PHYS 202 1nd EditionExam # 4 Study Guide Lectures: 1-23Lecture 1 (January 13)Based on the figure below, indicate whether the following statement is “True” or “False”: d a b c ea.A negative charge at point e will accelerate up. b. A negative charge at point d will accelerate down. c. A negative charge at point c will accelerate toward the lower-right. d. A positive charge at point a will accelerate toward the lower-left. e. A negative charge at point b will accelerate downAnswer:a. Trueb. Falsec. Falsed. Falsee. TrueExplanation:The main thing to remember here is that opposite charges attract, and like charges repel. At point a on the diagram, a negative charge would be attracted to and accelerate towards the lower right, and a positive charge would accelerate towards the upper right. At point b on the diagram, a negative charge would be attracted straight down and a positive charge would accelerate straight upwards. At point c on the diagram, a negative charge would be attracted to and accelerate towards the lower left, and a positive charge would accelerate towards the upper left. At point d on the diagram, since the positive charge is blocked by the negative charge, a negative charge at this point would be repelled straight upwards, and a positive chargewould accelerate straight down. On the opposite side, at point e on the diagram, a negative charge would be attracted to and accelerate straight upwards, but a positive charge would be repelled and move straight down, since the negative charge on the diagram would be blocked by the positive charge above point e.Lecture 2 (January 15)Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of -0.6638 N when separated by 50 cm, center-to-center. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.1820 N. What were the initial charges on the spheres (just enter the absolute value of both charges)?Answer: 2.6µC and 7.1µCExplanation:For this problem we will use two separate equations, one denoting the initial charges and one the final charges. Once we simplify these equations with our known values, we can use the quadratic formula to solve for our initial charges.Let q1 and q2 be the two initial charges;F=kq1q2/r2-.6638=(9x109)q1iq2i/.52q1q2=-1.84x10-11q2=-1.84x10-11/q1  Let this be equation 1 After they have been connected, each point has a charge q which is equal to the difference of the initial charges(q1 - q2).K((q1+q2)2/4)/.52 =0.182Nq1-q2 =4.497x10-6Subbing our equation 1 in for q2 we get q1+1.84x10-11/q1 = 4.497x10-6 . We can then rewrite thisas q12-4.497x10-6q1-1.84x10-11=0. We can then use quadratic formula ([-b±√(b2-4ac)]/2a) to solve for our initial charges. The quadratic formula will yield the absolute value of both initial charges.Lecture 3 (January 20)Determine whether the following statements are “True” or “False”:a. When a conductor is in equilibrium, the electric field everywhere inside the conductor is zero.b. There are charges within a conductor that are free to move about. When a conductor is in equilibrium, it means that there is a net motion of these charges within the conductor and on its surface.c. At equilibrium, any excess charge placed on a conductor resides entirely on its outer surface.d. The external electric field at the surface of a conductor is parallel to that surfaceAnswers:a. Trueb. Falsec. Trued. FalseExplanation:Within a conductor, charges are free to move without hindrance. When they are free to move around, charges will reside on the surface of the conductor. There cannot be any charge inside the conductor, meaning that Q inside the conductor is equal to 0. Since total flux through an enclosed surface is Q/E, the electric field inside the conductor is also zero. If there is no net electric field, there is no net movement. The external electric field at the surface of a conductor is perpendicular to that surface. If the field component was parallel to the surface then charges would move along the surface, which would violate the assumption of equilibrium.Lecture 4 (January 22)Three charges, Q1, Q2 and Q3 are located on a straight line. The charge Q3 is located 0.147 m to the right of Q2. The charges Q1 = 1.9 μC and Q2 = -3.03 μC are fixed at their positions, distance 0.279 m apart, and the charge Q3 = 3.33 μC could be moved along the line. For what position of Q3 relative to Q1 is the net force on Q3 due to Q1 and Q2 zero? Give your answer in meters, and use the plus sign for Q3 to the right of Q1.Answer: -1.0615mExplanation:Fnet=F1+F2=0  F1=-F2Kq1q3/r12 =-kq2q3/r22  (9x109)(1.9x10-6)(3.33x10-6)/r2=(-9x109)(-3.03x10-6)(3.33x10-6)/(r+.279)2.0569/r2=.0908/(r+.279)2  ((r+.279)/r)2=.0908/.0569 1+(.279/r) = √(.0908/.0569).279/r=√(.0908/.0569)-1Lecture 5 (January 27)The potential 0.010 m from a point charge is 0.10 V. What is the magnitude of the point charge?Answer:1.11x10-13CExplanation:V=kq/r.1=(9x109)q/.01Lecture 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 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