Question

An electron moving with a velocity of 5 x 104 ms-1 enters a uniform electric field and acquires a uniform acceleration of 104 ms-2 in the direction of its initial motion. Determine the time in which magnitude of its velocity would double.

Answer 1

Answer:

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

Explanation:

Thanku

Regards

,Rajat

Answer 2

Answer:

An electron moving with a velocity of 5 * 10 4 m/s

Explanation:

An electron moving with a velocity of 5 * 10 4 m/s

An electron moving with a velocity of 5×104ms-1 enters into a uniform electirc field and acquires a uniform acceleration of 104ms-2 in the direction of its initial motion.