Question

The electron gun in a TV picture tube accelerates electrons between two parallel plates 1.2 cm apart with a 30 kV potential difference between them. The electrons enter through a small hole in the negative plate, accelerate, then exit through a small hole in the positive plate. Assume that the holes are small enough not to affect the electric field or potential.
a)What is the electric field strength between the plates?

b)With what speed does an electron exit the electron gun if its entry speed is close to zero? Note: the exit speed is so fast that we really need to use the theory of relativity to compute an accurate value. Your answer to part B is in the right range but a little too big.

Answer 1

Answer:

E = 2.5*10⁶ V/m  

ve = 1.03*10⁸ m/s

Explanation:

a) Assuming no other external forces acting on the electron than the electric field between the plates, the work per unit charge -which is equal to the potential difference- that produces the displacement of the electron between the plates, is as follows:

$V = E*d$

where

V= 30 kV

d = 0.012 m

Solving for E, we get:

$E = \frac{30e3 V}{0.012m} =2.5e6 V/m$

b) If we can assume that the initial speed of the electron is zero, just due to conservation of energy, we can say as follows:

ΔK + ΔU = 0

⇒ ΔK = -ΔU

⇒ $\frac{1}{2}*me*ve^{2} = -(-e)*V$

where

me = mass of the electron= 9.1*10⁻³¹ kg.

e = elementary charge = 1.6*10⁻¹⁹ C.

V = potential difference between plates = 30 kV

Replacing by the values, we can solve for ve, as follows:

$ve =\sqrt{\frac{2*1.6e-19 C*30e3V}{9.1e-31kg} } = 1.03e8 m/s$

As this speed is an important fraction of c (speed of light) it would be needed to take into account the relativistic effects.