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Physics Question on Electric charges and fields

An electron initially at rest falls a distance of 1.5 cm1.5\, cm in a uniform electric field of magnitude 2×104 N/C2 \times 10^4\, N/C. The time taken by the electron to fall this distance is

A

1.3×102 s1.3 \times 10^2 \,s

B

2.1×10−12 s2.1 \times 10^{-12} \,s

C

1.6×10−10 s1.6 \times 10^{-10} \,s

D

2.9×10−9 s2.9 \times 10^{-9} \,s

Answer

2.9×10−9 s2.9 \times 10^{-9} \,s

Explanation

Solution

Here, the direction of field is upward.
So the negatively charged electron experiences a downward force.


∴\therefore The acceleration of electron is
ae=eEme ...(i)a_{e}=\frac{eE}{m_{e}}\,...\left(i\right)
The time required by the electron to fall through a distance hh is
te=(2hAae)=2hmeeE t_{e}=\sqrt{\left(\frac{2h}{Aa_{e}}\right)}=\sqrt{\frac{2hm_{e}}{eE}}\, (using (i))
=[2×1.5×10−2×9.11×10−311.6×10−19×2×104]1/2=\left[\frac{2\times1.5\times10^{-2}\times9.11\times10^{-31}}{1.6\times10^{-19}\times2\times10^{4}}\right]^{^{1/2}}
=2.9×10−9 s=2.9\times10^{-9}\,s