Question

An electron moving with a velocity of 5×103m/s enters into a uniform electric field and acquires a uniform acceleration of 103m/s2 in the direction of initial velocity.
(i) Find out the time in which electron velocity will be doubled?
(ii) How much distance electron would cover in this time?​

Answer 1

Answer:

(i) 5 s

(ii) 3.75 × 10⁴ m

Explanation:

Initial velocity(u) = 5 * 10³ m/s

Acceleration(a) = 10³ m/s²

(i):    Let the velocity is doubled in 't' seconds.

             Using equations of motion:

⇒ v = u + at

⇒ 2u = u + at          [final vel. is double]

⇒ u = at

⇒ 5 * 10³ = 10³ * t

⇒ 5 = t

         Required time = 5 s

(ii):   v² = u² + 2aS

⇒ (2u)² = u² + 2aS

⇒ u²(3/2a) = S

⇒ (5 * 10³)² * (3/2*10³) = S

⇒ 37.5 * 10³ = S

⇒ 3.75 × 10⁴ m = S

Answer 2

Given Data in Question :

• Intial Velocity (u) = 5 × 10³ m/s
• Acceleration (a) = 10³ m/s²

To Find :

1. Time in which electron velocity will be doubled .
2. Distance electron would cover in this time .

Solution :

1. Let the velocity doubled be 't' seconds.

also, v = 2u [.°. as velocity doubled]

formula : v = u + at

By applying the values ,

⇒ 2u = u + at

⇒ 2u - u = at

⇒ u = at

⇒ 5 * 10³ = 10³ * t

⇒ t = 5 seconds

• In 5 seconds, electron velocity will be doubled

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2. formula : v² - u² = 2as

⇒ v² = u² + 2as

⇒ (2u)² = u² + 2as

⇒ (3u²/2a) = s

⇒ (5 * 10³)² * (3/2 * 10³) = s

⇒ 25 * 10⁶ * (3/2) * 10³ = s

⇒ 37.5 * 10³ = s

⇒ 3.75 * 10⁴ = s

• Distance covered by the electron in 5 seconds is 3.75 * 10⁴ m