Phys 202 1nd EditionExam # 1 Study Guide Lectures: 1 - 6Lecture 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 negativecharge 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 1 (January 13)What is the equation for Coulomb’s Law? What is the equation for Gauss’s Law?Answer:Coulomb’s Law: F=kq1q2/r2Gauss’s Law: φelectric=Q/εExplaination:Coulomb's law states that the electrical force(F) between two charged objects is directly proportional to the product of the quantity of charge on the objects(q1 and q2) and inversely proportional to the square of the distance(r) between the two objects. k in this equation is a proportionality constant (Coulomb's law constant). The value of k depends on the medium that the object is in, but for air it is equal to 9.0 x 109 Nm2/C2.Gauss’s law states that that the net electric flux(φ) through any closed surface is equal to the charge(Q) inside that surface divided by the permittivity (ε0). Lecture 2 (January 15)Indicated whether the following statement is “True” or “False”:a. Electric Flux is the number of field lines that pass through a given surface because the magnitude of the electric field is inversely proportional to the number of lines of force in a given area.b. If a small surface area is oriented parallel to a uniform electric field direction, then the Electric Flux will equal Zero.c. If the electric field lines (E.F.L.) that emerge from a positive point charge, +q, pass through a spherical surface centered about the charge, then the number of field lines passing through the surface is dependent of the radius of the surface.Answer:a. Falseb. Truec. TrueExplanation: The electric field is proportional (not inversely proportional) to the number of electric field lines in a given area. Electric flux(φ) is the rate at which an electric field flows through a given area. The flux is proportional to the number of electric field lines going through a hypothetical surface. If the electric field is uniform, then the electric flux passing through a surface of vector area A is Δφ=|E||ΔA|Cosθ where E is the electric field, A is the area of the surface, and θ is the angle between the electric field lines and the normal (perpendicular) to A.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 =