Question

Electric field in a region is N / C towards west. The force experienced by an electron moving towards east with velocity 5*10^ 4 m/s is (charge of the electron is^ prime e^ prime ).​

Answer 1

Answer:

Explanation:

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

Answer 2

Given : (u) initial velocity = 5 x $10^{4}$ m/s

Assume that (a) acceleration = $10^{4}$ m/$s^{-2}$

To find the force of electron moving towards east

Final velocity (v) = 2 u

                       = 2 x 5 x $10^{4}$ m/s

v = 10 x $10^{4}$ m/s .

We need to find t

Here, v = at

or  t = $u-\frac{u}{a}$

therefore, 5 x $10^{4}$/$10^{4}$

t = 5s

Using s = $u t +\frac{1}{2}$ a$t^{2}$

s = (5 × 104) × 5 + 1/2 × 104 × (25)

s = 25 × 104 + (25 × 104) /2

s = 37.5×104 m

hence time required for electron move is 5s