Question

An electron moving with a velocity of 5×103m/s5×103m/s enters
into a uniform electric field and acquires a uniform acceleration
of 103m/s2103m/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:

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

Answer 2

Answer:

(1) Initial velocity of the electron is u =5×10

4

m/s

Acceleration of the electron is a=10

4

m/s

2

From first equation of motion,

v=u+at

2u=u+at

t=u/a=(5×10

4

)/(10

4

)=5s

(C) 5