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UVM PHYS 012 - Exam 1 Study Guide
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Physics 012 1st EditionExam # 1 Study Guide Lectures: 1 - 8Lecture 1 (January 12)What are the three types of charge transfer? Briefly describe each.1. By friction: transfer of electrons between the two objects that are rubbed together2. By conduction: transfer of charge without contact through polarization3. By induction: transfer of charge by contactLecture 2 (January 14) What is the net force on q1 in the diagram below? If q1 has a mass of 1.5 grams and it was free to move, what would be its acceleration?Fnet,x = F2,x + F3,x = [(k|+8μC||-5μC|)/(1.32)] + [(k|+8μC||+5μC|)/(1.32)] * cos (23°) = 0Fnet,y = F2,x + F3,x = [(k|+8μC||-5μC|)/(1.32)] + [(k|+8μC||+5μC|)/(1.32)] * sin (23°) = 1.663x10-3 NFnet = Fnet,y = 1.663x10-3 NLecture 3 (January 16)Find the direction and magnitude of the net electric field at the origin in the diagram below.E1 = (k|+6μC|)/(0.04m)2 * cos(0°) = E2 = (k|+6μC|)/(0.04m)2 * cos(180°) = E3 = (k|+3μC|)/(0.05m) 2 * sin(90°) =E4 = (k|-8μC|)/(0.07m) 2 * sin(90°) = Enet,x = E1 + E2 = 0Enet,y = E3 + E4 = Electric field is pointed along the y-axis.Lecture 4 (January 21)With q2 present, the electric field at P is twice what it is when only q1 is present. Given that q1 = +0.50μC, find q2 when it is a) positive and b) negative.a) Eq2 = Eq1(k|q2||P|)/4d2 = (k|q1||P|)/d24q1 = q2q2 = +2.00 μCb) Eq2 = -3Eq1(k|q2||P|)/4d2 = -3(k|q1||P|)/d2-12q1 = q2q2 = -6.00 μCLecture 5 (January 23)A long, thin wire of length L has a positive charge Q distributed uniformly along it. Use Gauss’ law to show that the electric field created by this wire at a radial distance r has a magnitude of E = λ/(2πε0r).λ = Q/L (charge density)For a cylinder, A = 2πr2 + 2πrLΦ at top and bottom of cylinder (cos(90°) = 0). Only area on sides (2πrL) matters for this problem.Φ = EAE(2 πrL) = q/ε0 = λL/ε0E = λL/2 πrLε0 = λ/(2πε0r)Lecture 6 (January 26)In the diagram below, all points have a charge of q0. Find the change in electric potential and electric potential energy between A and B, B and C, and A and C in terms of E, d, and q0. What is the relationship between the change in potential from A to C and from A to B?ΔEPEAB = -WAB = -Fscosθ = -q0Edcos(0°) = -q0EdΔVAB = ΔEPEAB/q0 = -EdΔEPEBC = -WBC = -Fscosθ = -q0Ehcos(90°) = 0ΔVBC = ΔEPEBC/q0 = 0ΔEPEAC = -WAC = -Fscosθ = -q0Escos(90°) = -q0EdΔVAC = ΔEPEAC/q0 = -EdΔVAC = ΔVABLecture 7 (January 28)Two positive point charges of equal magnitude reside along the x-axis. Where between the two points is the net electric field equal to zero? Where between the two points is the electric potential equal to zero?V = V1 + V2 = kq1/r + kq2/r ; V will never equal zero because it is the sum of two potentials that are both positive. Therefore, the electric potential will always be positive.E = E1 + E2 = F1/q1 + F2/q2 ; E will equal zero half-way between the two points since the force from each point will be equal and opposite.Two point charges of equal magnitude reside along the x-axis, one is positive and the other is negative. Where between the two points is the net electric field equal to zero? Where between the two points is the electric potential equal to zero?V will be zero half-way between the two points since the two potentials will be of equal but opposite magnitude at this point, thus cancelling each other out.E will never be zero between the two points. It was always point toward the negative charge with increasing magnitude as it approaches the negative charge.Lecture 8 (January 30)Points A and B line along a conducting wire. If VA > VB, what direction does the electric field point? What direction does the electric current flow? What direction do electrons move along the wire?The electric field points toward regions of lower electric potential, so the electric would point towards point B.Electric current flows in the direction positive charge would move, which is from higher to lowerelectric potential. This means electric current flows from point A to point B.Electrons would move in the opposite direction of the electric field, moving from point B towardpoint A.Lecture 9 (February 2)In the diagram below, there is a current of 0.5 A across the 8 Ω resistor. Ε is the electromotive force, which is equal to the maximum potential difference across a battery. Find: ΔV8, ΔV16, I16, I20, ΔV20, ε, P18, and Iε.ΔV8 = I8R8 = (0.5A)(8Ω) = 4.0VΔV16 = ΔV8 = 4.0VI16 = ΔV16/R16 = (4.0V)/(16Ω) = 0.25AI20 = I8 + I16 = (0.5A) + (0.25A) = 0.75AΔV20 = I20R20 = (0.75A)(20Ω) = 15Vε = Vbottom = Vtop = ΔV8,16 + ΔV20 = 4.0 V + 15 V = 19 VΔV9 = ε = 19VI9 = V9/R9 = (19V)/(9Ω) = 2.1AP18 = V182/R18 = V92/R18 = (19V)2/(18Ω) = 20.1WI18 = V18/R18 = (19V)/(18Ω) = 1.1AIε = I20 + I9 + I18 = 0.75A + 2.1A + 1.1A =


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UVM PHYS 012 - Exam 1 Study Guide

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