Question

A proton, initially traveling in the +x-direction with a speed of 5.05×10^5 m/s , enters a uniform electric field directed vertically upward. After traveling in this field for 3.90×10^−7 s , the proton’s velocity is directed 45° above the +x-axis. What is the strength of the electric field?

Answer 1

The strength of the electric field is
`1.35*10^(4)\ N/C`.

Step-by-step explanation:

Given that,

Speed
`v= 5.05*10^(5)\ m/s`

Time
`t= 3.90*10^(-7)\ s`

Angle = 45°

We need to calculate the acceleration

Using equation of motion

`v = u+at`

`5.05*10^(5)=0+a*3.90*10^(-7)`

`a =(5.05*10^(5))/(3.90*10^(-7))`

`a=1.29*10^(12)\ m/s^2`

We need to calculate the strength of the electric field

Using relation of newton's second law and electric force

`F= ma=qE`

`ma = qE`

`E=(ma)/(q)`

Put the value into the formula

`E=(1.67*10^(-27)*1.29*10^(12))/(1.6*10^(-19))`

`E=1.35*10^(4)\ N/C`

Hence, The strength of the electric field is
`1.35*10^(4)\ N/C`.