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

Answer:

$1.77\times 10^{-7}\ C/m^2$

0.000439077936334 m

Explanation:

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

E = Electric field = $2\times 10^{4}\ N/C$

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

d = Distance between plates = 2 cm (assumed)

m = Mass of electron = $9.11\times 10^{-31}\ kg$

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 = $8.85\times 10^{-12}\ F/m$

Electric flux is given by

$\phi=\epsilon_0E\\\Rightarrow \phi=8.85\times 10^{-12}\times 2\times 10^{4}\\\Rightarrow \phi=1.77\times 10^{-7}\ C/m^2$

The charge per unit area on the plates is $1.77\times 10^{-7}\ C/m^2$

Deflection is given by

$s=\dfrac{1}{2}\dfrac{qE}{m}(\dfrac{d}{v})^2\\\Rightarrow s=\dfrac{1}{2}\dfrac{1.6\times 10^{-19}\times 2\times 10^4}{9.11\times 10^{-31}}(\dfrac{0.02}{4\times 10^7})^2\\\Rightarrow s=0.000439077936334\ m$

The deflection is 0.000439077936334 m