Question

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

• A. $1.3\times {10}^{2}s$
• B. $2.1\times {10}^{-12}s$
• C. $1.6\times {10}^{-10}s$
• D. $2.9\times {10}^{-9}s$

Answer 1

The correct option is D $2.9\times {10}^{-9}s$

In figure the field is upward. So the negatively charged electron experiences a downward force.

$\therefore$ The acceleration of electron is

$a_e=\frac{eE}{m_e}$

The time requried by the electron to fall through a distance $h$ is

$t_e=\sqrt{\left(\frac{2h}{a_e}\right)}=\sqrt{\frac{2h{m}_e}{eE}}$ [using (i)]

$={\left[\frac{2\times 1.5\times {10}^{-2}\times 9.11\times {10}^{-31}}{1.6\times {10}^{-19}\times 2\times {10}^4}\right]}^{1/2}=2.9\times {10}^{-9}s$