SATHEE: Electrostatics Question 462

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 $e = 9 \times 10^{- 31} K g .$ 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 $q E = m a$

therefore $a = \frac{Q E}{m}$

$∴ a = \frac{1.6 \times 10^{- 19} \times 1 \times 10^{3}}{9 \times 10^{- 31}} = \frac{1.6}{9} \times 10^{15}$

$u = 5 \times 10^{6}$ and $v = 0$ From $v^{2} = u^{2} - 2 a s$

therefore $s = \frac{u^{2}}{2 a}$ Distance $s = \frac{{(5 \times 10^{6})}^{2} \times 9}{2 \times 1.6 \times 10^{15}}$

$= 7 c m . (approx)$

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Electrostatics Question 462

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