Question

An electron moving to the right at 7.5 x 10^5 m/s enters a uniform electric field parallel to its direction of motion. If the electron is to be brought to rest in the space of 7.0 cm . What is the strength of the field?

Answer 1

The strength of the field is 22.84 N/C.

Step-by-step explanation:

Given that,

Speed
`v= 7.5*10^(5)\ m/s`

Distance = 7.0 cm

We need to calculate the acceleration

Using equation of motion

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

Put the value in the equation

`0-(7.5*10^(5))^2=2* a*7.0*10^(-2)`

`a =-((7.5*10^(5))^2)/(2*7.0*10^(-2))`

`a =-4.017*10^(12)\ m/s^2`

We need to calculate the strength of the field

Using newton's second law and electric force

`F = ma = qE`

`-qE=-ma`

`E=(ma)/(q)`

Put the value into the formula

`E=(9.1*10^(-31)*4.017*10^(12))/(1.6*10^(-19))`

`E=22.84\ N/C`

Hence, The strength of the field is 22.84 N/C.