Question

an electron travelling with a speed of 10cube 5 *10cubem/s passes through an electric field with an acceleration of 10 to the power of 12 metre per second square how long will it take for electron to double it's speed what will be the distance covered by the electron in this time​

Answer 1

Answer:

The distance covered by the electron in this time will be 25\times10^{-6}\ m25×10−6 m

Explanation:

Given that,

Speed v=5\times10^{3}\ m/sv=5×103 m/s

Acceleration a=10^{12}\ m/s^2a=1012 m/s2

Let us consider initial velocity = u

Final velocity = 2 u

Using equation of motion

v= u+atv=u+at

t= \dfrac{v-u}{a}t=av−u

t= \dfrac{2u-u}{a}t=a2u−u

t= \dfrac{u}{a}t=au

t= \dfrac{5\times10^{3}}{10^{12}}t=10125×103

t= 5\times10^{-9}\ st=5×10−9 s

The distance is the product of the speed and time.

The distance covered by the electron in this time will be

D=v\times tD=v×t

D=5\times10^{3}\times 5\times10^{-9}D=5×103×5×10−9

D= 25\times10^{-6}D=25×10−6

Hence, The distance covered by the electron in this time will be 25\times10^{-6}\ m25×10−6 m