An electron moving with a velocity of 5×104ms-1 enters into a uniform electric field and acquires a uniform acceleration of 104ms-2 in the direction of its initial motion (i) Calculate the time in which the electron would acquire a velocity double of its initial velocity. (ii) How much distance the electron would cover in this time?

(i) The time in which the electron would acquire a velocity double of its initial velocity is 5s.

(ii) The distance the electron would cover in this time is 37.5×10⁴m.

Step 1: Given that

Acceleration acquired by the electron, a=10⁴ms^-2

Initial velocity of the electron, u=5×10⁴ms^-1

Step 2: Consider the formulae to calculate Distance covered

We can infer the following using the three motion equations:

Distance covered, s=ut+1/2at²,

Where u is the starting speed, t is the travel time, and an is the body’s acceleration.

We also know that,

time, ∆t=v-ua∵a=∆v∆t

Step 3: Now we have to find the time required to double the initial velocity

We require that the final velocity be twice as fast as the initial velocity, thus, v=2u=2×5×10⁴=10⁵ms^-1.

We require that the final velocity be twice as fast as the initial velocity is, t=10×10⁴-5×10⁴/10⁴=5s

Step 4: Now we have to find the distance covered

The distance travelled is determined by the first equation by

s=5×10⁴×5+1/2×10⁴×5²

On simplifying we get

s=37.5×10⁴m

Therefore,

Five seconds are needed to double the initial velocity.

distance travelled while double the initial velocity is 37.5×10⁴m.