An electron moving horizontally enters a region of constant

An electron moving horizontally enters a region of constant electric field produced by a parallel-plate capacitor with the positive plate on the bottom and the negative plate on the top. The electric field is directed perfectly vertically. The electron enters the Held with a speed of 4.55 x 10^6 m/s and exits the field 0.618 cm lower than it entered. The horizontal distance covered within the plates is 2.50 cm. Gravitation force is negligible.

Solution

The time take spend by the electron between the capacitor plates

                                                           t = distance /velocity

                                                              = 2.5 * 10 ^-2 m/ 4.55 * 10 ^ 6 m/ sec

                                                              t = 5.494 * 10 ^ - 7 sec

let Fe be the electric force experiencd by the electron in time t and covered a verticle distance d

                                                     d = at ^2 / 2        , a is the vertical acceleration due to E field

                                                    a =2d / t ^2

                                                          = 2 * 0.618*10^-2 / (5.494 * 10 ^ - 7 sec)^2

                                                         = 4.094*10^10 m / sec^2

      electric force F = qE = m * a       , m and q are mass and charge of electron

                                 E = m * a / q

                                    E = 0.232 N C^-1

           the final velocity V = sqrt (Vx ² + Vy ²)

                         Vx = 4.55 * 10^6 m/s         Vy = a * t = 4.094*10^10 m / sec^2 * 5.494 * 10 ^ - 7 sec

                                                                             = 0.022519*10^6

                                   V = sqrt (4.55 * 10^12 + 0.022519*10^12)

                                        = 4.572519 * 10^6 m/sec