Question

An electronic in a cathode ray tube is traveling horizontally at 2.10 × 10 to the power of 9 cm/s when deflection plates give it an upward acceleration of 5.30 × 10 to the power of 17cm/s2. (a) How long does it take for the electron to cover a horizontal distance of 6.20cm square? (b) what is its vertical displacement during that time?

Answer 1

Given data

*The given speed of the cathode ray tube is v = 2.10 × 10^9 cm/s

*The given acceleration is a = 5.30 × 10^17 cm/s^2

*The given horizontal distance is d = 6.20 cm

The formula for the time taken by the electron to cover a horizontal distance is given as

`t=(d)/(v)`

Substitute the known values in the above expression as

`\begin{gathered} t=\frac{6.20}{(2.10*10^9^{})} \\ =2.95*10^(-9)\text{ s} \\ =2.95\text{ ns} \end{gathered}`

Hence, the time taken by the electron to cover a horizontal distance is t = 2.95 ns

(b)

The formula for the vertical displacement is given by the equation of motion as

`d=(1)/(2)at^2`

Substitute the known values in the above expression as

`\begin{gathered} d=(1)/(2)(5.30*10^(17))(2.95*10^(-9))^2 \\ =2.31\text{ cm} \end{gathered}`

Hence, the vertical displacement during that time is d = 2.31 cm