The motion of the alpha particle can be analyzed by considering its kinetic and potential energies. At 1.75 cm, the electric potential is 3 volts, which means that the alpha particle experiences a repulsive force due to the electric field. As a result, the alpha particle will be accelerated in the direction of decreasing potential. Initially, the alpha particle is at rest and has no kinetic energy. Therefore, all of its energy is potential energy, which is equal to the product of its charge and the electric potential at its position. As it moves towards the left, the electric potential decreases, and its potential energy is converted into kinetic energy. The conservation of energy principle states that the sum of kinetic and potential energy is constant. Thus, as the potential energy decreases, the kinetic energy of the alpha particle increases. The alpha particle will continue to accelerate towards the left until it reaches a point where the potential energy is minimum, and the kinetic energy is maximum. At this point, the alpha particle will have its maximum speed, which is determined by the conservation of energy principle. After reaching the point of minimum potential energy, the alpha particle will begin to slow down as it moves towards the right. This is because the electric potential increases, and its kinetic energy is converted back into potential energy. Eventually, the alpha particle will come to a stop at a position where the potential energy is equal to the initial potential energy at 1.75 cm. At this point, the alpha particle will have zero kinetic energy and maximum potential energy. The motion of the alpha particle can be seen as a pendulum-like oscillation between the two turning points where the kinetic energy is zero and the potential energy is maximum. The period of this oscillation is determined by the mass and the potential energy of the alpha particle. Since the alpha particle has four times the mass of a proton, its oscillation period will be longer than that of a proton with the same initial conditions. In summary, the alpha particle will be accelerated towards the left, reach a maximum speed at the point of minimum potential energy, and then oscillate back and forth between the turning points, where the kinetic energy is zero, and the potential energy is maximum. The motion of the alpha particle is determined by the conservation of energy principle and the variation of the electric potential with position. The graph in the figure shows the variation of the V (x) electric potential V(x) in volts as a function of the position x in cm in a certain region of space. An alpha particle (which is a nucleus of helium) has twice the charge and four times the mass of the proton. The alpha particle is released from rest at 1.75cm (where the electric potential is 3 volts). Describe the motion of the alpha particle after its release in as much detail as possible. -2 -1.5 -1 -0.5 0.5 1 1.5 2xThe graph in the figure shows the variation of the V (x) electric potential V(x) in volts as a function of the position x in cm. [This is the same graph from question (2).] (a) Describe the electric field in this region in as much detail as possible. (b) Explain how your answer to (a) is consistent with your answer to question (2). -2 -1.5 -1 -0.5 0.5 1 1.5 2x