Question

A horizontal beam of electrons initially moving at 4.0×10^7 m/s is deflected vertically by the vertical electric field between oppositely charged parallel plates. The magnitude of the field is 2.00×10^4 N/C

(a) What is the direction of the field between the plates?
(b) What is the charge per unit area on the plates?
(c) What is the vertical deflection d of the electrons as they leave the plates?

Answer 1

`1.77* 10^(-7)\ C/m^2`

0.000439077936334 m

Step-by-step explanation:

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

E = Electric field =
`2* 10^(4)\ N/C`

`\epsilon_0` = Permittivity of free space =
`8.85* 10^(-12)\ F/m`

d = Distance between plates = 2 cm (assumed)

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

The beam consists of electrons which means it has negative charge this means the upper plates will be positive and the lower plate will be negative.

The direction is upper to lower lower plate.

`\epsilon_0` = Permittivity of free space =
`8.85* 10^(-12)\ F/m`

Electric flux is given by

`\phi=\epsilon_0E\\\Rightarrow \phi=8.85* 10^(-12)* 2* 10^(4)\\\Rightarrow \phi=1.77* 10^(-7)\ C/m^2`

The charge per unit area on the plates is
`1.77* 10^(-7)\ C/m^2`

Deflection is given by

`s=(1)/(2)(qE)/(m)((d)/(v))^2\\\Rightarrow s=(1)/(2)(1.6* 10^(-19)* 2* 10^4)/(9.11* 10^(-31))((0.02)/(4* 10^7))^2\\\Rightarrow s=0.000439077936334\ m`

The deflection is 0.000439077936334 m