Question

The electron gun in a television tube is used to accelerate electrons with mass 9.109 × 10−31 kg from rest to 3 × 107 m/s within a distance of 6.3 cm. Assuming the electric field is constant, what electric field is required in the tube? The fundamental charge is qe .

Answer 1

Answer:

Electric field, E = 40608.75 N/C

Explanation:

It is given that,

Mass of electrons, $m=9.1\times 10^{-31}\ kg$

Initial speed of electron, u = 0

Final speed of electrons, $v=3\times 10^7\ m/s$

Distance traveled, s = 6.3 cm = 0.063 m

Firstly, we will find the acceleration of the electron using third equation of motion as :

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(3\times 10^7)^2}{2\times 0.063}$

$a=7.14\times 10^{15}\ m/s^2$

Now we will find the electric field required in the tube as :

$ma=qE$

$E=\dfrac{ma}{q}$

$E=\dfrac{9.1\times 10^{-31}\times 7.14\times 10^{15}}{1.6\times 10^{-19}}$

E = 40608.75 N/C

So, the electric field required in the tube is 40608.75 N/C. Hence, this is the required solution.