**Question 1**

An electron, initially at rest, is accelerated through a potential difference of 550 V. What is the speed of the electron due to this potential difference?

Question 2

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.

Question 3

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 traveling. 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.

Considering the constant velocity of the particle in +ve x-direction, we have the time it takes the particle to reach point P(x,y)P(x,y) expressed as:

Answer to question 1

KE = PE

0.5*(9.1*10^-31)*V^2 = (1.6*10^-19)*550

V = 13907069.9073 = 1.39*10^7 m/s

Work done by moving electron through 550 V p.d. = kinetic energy gained by electron.

qV = eV =

Answer to question 2

(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 v^{2} 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 v^{2}

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

v = (100.258 * 10^{12})

v = = 10.01 * 10^{6} 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

Answer to question 3

Part (a)

Part (b)