Question

An electron of mass 9.11×10-31kg 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 1.30cm away. It reaches the grid with a speed of 2.50×106m/s . The accelerating force is constant.

a)Find the acceleration.
b)Find the time to reach the grid.
c)Find the net force. (You can ignore the gravitational force on the electron).

Answer 1

(A) Acceleration will be
`240.3846* 10^(12)m/sec^2`

(b) Time taken will be
`1.4* 10^(-8)sec`

(c) Force will be
`2189.9* 10^(-19)N`

Step-by-step explanation:

We have given that electron starts from rest so initial velocity u = 0 m/sec

Final velocity
`v=2.50* 10^6m/sec`

Mass of electron
`m=9.11* 10^(-31)kg`

Distance traveled by electron
`s=1.30cm =0.013m`

From third equation of motion we know that
`v^2=u^2+2as`

(a) So
`(2.5* 10^6)^2=0^2+2* a* 0.013`

`a=240.3846* 10^(12)m/sec^2`

(b) From first equation of motion we know that v = u+at

So
`2.50* 10^6=0+240.3846* 10^(12)t`

`t=0.014* 10^(-6)=1.4* 10^(-8)sec`

(c) From newton's law we know that force

`F=ma=9.11* 10^(-31)* 240.3846* 10^(12)=2189.9* 10^(-19)N`