Question

Q6. An electron moving with a velocity of 5x10³m/s enters into a uniform electric field and
acquires a uniform acceleration of 10³ m/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:

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Explanation:

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QUESTION:- An electron moving with a velocity of 5×10^4 m/s enters into a uniform electric field and acquiers a uniform acceleration of 10^4 m/s^2 in the direction of its initial motion

1. Calculate the time in which the electron would acquire a velocity doubke of its initial velociry and

2) how much distance the electron would cover in this time

ANSWER:-

First of all we can write the given information

An electron moving with velocity of 5X10^4m/s

U= 5X10 ^4 m/s

And here the acceleration of the electron is

Given by

a= 10^4m/s

(1):- V=U+at

2U=U+at

=> t=u/a

So, => (5x10^4)/ (10^4)

By cutting 10^4 and 10^4

We will get the answer as 5sec

(2). So the second part is the distance traveled by the electron

S=U+1/2 at

=> s= (5x10^4)+5x 1/2 x 10^4X 5^2

=> s= 3,57,000 m

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