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Question: An electron initially at rest falls a distance of 1.5 cm in a uniform electric field of magnitude \(...

An electron initially at rest falls a distance of 1.5 cm in a uniform electric field of magnitude 2×1042 \times 10^{4} N/C The time taken by the electron to fall this distance is.

A

1.3×102s1.3 \times 10^{2}s

B

2.1×10−12s2.1 \times 10^{- 12}s

C

1.6×10−10s1.6 \times 10^{- 10}s

D

2.9×10−9s2.9 \times 10^{- 9}s

Answer

2.9×10−9s2.9 \times 10^{- 9}s

Explanation

Solution

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

∴\thereforeThe acceleration of electron is

=ae=eEme= a_{e} = \frac{eE}{m_{e}}

The time required by the electron to fall through a distance h is.,

te =(2hae)=2hmeeE= \sqrt{\left( \frac{2h}{a_{e}} \right)} = \sqrt{\frac{2hm_{e}}{eE}} (using (i))

=[2×1.5×10−2m×9.11×10−311.6×10−19C×2×104]= \left\lbrack \frac{2 \times 1.5 \times 10^{- 2}m \times 9.11 \times 10^{- 31}}{1.6 \times 10^{- 19}C \times 2 \times 10^{4}} \right\rbrack

=2.9×109s= 2.9 \times 10^{9}s