Question

In the uniform electric field of $E=1\times {10}^{4}N{C}^{-1}$, an electron is accelerated from rest. The velocity of the electron when it has travelled a distance of $2\times {10}^{-2}m$ is nearly ________ $m{s}^{-1}$ $(\frac{e}{m}$ of electron $\simeq 1.8\times {10}^{11}C{kg}^{-1}$)

• A. $1.6\times {10}^6$
• B. $0.85\times {10}^6$
• C. $0.425\times {10}^6$
• D. $8.5\times {10}^6$

Answer 1

The correct option is D $8.5\times {10}^6$

Given : Distance covered $S=2\times {10}^{-2}$ m

Acceleration of the electron $a=\frac{eE}{m}$

$\therefore$ $a=1.8\times {10}^{11}\times {10}^4=1.8\times {10}^{15}$ $m/s^2$

Initial speed of the electron is zero i.e. $u=0$

Using $v^2-u^2=2aS$

$\therefore$ $v^2-0=2(1.8\times {10}^{15})(2\times {10}^{-2})$ $⟹v=8.5\times {10}^6$ $m/s$