Two plane-parallel plates `l=2 cm` long serve as the control electrodes of a cathode-ray tube. The distance from the control electrodes to the tube's screen is L = 30 cm. An electron beam enters midway between the plates parallel to them at a velocity of `v_(0) = 2 xx 10^(7) m//s`. What is the electric field between the electrodes if the beam's displacement on the screen is `d=12cm`?

Answer

`4.3 xx 10^(4)N//C`

Explanation

No forces act on the electron in the direction of the x-axis (Fig.), and the projection of its velocity on this axis does not change with time: `v_(x)=v_(0)=` const. While the electron moves in the field, a force `F_(y)= -eE` acts on it in the direction of the y-axis. This causes a displacement of the electron
`h=(a l^(2))/(2)= (e E l^(2))/(2mv_(0)^(2))`
its speed in the y-direction being
`v_(y) = at = e El//mv_(0)`
The electron leaves the field at an angle determined from the condition
`tan alpha= (v_(y))/(v_(x)) = (e El)/(mv_(0)^(2))`
its subsequent motion being inertial. As may be seen from the figure,
`d=h+L tan alpha, or d=(eEl^(2))/(2mv_(0)^(2)) +(eElL)/(mv_(0)^(2))`
from which
`E=(mv_(0)^(2)d)/(el(L+l//2))`