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 :

Answer:

(A) Acceleration will be [tex]240.3846\times 10^{12}m/sec^2[/tex]

(b) Time taken will be [tex]1.4\times 10^{-8}sec[/tex]

(c) Force will be [tex]2189.9\times 10^{-19}N[/tex]              

Explanation:

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

Final velocity [tex]v=2.50\times 10^6m/sec[/tex]

Mass of electron [tex]m=9.11\times 10^{-31}kg[/tex]

Distance traveled by electron [tex]s=1.30cm =0.013m[/tex]

From third equation of motion we know that [tex]v^2=u^2+2as[/tex]

(a) So [tex](2.5\times 10^6)^2=0^2+2\times a\times 0.013[/tex]

[tex]a=240.3846\times 10^{12}m/sec^2[/tex]

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

So [tex]2.50\times 10^6=0+240.3846\times 10^{12}t[/tex]

[tex]t=0.014\times 10^{-6}=1.4\times 10^{-8}sec[/tex]

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

[tex]F=ma=9.11\times 10^{-31}\times 240.3846\times 10^{12}=2189.9\times 10^{-19}N[/tex]

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