Q. An electron moving with the speed $5 \times 10^{6}\ m$ per sec is shooted parallel to the electric field of intensity $1 \times 10^{3}N/C.$ Field is responsible for the retardation of motion of electron. Now evaluate the distance travelled by the electron before coming to rest for an instant (mass of $e = 9 \times 10^{- 31}\ Kg.\ charge = 1.6 \times 10^{- 19}C$)
A) $7\ m$
B) $0.7\ mm$
C) $7\ cm$
D) $0.7\ cm$
Explanation
(c) Electric force $qE = ma \Rightarrow a = \frac{qE}{m}$
$\therefore a = \frac{1.6 \times 10^{- 19} \times 1 \times 10^{3}}{9 \times 10^{- 31}} = \frac{1.6}{9} \times 10^{15}$
$u = 5 \times 10^{6}$ and $v = 0$
$\therefore From\ v^{2} = u^{2} - 2as \Rightarrow s = \frac{u^{2}}{2a}$
$\therefore$ Distance $s = \frac{\left( 5 \times 10^{6} \right)^{2} \times 9}{2 \times 1.6 \times 10^{15}} = 7\ cm.$ (approx)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.