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, 1.3 cm away, in a time 7.2 times 10-7 s. What is the magnitude of the electric field?

Answer 1

E=0.284 N/C

Step-by-step explanation:

Given that

Distance ,d= 1.3 cm = 0.013 m

Time ,t

`t=7.2* 10^(-7)\ s`

Initial velocity of electron u=0 m/s

We know that

`d=ut+(1)/(2)at^2`

`0.013=0+(1)/(2)* a*(7.2* 10^(-7)) ^2`

`a=5.01* 10^(10)\ m/s^2`

We know that

mass of electron,m

`m=9.1* 10^(-31)\ kg`

Charge on electron

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

F= m a=E q

So

`E=(ma)/(q)\ N/C`

`E=(9.1* 10^(-31)* 5.01* 10^(10))/(1.6* 10^(-19))\ N/C`

E=0.284 N/C

Electric field will be 0.284 N/C.