Question

An electron moving with the speed $5\times 10^6$ 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. 7cm
• D. 0.7cm

Answer 1

Answer: (C)

Electric force, $qE=ma$
$\Rightarrow a=\frac{qE}{m}$
$\therefore a=\frac{1.6\times 10^{-19}\times 1\times 10^3}{9\times 10^{-31}}$
$=\frac{1.6\times 10^5}{9}$
$u=5\times 10^6$ and $v=0$
$\therefore$ From $v^2=u^2-2as$
$\Rightarrow s=\frac{u^2}{2a}$
$\therefore$ Distance, $s=\frac{{\left(5\times 10^6\right)}^2\times 9}{2\times 1.6\times 10^{15}}\simeq 7cm$