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

$t=\frac{d}{v}$

Substitute the known values in the above expression as

$\begin{gathered} t=\frac{6.20}{(2.10\times10^9^{})} \\ =2.95\times10^{-9}\text{ s} \\ =2.95\text{ ns} \end{gathered}$

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

$d=\frac{1}{2}at^2$

Substitute the known values in the above expression as

$\begin{gathered} d=\frac{1}{2}(5.30\times10^{17})(2.95\times10^{-9})^2 \\ =2.31\text{ cm} \end{gathered}$

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