An electron initially at rest falls a distance of 1.5 cm in a uniform electric field of magnitude 2$ imes 10^{4}$ N/C. The time taken by electron to fall this distance is

Question

An electron initially at rest falls a distance of 1.5 cm in a uniform electric field of magnitude 2$	imes 10^{4}$ N/C. The time taken by electron to fall this distance is

Dr. Abubacker · Accepted Answer

(d) As the electron is moving in a straight line and starts from rest.
Thus, from
$s = ut + \frac{1}{2}at^{2}$
We get
$h = (0)t + \frac{1}{2}a_{e}t^{2}$
$t = \sqrt{\frac{2h}{a_{e}}}$
$= \sqrt{\frac{2hm_{e}}{eE}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( a_{e} = \frac{eE}{m_{e}} ight)$
$= \sqrt{\frac{2 	imes 1.5 	imes 10^{- 2} 	imes 9.1 	imes 10^{- 31}}{1.6 	imes 10^{- 19} 	imes 2 	imes 10^{4}}}$
$= 2.9 	imes 10^{- 9}s$

Q. An electron initially at rest falls a distance of 1.5 cm in a uniform electric field of magnitude 2$\times 10^{4}$ N/C. The time taken by electron to fall this distance is

Explanation

Reviewed by Dr. Abubacker

Want more practice?