Question

As the electrons are accelerated through the second anode, the gain in kinetic energy is 2.0 x 10-15 J, and the speed of the electrons as they enter the region between the plates is 6.6 x 107 m/s. The electrons are moving to the right as they pass between the plates. The plates are 2.0 cm long, 1.0 mm apart, and as the electrons pass between the plates, the potential difference is 450 V. Determine the time it takes to pass the region between the plates. Determine the electric field strength between the plates. Determine the acceleration of the electron in the region between the deflection plates. Determine the vertical component of the velocity of the electrons when they emerge from the region between the plates. Determine the vertical displacement of the electrons as they are deflected.

Answer 1

a)

The time it takes for the electrons to pass between the plates is obtained by using: S = Vx t

Where: S = plate length of 2 cm, and V, is the velocity of the electrons as they enter the region betwen the plates. This gives us:-

t = S/V

t = 2 x 10^-2/6.6 x 10^7

t = 3.02 x 10^-10 s

b)

The electric field between the plates is given by: E = V/d

Where: V is the voltage applied between the plates and d, is the plate separation.

E = V/d

E = 450/1x 10^-3

E = 4.5 x 10^5 Vm-1 (or NC-1)

c)

To determine the accleration on the electrons as thay pass between the plates we use:-

F = qE and F = ma

Where: F is the force eleperience by the electon (N); Q is the electronic charge: 1.6 x 10^-19 C;

m is the mass of the electron (kg) and a is the acceleration of the electron (ms-2)

Equating the two expressions gives us: ma = qE hence:-

a = qE/m

a = (1.6 x 10^-19 x 4.5 x 10^5)/9.1 x 10^-31

a = 7.9 x 10^16 ms-2

d)

Use v = u +at

Since tthe electron beam enters horizontally between the plates then there is no initial vertical component of velocity, in other words: u = 0.

v = at

v = 7.9 x 10^16 x 3.02 x 10^-10

v = 2.39 x 10^7 ms-1

e)

The vertical displacement of the electrons as they leave the plate deflected is derived by making use of: S = ut +0.5 at^2

But, u = 0. This means that S = 0.5 at^2

Where: S, is the deflection on emerging from the plates t = 3.01 x 10^-10 s later.

S = 0.5 x 7.9 x 10^16 x (3.02 x 10^-10)^2

S = 0.0036 m

S = 3.6 mm