Question

A uniform electric field exists in a region between two oppositely-charged plates. An electron is released from rest at the surface of the negatively-charged plate and strikes the surface of the opposite plate, d = 2 cm away, 20 ns later. a) What is the speed of the electron as it strikes the second plate? b) What is the magnitude of the electric field? c) If instead, a proton is released from rest at the positively charged plate, how long would it take to get to the opposite plate?

Answer 1

2000000 m/s

569.375 N/C

`8.5630610979* 10^(-7)\ s`

Step-by-step explanation:

m = Mass of electron =
`9.11* 10^(-31)\ kg`

t = Time taken = 20 ns

u = Initial velocity

v = Final velocity

s = Displacement = 2 cm

a = Acceleration

`s=ut+(1)/(2)at^2\\\Rightarrow 2* 10^(-2)=0* t+(1)/(2)* a* (20* 10^(-9))^2\\\Rightarrow a=(2* 10^(-2)* 2)/((20* 10^(-9))^2)\\\Rightarrow a=1* 10^(14)\ m/s^2`

`v=u+at\\\Rightarrow v=0+1* 10^(14)* 20* 10^(-9)\\\Rightarrow v=2000000\ m/s`

The speed of the electron is 2000000 m/s

Electric field is given by

`E=(ma)/(q)\\\Rightarrow E=(9.11* 10^(-31)* 1* 10^(14))/(1.6* 10^(-19))\\\Rightarrow E=569.375\ N/C`

The electric field is 569.375 N/C

m = Mass of proton =
`1.67* 10^(-27)\ kg`

`E=(ma)/(q)\\\Rightarrow a=(Eq)/(m)\\\Rightarrow a=(569.375* 1.6* 10^(-19))/(1.67* 10^(-27))\\\Rightarrow a=5.4550898204* 10^(10)\ m/s^2`

`s=ut+(1)/(2)at^2\\\Rightarrow 0.02=0t+(1)/(2)* 5.4550898204* 10^(10)* t^2\\\Rightarrow t=\sqrt{(0.02* 2)/(5.4550898204* 10^(10))}\\\Rightarrow t=8.5630610979* 10^(-7)\ s`

The time taken is
`8.5630610979* 10^(-7)\ s`