The electrostatic deflection system of a cathode-ray oscilloscope is depicted in Fig. 3-2. Electron, from a heated cathode are given an initial velocity { u }_{ 0 }={ a }_{ z }{ u }_{ 0 } by a positively charged anode (not shown). The electrons enter at z = 0 into a region of deflection plates where a uniform electric field { E }_{ d }=-{ a }_{ y }{ E }_{ d } is maintained over a width w. Ignoring gravitational effects, find the vertical deflection of the electron on the fluorescent screen at z = L.
Question : The electrostatic deflection system of a cathode-ray oscillo...
Since there is no force in the z-direction in the z > 0 region, the horizontal velocity { u }_{ 0 } is maintained. The field { u }_{ 0 } exerts a force on the electrons each carrying a charge – e. causing a deflection in the y-direction:
F=\left( -e \right) { E }_{ d }={ a }_{ y }e{ E }_{ d }
From Newton’s second law or motion in the vertical direction we have
m\frac { d{ u }_{ y } }{ dt } =e{ E }_{ d }
where m is the mas or an electron. Integrating both ides, we obtain
{ u }_{ y }=\frac { dy }{ dt } =\frac { e }{ m } { E }_{ d }t
where the constant of integration is set to zero because { u }_{ y } = 0 at t = 0. Integrating again, we have
y=\frac { e }{ 2m } { E }_{ d }{ t }^{ 2 }
The constant of integration is again zero because y = 0 at t = 0. Note that the electrons have a parabolic trajectory between the deflection plate . At the exit from the deflection plate , t = w/{ u }_{ 0 } ,
{ d }_{ 1 }=\frac { e{ E }_{ d } }{ 2m } { \left( \frac { w }{ { u }_{ 0 } } \right) }^{ 2 }
and
{ u }_{ y1 }={ u }_{ y }\left( at t=\frac { w }{ { u }_{ 0 } } \right) =\frac { e{ E }_{ d } }{ m } \left( \frac { w }{ { u }_{ 0 } } \right)
When the electrons reach the screen, they have traveled a further horizontal distance of (L – w), which takes them (L-w)/{ u }_{ 0 } seconds. During that time there is an additional vertical deflection
{ d }_{ 2 }={ u }_{ y1 }\left( \frac { L-w }{ { u }_{ y } } \right) =\frac { e{ E }_{ d } }{ m } \frac { w\left( L-w \right) }{ { u }_{ 0 }^{ 2 } }
Hence the deflection at the screen is
{ d }_{ 0 }={ d }_{ 1 }+{ d }_{ 2 }=\frac { e{ E }_{ d } }{ m{ u }_{ 0 }^{ 2 } } w\left( L-\frac { w }{ 2 } \right)