P Deflection in a CRT. Cathode­ ray tubes (CRTs) were often found in oscilloscopes and computer monitors. In Fig. P23.63 an electron with an initial speed of is projected along the axis midway between the deflection plates of a cathode ­ray tube. The potential difference between the two plates is and the lower plate is the one at higher potential. (a) What is the force (magnitude and direction) on the electron when it is between the plates? (b) What is the acceleration of the electron (magnitude and direction) when acted on by the force in part (a)? (c) How far below the axis has the electron moved when it reaches the end of the plates? (d) At what angle with the axis is it moving as it leaves the plates? (e) How far below the axis will it strike the fluorescent screen S?

Given data:

$\begin{array}{r}{\mathrm{v}}_{\mathrm{x}}=6.50\times {10}^6\mathrm{m}/\mathrm{s}\\ \mathrm{V}=22.0\mathrm{V}\\ \mathrm{x}=6.0\mathrm{cm}\\ \mathrm{d}=2.0\mathrm{cm}\\ {\mathrm{x}}_2=12.0\mathrm{cm}\end{array}$

The force is

$\mathrm{F}=\mathrm{Ee}$

And the potential for parallel plates is

$\mathrm{V}=\mathrm{Ed}$

Thus

$\mathrm{F}=\frac{\mathrm{Ve}}{\mathrm{d}}$

Putting values

$\begin{array}{r}F=\frac{22\times 1.6\times {10}^{-19}}{2.0\times {10}^{-2}}\\ =1.76\times {10}^{-16}\mathrm{N}\end{array}$

Hence, the magnitude and direction of the electron is $1.76\times {10}^{-16}\mathrm{N}$from upward to downwards direction

So, the magnitude of the acceleration is;

$\begin{array}{r}\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{1.76\times {10}^{-16}}{9.11\times {10}^{-31}}\\ =1.93\times {10}^{14}{\mathrm{m}}^{-2}{\mathrm{s}}^2\end{array}$

As the acceleration is in the of force so it’s towards downwards.

Hence, the acceleration of the electron is$1.93\times {10}^{1}4m.s^{-2}$from upwards to the downward direction.

Thus, the time is taken to move (r)

$\begin{array}{r}{\mathrm{v}}_{\mathrm{x}}=\frac{\mathrm{x}}{\mathrm{t}}\\ \mathrm{t}=\frac{6.0\times {10}^{-2}}{6.5\times {10}^6}\\ =9.23\times {10}^{-9}\mathrm{s}\end{array}$

The distance in the vertical direction

$\begin{array}{r}y=v_{y}t-\frac{1}{2}a{t}^2\\ y=\left|0-\frac{1}{2}\times 1.93\times {10}^{14}\times {\left(9.23\times {10}^{-9}\right)}^2\right|\\ =8.2\times {10}^{-3}\mathrm{m}\end{array}$

Therefore, Electros moved to$8.2\times {10}^{-3}\mathrm{m}$ .

$\theta =\frac{{\mathrm{v}}_{\mathrm{y}}}{{\mathrm{v}}_{\mathrm{i}}}$

As ${\mathrm{v}}_{\mathrm{y}}$is the final velocity so ${\mathrm{v}}_{\mathrm{y}}$is given by

$\begin{array}{r}{v}_y=\left|v_{yi}-at\right|\\ =9.23\times {10}^{-9}\times 1.93\times {10}^{14}\\ =1.78\times {10}^6\mathrm{m}/\mathrm{s}\end{array}$

So, the angle is given by

$\begin{array}{r}\theta =\arctan \left(\frac{{\mathrm{v}}_{\mathrm{y}}}{{\mathrm{v}}_{\mathrm{x}}}\right)\\ =\arctan \left(\frac{1.78\times {10}^6}{6.5\times {10}^6}\right)\\ \theta ={15.3}^{\circ }\end{array}$

Hence, the angle at which it leaves the plate is$\theta =15.3^o$

So, the time taken by${\mathrm{X}}_2$

$\begin{array}{r}{\mathrm{v}}_{\mathrm{x}}=\frac{{\mathrm{x}}_2}{\mathrm{t}}\\ \mathrm{t}=\frac{12.0\times {10}^{-2}}{6.5\times {10}^6}\\ =1.85\times {10}^{-9}\mathrm{s}\end{array}$

Vertical displacement is

role="math" localid="1668262789285" $\begin{array}{r}y=y_0-v_{y\mathrm{y}}t-\frac{1}{2}{g{t}^2}^{}\\ y=\left|-8.2\times {10}^{-3}-1.78\times {10}^6\times 1.84\times {10}^{-8}-\frac{1}{2}\times 9.8\times {\left(1.84\times {10}^{-8}\right)}^2\right|\\ =0.041\mathrm{m}\end{array}$

Therefore, It strikes the fluorescent screen S at $\mathrm{y}=0.041\mathrm{m}$ and below the axis.