1) q 1 = q 2 C 2 2) q 2 = q 3 3) V 2 = V 3 4) ε = V 1 ε +q 2 + - V 1 +q 1 -q 1 C 1 -q 2 V 2 +q 3 V 3 -q 3 C 3 5) V 1 < V 2 6) Ceq > C 1 43

ACT/Preflight 4. 4: Which is true? A circuit consists of three initially uncharged capacitors C 1, C 2, and C 3, which are then connected to a battery of emf ε. 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 ε +q 2 + - V 1 +q 1 -q 1 C 1 -q 2 V 2 +q 3 V 3 -q 3 1) q 1 = q 2 Not necessarily C and C are NOT in series. 1 2 2) q 2 = q 3 Yes! C 2 and C 3 are in series. C 3 45

ACT/Preflight 4. 4: Which is true? A circuit consists of three initially uncharged capacitors C 1, C 2, and C 3, which are then connected to a battery of emf ε. 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 10 V ε 7 V? ? +q 2 + V 1 - +q 1 C 1 -q 2 V 2 -q 1 +q 3 V 3 -q 3 C 3 0 V 1) V 2 = V 3 Not necessarily, only if C 1 = C 2 2) ε = V 1 Yes! Both ends are connected by wires 48

ACT/Preflight 4. 4: Which is true? A circuit consists of three initially uncharged capacitors C 1, C 2, and C 3, which are then connected to a battery of emf ε. 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 10 V ε +q 2 + V 1 - +q 1 C 1 7 V? ? -q 2 V 2 -q 1 +q 3 V 3 -q 3 C 3 0 V 1) V 1 < V 2 Nope, V 1 > V 2. (E. g. V 1 = 10 -0, V 2 =10 -7) 2) Ceq > C 1 Yes! C 1 is in parallel with C 23 48

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