Question

An electron with velocity v = 1.0 ´ 106 m/s is sent between the plates of a capacitor where the electric field is E = 500 V/m. If the distance the electron travels through the field is 1.0 cm, how far is it deviated (Y) in its path when it emerges from the electric field? (me = 9.1 ´ 10 - 31 kg, e = 1.6 ´ 10 - 19 C)

Answer 1

The deviation in path is
`4.39 * 10^(-3)`

Step-by-step explanation:

Given:

Velocity
`v = 1 * 10^(6)`
`(m)/(s)`

Electric field
`E = 500`
`(V)/(m)`

Distance
`x = 1 * 10^(-2)` m

Mass of electron
`m = 9.1 * 10^(-31)` kg

Charge of electron
`q = 1.6 * 10^(-19)` C

Time taken to travel distance,

`t = (x)/(v)`

`t = (1 * 10^(-2) )/(1 * 10^(6) )`

`t = 10^(-8)` sec

Acceleration is given by,

`F = qE`

`ma = qE`

`a = (qE)/(m)`

`a = (1.6 * 10^(-19) * 500)/(9.1 * 10^(-31) )`

`a = 8.77 * 10^(13)`
`(m)/(s^(2) )`

For finding the distance, we use kinematics equations.

`y = vt + (1)/(2) at^(2)`

Where
`v = 0` because here initial velocity zero

`y = (1)/(2) at^(2)`

`y = (1)/(2) * 8.77 * 10^(13 ) * (10^(-2) )^(2)`

`y = 4.39 * 10^(-3)` m

Therefore, the deviation in path is
`4.39 * 10^(-3)`