Question

An electron cathode ray tube is accelerated when it passes between two parallel plates separated by 5.0 cm. If the electron is initially moving with speed vi = 1000 m/s to the right and it has a speed of vf = 5500 m/s after leaving the acceleration region, what is the potential difference deltaV between the plates and which plate, the right or the left, is at the highest potential? Justify your answer.

Answer 1

Sepration of the two parallel plates d = 5 cm = 0.05 m

Initial velocity of the electron v = 1000 m/s

Final velocity of the electron V = 5500m/s

From the relation V ² - v ² = 2ad

Accleration of the electron a = [V ² - v ²] / 2d

                                          = 292.5 x10 ⁶ m/s ²

We know a = Eq / m

Where m = mass of electron = 9.1 x10 ^-31 kg

            q = charge of electron = (1.6 x10 ^-19 )C

From this electric field E = ma / q

                                    = 1.663 x10 ^-3 N/C

Potential difference = E d

                            = 1.663 x10 -3 x 0.05

                             = 8.31 x10 ^-5 volt

Right plate is at highest potential.Since it moves with right and it speed is increases