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.

Respuesta :

Answer: a) 2.5 * 10 ^6 N/C; b)  102.71* 10^6 m/s

Explanation:  In order to explain this problem we have to use the following expressions:

E= V/d the relationship of the electric field and the voltage. d is the separation distance between the plates.

Also we know,

e*V= (m*v^2)/2 ;  this means that the change in potencial energy is gained by the electrons and converted in kinetic energy.

v= (2*e*V/m)^1/2= (2*1.6*10^-19*30000/9.1*10^-31)^1/2=102.71 * 10^6 m/s