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 $\left.=1.6 \times 10^{-19} \,C \right)$

• A. 7 m
• B. 0.7 mm
• C. 7cm
• D. 0.7cm

Answer 1

Answer: (C)

Electric force, $q E=m a$
$\Rightarrow a=\frac{q E}{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}-2 a s $
$\Rightarrow s=\frac{u^{2}}{2 a}$
$\therefore $ Distance, $s=\frac{\left(5 \times 10^{6}\right)^{2} \times 9}{2 \times 1.6 \times 10^{15}} \simeq 7 \,cm$