Can you please explain how the calculations were done

Example 22.1 Conservation of energy A 2.0 cm x 2.0 cm parallel-plate capacitor with a 2.0 mm spacing is charged to 11.0 nC. First a proton and then an electron are released from rest midway between the capacitor plates. Assume the motion takes place in a vacuum. a. What is each particle's energy? b. What is each particle's speed as it reaches the plate? SOLVE 1 a. Both charged particles have xi = 5d, where d = 2.0 mm is the plate separation. The proton (q = 3) begins with only potential energy, so its mechanical energy is 1 Similarly, the electron (q = e) has 1 Emeche = Ki + Ui = 0 eExi = eEd The electric eld inside the parallelplate capacitor, from Chapter 21 ID, is found to be E: i = 2.82x105 N/C 60A Thus the particles' energies can be calculated to be 5mm = 4.5 x 1017 J and Emeche = 4.5 x 1017 J b. Conservation of mechanical energy requires Kf + Uf = Ki + Uj = Emech . The proton collides with the negative plate, so Uf = 0, and the final kinetic energy is Kf = mpV? = Emech p. Thus the proton's impact speed is 2Emech p (vf) p .= 2.3 x 10' m/s mp Similarly, the electron collides with the positive plate at * = d, where Uf = qEd = -eEd = 2Emeche. Thus energy conservation for the electron is Kf = -meVF = Emeche - Uf = Emeche - 2Emeche = -Emeche We found the electron's mechanical energy to be negative, so Kf is positive. The electron reaches the positive plate with speed (ve)e = -2Emech e = 1.0 x 10' m/s me ESS Both particles have mechanical energy with the same magnitude, but the electron a much greater final speed due to its much smaller mass