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}) $



NCERT Chapter Video Solution

Dual Pane