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## Q. 3.6

Two plane parallel plates A and B are placed 8 mm apart and plate B is 300 V more positive than plate A. The electron travels from plate A to plate B with an initial velocity of $1 × 10^{6}$ m/s. Calculate the time of travel.

## Verified Solution

The speed acquired by the electron due to the applied voltage is,

$v=\sqrt{v^{2}_{initial}+\frac{2qV}{m} }$

$=\sqrt{(1\times 10^{6} )^{2}+\frac{2(1.602 \times 10^{-19} )300}{9.1\times 10^{-31} } }$

$= 10.33 × 10^{6} m/s$

The average velocity,

$v_{average}=\frac{v_{initial} + v_{final}}{2}=\frac{1\times 10^{6}+10.33\times 10^{6} }{2}=5.665\times 10^{6} m/s$

Therefore, the time for travel

$=\frac{separation between plates}{v_{average} }=\frac{8\times 10^{-3} }{5.665\times 10^{6} }=1.4\times 10^{-9} s$