Question

An electron and a proton are each placed at rest in a uniform electric field of magnitude 554 N/C. Calculate the speed of each particle 52.0 ns after being released. electron m/s proton m/s Need Help?

Answer 1

Step-by-step explanation:

It is given that,

Electric field, E = 554 N/C

Time,
`t=52\ ns=52* 10^(-9)\ s`

Electric force, F = qE

For both electron and proton,
`F=1.6* 10^(-19)\ C* 554\ N/C`

`F=8.86* 10^(-17)\ N`

For electron,
`F=m_ea_e`

`a_e=(F)/(m_e)`

`a_e=(8.86* 10^(-17)\ N)/(9.1* 10^(-31)\ kg)`

`a_e=9.73* 10^(13)\ m/s^2`

Using first equation of motion as :

`v=u+at`

u = 0

`v=9.73* 10^(13)\ m/s^2* 52* 10^(-9)\ s`

v = 5059600 m/s

or

v = 5.05 × 10⁶ m/s

For proton :

`F=m_pa_p`

`a_p=(F)/(m_e)`

`a_p=(8.86* 10^(-17)\ N)/(1.67* 10^(-27)\ kg)`

`a_p=5.3* 10^(10)\ m/s^2`

Using first equation of motion as :

`v=u+at`

u = 0

`v=5.3* 10^(10)\ m/s^2* 52* 10^(-9)\ s`

v = 2756 m/s

Hence, this is the required solution.