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

Answer:

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

(b) Time taken will be $1.4\times 10^{-8}sec$

(c) Force will be $2189.9\times 10^{-19}N$              

Explanation:

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

Final velocity $v=2.50\times 10^6m/sec$

Mass of electron $m=9.11\times 10^{-31}kg$

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

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

(a) So $(2.5\times 10^6)^2=0^2+2\times a\times 0.013$

$a=240.3846\times 10^{12}m/sec^2$

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

So $2.50\times 10^6=0+240.3846\times 10^{12}t$

$t=0.014\times 10^{-6}=1.4\times 10^{-8}sec$

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

$F=ma=9.11\times 10^{-31}\times 240.3846\times 10^{12}=2189.9\times 10^{-19}N$