Question

An electron moving parallel to a uniform electric field increases its speed from 2.0 × 107 m/s to 4.0 × 107 m/s over a distance of 1.9 cm. You may want to review (Pages 646 - 648) . Part A What is the electric field strength?

Answer 1

`1.8* 105 N/C`

Step-by-step explanation:

We are given that

`u=2* 10^7 m/s`

`v=4* 10^7 m/s`

d=1.9 cm=
`(1.9)/(100)=0.019 m`

Using 1m=100 cm

We have to find the electric field strength.

`v^2-u^2=2as`

Using the formula

`(4* 10^7)^2-(2* 10^7)^2=2a(0.019)`

`16* 10^(14)-4* 10^(14)=0.038a`

`0.038a=12* 10^(14)`

`a=(12)/(0.038)* 10^(14)=3.16* 10^(16)m/s^2`

`q=1.6* 10^(-19) C`

Mass of electron,m
`=9.1* 10^(-31) kg`

`E=(ma)/(q)`

Substitute the values

`E=(9.1* 10^(-31)* 3.16* 10^(16))/(1.6* 10^(-19))`

`E=1.8* 105 N/C`