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ff;'Summary If work is done to move a positive charge against an electric field, then the work done is exactly equal to the increase in potential energy. W = APE = PEg - PE, If the charge is now released and allowed to return to the original height, the gain in kinetic energy will be equal to the magnitude of the loss in potential energy KE = APE The speed of the charged particle can be determined by 2 KE = -mv 2 The change in potential energy could be defined as APe =qgAV The work done by an external force to move a charge from a region of low potential to a region of high potential is W=qAV Thus when a charged particle is released and its potential energy becomes kinetic energy, the speed of the particle can be found using 1 2 _ Zmv = gAV The electric field strength between two charged plates, the voltage applied to the plates, and the distance between the plates are related by V E= d These principles can be applied to practical devices that make use of charged particles moving through electric fields. One such device is the ink-jet printer. For a charged particle moving through an electric field where it experiences a constant force in the vertical direction while initially moving in the horizontal direction, the kinematics equations that we used for projectile motion could also be used in this context. The equations for horizontal motion are Vx = Vxq X =Vt The equations for vertical motion are v, =v, +tat d=1(v +v,)t =sn+v, d=vt+Liars =2 . A d=v,t 2at V2 =v?+2ad Also keep in mind that the force acting on a charge in an electric field is given by F = qE. Also, the acceleration of a mass when a force is acting on it can be found using Newton' second law, F=ma. Question(s): The physics of a cathode ray tube: Part 11 In the lesson assignment, the functioning of a cathode ray tube was described. For the purposes of this question, let us assume the following initial conditions. As the electrons are accelerated through the second anode, the gain in kinetic energy is 2.0 x 1071 J, and the speed of the electrons as they enter the region between the plates is 6.6 x 107 m/s. The electrons are moving to the right as they pass between the plates. The plates are 2.0 cm long, 1.0 mm apart, and as the electrons pass between the plates, the potential difference is 450 V. 1. Determine the time it takes to pass the region between the plates. 2. Determine the electric field strength between the plates. 3. Determine the acceleration of the electron in the region between the deflection plates. 4. Determine the vertical component of the velocity of the electrons when they emerge from the region between the plates. 5. Determine the vertical displacement of the electrons as they are deflected