Products Rewards
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY

HOLOOLY
TABLES

All the data tables that you may search for.

HOLOOLY
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY
HELP DESK

Need Help? We got you covered.

## 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

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