Question

An electron, initially at rest, is accelerated through a potential difference of 285 V. It then passes midway between two parallel plates providing a uniform electric field perpendicular to the direction in which it is travelling. The plates are 50 mm long and 25 mm apart and there is a potential difference of 71 V between them. Find

(a) The speed of the electron after its initial acceleration and
(b) The transverse deflection experienced by the electron as it emerges from between the plates.

Answer 1

(a) Potential energy of the electron when at rest is

P.E = e * V = 1.6022 * 10^-19 * 285 = 456.627 * 10^-19 J

After acceleration, kinetic energy of the electron is,

K.E = 1/2 m v² where m is the mass of the electron and v the speed.

From conservation of energy,

P.E = K.E

456.627 * 10^-19 = 1/2 m v²

v = (456.627 *10^-19 * 2/ m)^1/2 = (456.627 *10^-19 * 2/ 9.109 *10^-31)^1/2

v = (100.258 * 10¹²)

v = = 10.01 * 10⁶ m/s

(b) Transverse deflection is given by,

x = L D/ 2d * V_x/V, where

D is the deflection length, D = 50 * 10^-3 m

L is the distance from the middle of the deflection plates to the screen and in this case, L = D/2 =25 *10^-3 m

d is the distance between two plates, d = 25 * 10^-3 m

V_x is the potential difference between two plates, V_x = 71V

and V = 285 V

Thus,

x = 25 *10^-3 * 50 * 10^-3 * 71 / ( 2 * 25 *10^-3 *285 )

x = 6.228 * 10^-3 m = 6.228 mm