Deflection of electrons in an oscilloscope?

Question

Using an appropriate equation for the force acting on the electrons as they pass between the parallel plates to the screen, explain how the deflection of the spot on the screen is proportional to the applied voltage.

Answer 1

U=the voltage between the vertical deflection plates

d=the distance between the plates

l=the length of the plates

m=mass of the electron

q=charge of the electron

y=vertical deflection

v=horizontal speed of the electron at the moment it enters between the plates

The electric field between the plates is homogeneous (E=constant):

E=U/d and its direction is vertical --> this will cause vertical acceleration of the electrons a[m/s^2], while their horizontal speed between the plates will remain constant.

The electric field force on the electron is: F=qE=qU/d

On the other hand F=ma -->

ma=qU/d

a=qU/(md) is the vertical acceleration which deflects the electron vertically.

The time the electron travels between the plates is t=l/v.

The vertical deflection during time t is y=0.5at^2

y=0.5*[qU/(md)]*(l/v)^2

y=[ql^2/(2mdv^2)]*U -->

The deflection is linear function of the voltage between the plates.