A proton enters a parallelplate capacitor traveling to the r

A proton enters a parallel-plate capacitor traveling to the right at a speed of 1.277 10-5 m/s, as shown in the figure. The distance between the two plates is 1.64 cm. The proton enters the capacitor halfway between the top plate and the bottom plate; that is, a distance r = 0.820 cm from each plate, as shown in the figure. The capacitor has a 2.70 10-4 N/C uniform electric field between the plates that points downward from the top plate to the bottom plate. Neglecting gravitational forces, what horizontal distance does the proton traverse before the proton hits the bottom plate?

Solution

vertical acceleration created by E field between plates of capacitor

                                      a = qE / m

                              = 1.602 *10^-19 C * 2.70 10-4 N/C / 9.109*10^-31 kg

                            a = 4.748*10^7 m/s2

let t be the time taken by electron to reach the plate,

                                                                t = sqrt (2d /a )          d be the distance to target = 0.00820 m ,

                                                                                                  a = acceleration due to E-field = 4.748*10^7 m/s2

                                                                 t = 1.858 * 10^-5 sec          before hitting plate

horizantal distance travelled by electron before hitting the capacitor plates

                                                            distance = time * velocity

                                                                             = 1.858 * 10^-5 sec * 1.277*10^-5

                                                                          = 2.372*10^-10 m