Question

An electron with speed v0 = 27.5 X 106 m/s is traveling parallel to a uniform electric field of magnitude E = 11.4 X 103 N/C. How far will the electron travel before its velocity is zero?

Answer 1

Answer:

Electron will travel to a distance of 0.189 m before its velocity is 0    

Explanation:

We have initial speed of the electron $u=27.5\times 10^6m/sec$

Electric field $E=11.4\times 10^3N/C$

Charge on electron will be $q=1.6\times 10^{-19}C$

Mass of electron $m=9.1\times 10^{-31}kg$

We know that Columbus force is equal to F = qE and according to newtons law force is equal to F = ma , here m is mass and a is acceleration

These two forces will be equal

So ma = qE

$a=\frac{qE}{m}=\frac{1.6\times 10^{-19}\times 11.4\times 10^3}{9.1\times 10^{-31}}=2\times 10^{15}m/sec^2$

According to third equation of motion

$v^2=u^2+2as$

So $0^2=(27.5\times 10^{6})^2+2\times 2\times 10^{15}\times s$

s = 0.189 m

So electron will travel to a distance of 0.189 m before its velocity is 0