4.6) An electron of mass 9.11×10 31 kg leaves one end of a TV picture tube with

Question

4.6)

An electron of mass 9.11×1031 kg leaves one end of a TV picture tube with zero initial speed and travels in a straight line to the accelerating grid, which is a distance 1.95 cm away. It reaches the grid with a speed of 3.10×106 m/s . The accelerating force is constant.

Find the acceleration.

Find the time to reach the grid.

Find the net force. (You can ignore the gravitational force on the electron).

Find the time to reach the grid.

Find the net force. (You can ignore the gravitational force on the electron).

Explanation / Answer

Here,

mass of electron, m = 9.11 *10^-31 Kg

distance , d = 1.95 cm = 0.0195 m

speed of electron , u = 3.1 *10^6 m/s

let the acceleration of the electron is a

Using third equation of motion

v^2 - u^2 = 2 * a * d

(3.1 *10^6)^2 = 2 * a * 0.0195

a = 2.464 *10^14 m/s^2

the acceleration is 2.464 *10^14 m/s^2

let time to reach the grid is t

Using first equation of motion

3.1 *10^6 = 2.464 *10^14 * t

t = 1.258 *10^-8 s

the time to reach the grid is 1.258 *10^-8 s

----

the net force on electron = mass * acceleartion

net force on electron = 2.464 *10^14 * 9.11 *10^-31

net force on electron = 2.244 *10^-16 N

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