# Lecture04bd - Physics 102 Lecture 04 Capacitors&...

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Unformatted text preview: Physics 102: Lecture 04 Capacitors (& batteries) Physics 102: Lecture 4, Slide 1 Physics 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 Physics 102: Lecture 4, Slide 2 Recall from last lecture..... Electric Fields, Electric Potential Physics 102: Lecture 4, Slide 3 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 Physics 102: Lecture 4, Slide 4 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 + 150 E-field 100 + 50 Equipotential lines 0 0 50 100 150 DUAB = q DVAB Physics 102: Lecture 4, Slide 5 Important Special Case: Uniform Electric Field Two large parallel conducting plates of area A +Q on one plate -Q on other plate 150 Then E is uniform between the two plates: E=4kQ/A zero everywhere else This result is independent of plate separation 100 50 + + + +A + + + + 0 +Q E 0 d 50 Q A 100 150 This is called a parallel plate capacitor Physics 102: Lecture 4, Slide 6 Parallel Plate Capacitor: Potential Difference Charge Q on plates V =VA VB = +E0 d =4 k Q d / A Voltage is proportional to the charge! + + + + + d Physics 102: Lecture 4, Slide 7 PhET Simulation E=E0 - A B Capacitance: The ability to store separated charge CQ/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 + E Stores Energy U =() Q V Units: 1 Coulomb/Volt = 1 Farad (F) Physics 102: Lecture 4, Slide 8 + + + + d Why Separate Charge? A way to store and release energy! Camera Flash Defibrillator AC DC Tuners / resonant circuits Radio Cell phones Cell membranes Physics 102: Lecture 4, Slide 9 Capacitance of Parallel Plate Capacitor V = Ed E=4kQ/A (Between two large plates) V +Q A E So: V = 4kQd/A Recall: CQ/V So: C = A/(4kd) Recall: e0 =1/(4k)=8.85x10-12 C2/Nm2 C =e0A/d Q A d Parallel plate capacitor Physics 102: Lecture 4, Slide 10 Dielectric Placing a dielectric (insulator) between the plates increases the capacitance. Dielectric constant (k > 1) + + + + + + + + + + -+ -+ -+ -+ -+ -+ -+ -+ -+ -+ -+ -+ d - C = k C0 Capacitance with dielectric Physics 102: Lecture 4, Slide 11 Capacitance without dielectric For same charge Q, E (and V) is reduced so C = Q/V increases 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? A) Increases B) Constant C) Decreases Remember charge is real/physical. There is no place for the charges to go. Physics 102: Lecture 4, Slide 12 -Q - + CheckPoint 4.1 d + + + +Q pull 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 87% 92% 19% True False C = 0A/d E= Q/(0A) V= Ed Physics 102: Lecture 4, Slide 13 C decreases! True False 2)The electric field increases Constant True False 3)The voltage between the plates increases ACT/CheckPoint 4.1 pull - + -Q 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= QV B) constant Q constant, V increased C) decreases Plates are attracted to each other, you must pull them apart, so the potential energy of the plates increases. Physics 102: Lecture 4, Slide 14 Capacitors are used in circuits! 