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

Answer:

E=0.284 N/C

Explanation:

Given that

Distance ,d= 1.3 cm = 0.013 m

Time ,t

$t=7.2\times 10^{-7}\ s$

Initial velocity of electron u=0 m/s

We know that

$d=ut+\dfrac{1}{2}at^2$

$0.013=0+\dfrac{1}{2}\times a\times(7.2\times 10^{-7}) ^2$

$a=5.01\times 10^{10}\ m/s^2$

We know that

mass of electron,m

$m=9.1\times 10^{-31}\ kg$

Charge on electron

$q=1.6\times 10^{-19}\ C$

F= m a=E q

So

$E=\dfrac{ma}{q}\ N/C$

$E=\dfrac{9.1\times 10^{-31}\times 5.01\times 10^{10}}{1.6\times 10^{-19}}\ N/C$

E=0.284 N/C

Electric field will be 0.284 N/C.