Electrostatics Question 462
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) $ [MP PMT 2003]
Options:
A) 7 m
B) 0.7 mm
C) 7 cm
D) 0.7 cm
Show Answer
Answer:
Correct Answer: C
Solution:
Electric force $ qE=ma $
therefore $ 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 $From $ v^{2}=u^{2}-2as $
therefore $ s=\frac{u^{2}}{2a} $Distance $ s=\frac{{{(5\times 10^{6})}^{2}}\times 9}{2\times 1.6\times 10^{15}} $
$ =7cm.(\text{approx}) $
BETA