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 C1 VC1= 5-0 V= 5 V Physics 102: Lecture 4, Slide 15 Vwire 2= 5 V C2 Vwire 3= 12 V C3 Vwire 4= 15 V VC2= 12-5 V= 7 V VC3= 15-12 V= 3 V To understand complex circuits... C2 + - C1 C3 ...treat capacitors in series and parallel as a fictitious equivalent capacitor! Physics 102: Lecture 4, Slide 16 + Ceq Capacitors in Parallel Both ends connected together by wire Same voltage: V1 = V2 = Veq Share Charge: Qeq = Q1+Q2 Equivalent C: Ceq = C1+C2 Add areas remember C=e0A/d 15 V C1 The 10 V Physics 102: Lecture 4, Slide 17 15 V C2 10 V 15 V Ceq 10 V the pair acts just like this one! Parallel Practice A 4 mF capacitor and 6 mF capacitor are connected in parallel and charged to 5 volts. Calculate Ceq, and the charge on each capacitor. Ceq = C4+C6 = 4 mF+6 mF = 10 mF Q4 = C4 V4 = (4 mF)(5 V) = 20 mC Q6 = C6 V6 = (6 mF)(5 V) = 30 mC Qeq = Ceq Veq = (10 mF)(5 V) = 50 mC = Q4+Q6 5V C4 0V Physics 102: Lecture 4, Slide 18 5V C6 0V V=5V 5V Ceq 0V Capacitors in Series Connected end-to-end with NO other exits Same Charge: Q1 = Q2 = Qeq Share Voltage:V1+V2=Veq Equivalent C: +Q -Q +Q -Q 5V + + - 1 1 1 = + Add d remember C=e0A/d 1 2 C1 C2 0V +Q + 5V Ceq -Q - 0V Physics 102: Lecture 4, Slide 19 Series Practice A 4 mF capacitor and 6 mF capacitor are connected in series and charged to 5 volts. Calculate Ceq, and the charge on the 4 mF capacitor. 1 1 1 1 = + = 2.4 Ceq = + 4 6 C4 C6 Q = CV Q4 = Q6 = Qeq = = 2.4 5 = 12 +Q + -1 -1 5V + - -Q +Q -Q C4 C6 0V +Q + 5V - -Q Ceq 0V Physics 102: Lecture 4, Slide 20 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). C1 C1 C2 Physics 102: Lecture 4, Slide 21 C2 Electromotive Force Battery Maintains constant potential difference V (electromotive force emf ) Does NOT produce or supply charges, just "pushes" them. Like a pump for charge! Physics 102: Lecture 4, Slide 22 + - Usually "0V" by convention CheckPoint 4.4 A circuit consists of three initially uncharged capacitors C1, C2, and C3, which are then connected to a battery of emf . The capacitors obtain charges q1, q2, q3, and have voltages across their plates V1, V2, and V3. Ceq is the equivalent capacitance of the circuit. Which of these are true? 1) q1 = q2 2) q2 = q3 3) V2 = V3 4) = V1 5) V1 < V2 Physics 102: Lecture 4, Slide 23 C2 + V1 +q1 -q1 C1 +q2 -q2 V2 +q3 V3 -q3 C3 ACT/CheckPoint 4.4: Which is true? A circuit consists of three initially uncharged capacitors C1, C2, and C3, which are then connected to a battery of emf . The capacitors obtain charges q1, q2, q3, and have voltages across their plates V1, V2, and V3. Ceq is the equivalent capacitance of the circuit. C2 + V1 +q1 -q1 C1 +q2 -q2 V2 +q3 V3 -q3 C3 1) q1 = q2 Not necessarily. C1 and C2 are NOT in series. Yes! C2 and C3 are in series. 2) q2 = q3 Physics 102: Lecture 4, Slide 24 ACT/CheckPoint 4.4: Which is true? A circuit consists of three initially uncharged capacitors C1, C2, and C3, which are then connected to a battery of emf . The capacitors obtain charges q1, q2, q3, and have voltages across their plates V1, V2, and V3. Ceq is the equivalent capacitance of the circuit. 10V + V1 +q1 -q1 C1 C2 ?? V C3 +q2 -q2 V2 +q3 V3 -q3 0V 1) V2 = V3 2) = V1 Not necessarily, only if C2 = C3 Yes! Both ends are connected by wires Physics 102: Lecture 4, Slide 25 Recap of Today's Lecture Capacitance C = Q/V Parallel Plate: C = e0A/d Capacitors in parallel: Ceq = C1+C2 Capacitors in series: 1/Ceq = 1/C1+1/C2 Batteries provide fixed potential difference Physics 102: Lecture 4, Slide 26 ...
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