PHYS 202 1nd EditionExam # 2 Study Guide Lectures: 6 - 10Lecture 6 (February 1)A parallel-plate air capacitor of area A= 17.0 cm2 and plate separation d= 3.80 mm is charged by a battery to a voltage 50.0 V. If a dielectric material with κ = 3.20 is inserted so that it fills the volume between the plates (with the capacitor still connected to the battery), how much additional charge will flow from the battery onto the positive plate?Answer:4.36x10-10CExplanation:Free space permiativity=8.85x10-12k=8.85x10-12k’=3.2C=k(A/d)C’=k’(A/d)Q=CVQ’=C’VQ-Q’=4.36x10-10CLecture 6 (February 1)Indicate whether the following statements are true or false:A) In an isolated capacitor, the distance(d) between the plates becomes d/2. This implies that Q=Q/2.B) In an isolated capacitor, the distance(d) between the plates becomes d/2. This implies that the voltage will increase.C) In an isolated capacitor, the distance(d) between the plates becomes d/2. This implies that U will increase.Answer:A) FalseB) FalseC) FalseExplanation: A) For an isolated system, Q will always stay the same.B) V=Ed, so if d is halved, then the value of V decreases.C) U=QV and V=Ed so once again, if d is halved, V decreases. If the value of V decreases, then the value of U also decreases.Lecture 7 (February 3)A slab of copper of thickness b = 7.246×10-1 mm is thrust into a parallel-plate capacitor of C = 5.00×10-11 F of gap d = 5.0 mm; it is centered exactly halfway between the plates. A) If a charge q = 4.00×10-6 C is maintained on the plates, what is the ratio of the stored energy before to thatafter the slab is inserted? B) How much work is done on the slab as it is inserted? C) Is the slab pulled in or must it be pushed in?Answer:A) 1.17B) -.0232JC) PulledExplanation:A) C = kA/dC' = kA/(d-b)C' = C(d/(d-b)) = 5x10-11(.005/(.005-.0007246)) = 5.847x10-11 FU = 0.5 q2/C = 0.5(4x10-6)2/5x10-11 = 0.16 JU' = 0.5 q2/C' = 0.5((4x10-6)2/5.847x10-11) = 0.136 JU/U' = 0.16/0.136 = 1.176B) U’ - U = 0.136 - 0.16 = -0.024 JC) Because the amount of work done on the slab is negative, the slab is being pulled in.Lecture 8 (February 12)Three capacitors (3.88, 6.08, and 11.9 μF) are connected in series across a 53.0 V battery. Calculate the voltage across the 3.88 μF capacitor.Answer:27VExplanation:Q=CVC=1/((1/3.88)+(1/6.08)+(1/11.9))=1.975µFQ=(1.975x10-6)(53)=1.04x10-4V=Q/C=1.04x10-4/3.88x10-6=26.8VLecture 9 (February 17) RA E B R EC R R EFor the three circuits above, assume that all R’s displayed are equal to one another and all E’s displayed are equal to one another. Based on these assumptions and the three circuits above, state whether the following are true or false:A) The potential difference across one resistor in figure C is equal to two times the potential difference across one resistor in figure B.B) To total power in circuit C is equal to two times the power in circuit B.C) The power in circuit A is equal to two times the power in circuit BD) The current through one resistor in circuit A is equal to the current through one resistor in BRE) The current through circuit A is equal to the current through one resistor in circuit CAnswer:A) TrueB) FalseC) TrueD) FalseE) TrueExplanation:P=IV=I2RR+R=2RE/2R=IB =.5IAPB=IB2(2R)=(IA2/4)2R=.5IA2R=.5PARR/(R+R)=R/2E/(R/2)=2E/R=IC=2IA=4IBPC=I2C(R/2)=(4IA2R)/2=2IA2R=2PALecture 9 (February 17) C1 C3 S2C2 C4 S1 VIn the figure, battery V supplies 24 V. Take C1 = 1.6 μF, C2 = 2.8 μF, C3 = 3.1 μF, and C4 = 4.6 μF. A) Find the charge Q3 on capacitor C3 when ONLY switch S1 is closed. B) Find the charge Q3 on capacitor C3 when switch S2 is ALSO closed. C) Find the charge Q4 on capacitor C4 when switch S2 is ALSO closed.Answer:A). 2.53X10-5CB). 2.71x105CC) 4.01x10-5CExplanation:A)C1 is in series with C3= C13=C1C3/C1+C3=1.055µFV13=V24=V1324=24VQ13=(V13)(C13)=2.53x10-5C =Q1=Q3B and C) C1 is parallel to C2, so C12=C1+C2=4.4C3 is parallel to C4, C34=C3+C4=7.7C12 is in series with C34= C1234 = C12C34/C12+C34=2.8Q1234=(V1234)(C1234)=(24)(2.8)=67.2 Q1234=Q12=Q34Q34=(V34)(C34)67.2=V34(7.7)=8.7273VV34=V3=V4Q3=V3C3=(8.727)(3.1)=27.05uCQ4=V4C4=(8.727)(4.6)=40.1uCLecture 10 (February 19)In the figure, calculate the potential difference between points c and d by as many paths as possible. Assume that ε1 = 4.9 V, ε2 = 1.0 V, R1 = R2 = 10 Ω, and R3 = 5 Ω.Answer:.475VExplanation:Vcd=I2R2I1+I3=I2  Equation 1Loop badb: +E1-I1R1+I3R3 => 4.9-I110+I35=0 => I1=.49+.5I3 Equation 2Loop bdcb: -I3R3-I2R2-E2 => -5I3-10R2-1=0 +> -.5I3-.1=I2 Equation 3Plug equations 2 and 3 into equation 1.49+.5I3 +I3=-.5I3-.1 => 1.5I3=-.5I3-.59 => 2I3=-.59I3=-.295Plug this value into either equation 2 or 3 and solve for I2 algebraically, then multiply by