- Part A An electron is moving east in a uniform electric field of 1.55 N/C directed to the west. At point A, the velocity of the electron is 4,52x105 m/s pointed toward the east, What is the speed of the electron when it reache: point B, which is a distance of 0.365 m east of point A? Express your answer in meters per second. ν ΑΣφ ve = m/s - Part B A proton is moving in the uniform electric field of part A. At point A, the velocity of the proton is 1.88x10 m/s , again pointed towards the east. What is the speed of the proton at point B? Express your answer in meters per second. Vp = m/s

B5

Hey I am doing some physics homework and am stumped on this question.  Any help will be great! Thanks!

In the ever popular game from the future, Electrodisc, players can score double points by pushing a disc into another disc with the electric field generated by a third disc. The discs are by regulation 1 kg. All discs have the same shape and are made of the same conducting material. In the figure below, disc 1 is held in place, and disc 2 is held stationary 4.1 m from disc 1 by a player. The player then releases disc 2. 10 m. F12 What is the initial magnitude of the acceleration imposed on the second disc? The charge on the first disc is 1.7 mC. The charge on the second disc is 1.7 mC. The charge on the third disc is neutral. m/s² Later, discs 2 and 3 touch and charge is exchanged. How much charge is exchanged once the charges on each disc reach an equilibrium? mC After charge is exchanged, what is the direction of the force on disc 3 due to disc 2? O toward the left O toward the right O upward O downward