Question

In a cathode ray tube, electrons are accelerated from rest by a constant electric force of magnitude 6.40 × 10−17 N during the first 6.40 cm of the tube’s length; then they move at essentially constant velocity another 45.0 cm before hitting the screen.Find the speed of the electrons when they hit the screen.How long does it take them to travel the length of the tube?

Answer 1

Given data

*The magnitude of the electric force is F = 6.40 × 10^-17 N

*The given length is d_1 = 6.40 cm = 0.064 m

The acceleration of the electrons is calculated as

`\begin{gathered} a=\frac{F_{}}{m} \\ =(6.40*10^(-17))/(9.1*10^(-31)) \\ =7.033*10^(13)\text{ m/s\textasciicircum{}2} \end{gathered}`

The formula for the speed of the electron after it travels 6.40 cm is given by the third kinematic equation of motion as

`v^2=u^2+2ad_1^{}^{}`

Substitute the known values in the above expression as