Capacitance: The ability to store separated charge C=Q/V. Capacitors! Capacitor. Capacitance Practice SPH4UW 24/08/2010 Q = CV
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1 SPH4UW Capacitors! Capacitance: The ability to store separate charge C=Q/V Charge Q on plates V = V V B = E 0 Charge 2Q on plates V = V V B =2E 0 E=E 0 B E=2E 0 B Physics 102: Lecture 4, Slie 1 Potential ifference is proportional to charge: Double Q Double V Q = CV Capacitor ny pair conuctors separate by a small istance. (e.g. two metal plates) Capacitor stores separate charge Positive Q on one conuctor, negative Q on other Net charge is zero Stores Energy U =(½) Q V Q=CV E Capacitance Practice How much charge is on a 0.9 F capacitor which has a potential ifference of 20olts? Q = CV= (0.9)(200) = 180 Coulombs How much energy is store in this capacitor? U = ½ Q V= ½ (180) (200) = 18,000 Joules! 1
2 Capacitance of Parallel Plate Capacitor V = E ND E = Q/(e 0 ) (Between two large plates) So: V = Q/(e 0 ) Recall: C Q/V C = e 0 / So: (For parallel plate capacitor) e 0 =1/(4pk)=8.85x10 12 /Nm 2 E V Parallel Plate Capacitor Calculate the capacitance of a parallel plate capacitor mae from two large square metal sheets 1.3 m on a sie, separate by 0.1 m. C C = e 0 / 12 ( )(1.3)(1.3) (0.1) F Dielectric Placing a ielectric between the plates increases the capacitance. Dielectric constant (k > 1) Unerstaning: Parallel Plates q q parallel plate capacitor is given a charge q. The plates are then e a small istance further apart. What happens to the charge q on each plate of the capacitor? Capacitance with ielectric C = k C 0 Capacitance without ielectric The ielectric allows the potential ifference between the plates to obtain a higher value an thus store more energy, therefore the capacitance is greater 1) Increases 2) Constant 3) Decreases Remember charge is real/physical. There is no place for the charges to go. 2
3 Unerstaning q q Unerstaning Two ientical parallel plate capacitors are shown in enview in ) of the figure. Each has a capacitance of C. parallel plate capacitor is given a charge q. The plates are then e a small istance further apart. Which of the following apply to the situation after the plates have been move? 1)The capacitance increases True False C = e 0 / C ecreases! 2)The electric fiel increases True False E= Q/(e 0 ) Constant 3)The voltage between the plates increases True False V= E 4)The energy store in the capacitor increases True False U = ½QV ) B) If the two are joine as in (B) of the figure, forming a single capacitor, what is the final capacitance? 1) 2C 2) C 3) C/2 e C 0 Voltage in Circuits Elements are connecte by wires. ny connecte region of wire has the same potential. The potential ifference across an element is the element s voltage. V wire 1 = V wire 2 = V wire 3 = 12 V V wire 4 = 1 C1 C2 C3 V C1 = V C2 = 7 V V C3 = 3 V Capacitors in Parallel Both ens connecte together by wire Same voltage: V 1 = V 2 = V eq Share Charge: Q eq = Q 1 Q 2 Total Cap: C eq = (Q 1 Q 2 )/V = C1 C2 Ceq
4 Parallel Practice 4 mf capacitor an 6 mf capacitor are connecte in parallel an charge to 5 volts. Calculate C eq, an the charge on each capacitor. C eq = C 4 C 6 Q 4 = C 4 V 4 Q 6 = C 6 V 6 = 4 mf6 mf = 10 mf = (4 mf)() = 20 mc = (6 mf)() = 30 mc Capacitors in Series Connecte entoen with NO other exits Same Charge: Q 1 = Q 2 = Q eq Share Voltage: V 1 V 2 =V eq C C C eq 1 2 Q eq = C eq V eq = (10 mf)() = 50 mc = Q 4 Q 6 C4 C6 V = Ceq Q Q Q Q C1 C2 Q Q Ceq Physics 102: Lecture 4, Slie 13 Physics 102: Lecture 4, Slie 14 Series Practice 4 mf capacitor an 6 mf capacitor are connecte in series an charge to 5 volts. Calculate C eq, an the charge on the 4 mf capacitor. C C 4mF 6mF 4 C6 2.4mF 1 eq Q = CV Q4 Q6 Qeq CV eq ( 2.4mF ( 5V 12mC Comparison: Series vs. Parallel Series Can follow a wire from one element to the other with no branches in between. Parallel Can fin a loop of wire containing both elements but no others (may have branches). Q Q Q Q C4 C6 Physics 102: Lecture 4, Slie 15 Q Q Ceq C1 C2 C1 Physics 102: Lecture 4, Slie 16 C2 4
5 Comparison: Capacitors vs. Resistors Capacitors store energy as separate charge: U=1/2QV Capacitance: ability to store separate charge: C = ke 0 / Voltage etermines charge: V=Q/C Resistors issipate energy as power: P=VI Resistance: how ifficult it is for charges to get through: R = r L / Voltage etermines current: V=IR Don t mix capacitor an resistor equations! Battery Electromotive Force Maintains potential ifference V Not constant power Not constant current Does NOT prouce or supply charges, just pushes them. Unerstaning circuit consists of three initially uncharge capacitors,, an, which are then connecte to a battery of emf E. The capacitors obtain charges q 1, q 2,q 3, an have voltages across their plates V 1, V 2, an V 3. Which Unerstaning circuit consists of three initially uncharge capacitors,, an, which are then connecte to a battery of emf E. The capacitors obtain charges q 1, q 2,q 3, an have voltages across their plates V 1, V 2, an V 3. Which 1) q 1 = q 2 2) q 2 = q 3 3) V 2 = V 3 4) E = V 1 5) V 1 < V 2 E q1 q1 q2 q2 q3 q3 E q1 q1 q2 q2 q3 q3 6) C eq > 1) q 1 = q 2 Not necessarily C1 an are NOT in series. 2) q 2 = q 3 Yes! C2 an are in series. 5
6 Unerstaning Unerstaning circuit consists of three initially uncharge capacitors,, an, which are then connecte to a battery of emf E. The capacitors obtain charges q 1, q 2,q 3, an have voltages across their plates V 1, V 2, an V 3. Which circuit consists of three initially uncharge capacitors,, an, which are then connecte to a battery of emf E. The capacitors obtain charges q 1, q 2,q 3, an have voltages across their plates V 1, V 2, an V 3. Which 10V 7V?? 10V 7V?? E 0V q1 q1 q2 q2 q3 q3 E 0V q1 q1 q2 q2 q3 q3 3) V 2 = V 3 Not necessarily, only if = 5) V 1 < V 2 Nope, V 1 > V 2. (E.g. V 1 = 100, V 2 =107 4) E = V 1 Yes! Both ens are connecte by wires 6) C eq > C1 Yes! is in parallel with 3 (C eq = 3 ) 6
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