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

Question

An electron ( 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 1.77 cm away. It reaches the grid with a speed of 2.97×106 m/s . (You can ignore the gravitational force on the electron.)

If the accelerating force is constant, compute the acceleration of the electron.

If the accelerating force is constant, compute the acceleration of the electron.

If the accelerating force is constant, compute the net force that is accelerating the electron, in newtons.

Explanation / Answer

Given that

An electron of mass (me) = 9.11×1031 kg )

The intial speed of the electron leaving one end of a TV picture tube with zero initial speed (u) =0

Distance travelled by the electron (d) =1.77 cm =0.0177m

The final speed (v) = 2.97×106 m/s

The acceleration of the electron is given by

v2-u2 =2as

v2-0 =2as ===>v2 =2as

a =v2/2s =( 2.97×106 m/s)2/2*0.0177m=8.8209*1012/0.0354=249.177*1012m/s2=2.49*1014m/s2

Time taken for theelectron to reach the grid is

v =u+at ===> v =at then time t =v/a = 2.97×106 m/s /2.49*1014m/s2=1.19*10-8s=11.9ns

If the accelerating force is constant, compute the net force that is accelerating the electron, in newtons.

then Fnet =ma

              =(9.11*10-31)(2.49*1014m/s2)=22.68*10-17N =2.268*10-16N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.