Question

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?

Short answer

(a) The magnitude and direction of the electron is $\mathrm{F}=1.76\times {10}^{-16}\mathrm{N}$from upwards to downwards direction.

(b) Acceleration of the electron is$\mathrm{a}=1.93\times {10}^{14}{\mathrm{ms}}^{-2}$ from upwards to the downward direction.

(c) Electros moved to $\mathrm{y}=8.2\times {10}^{-3}\mathrm{m}$

(d) The angle at which it leaves the plate is $\theta =15.3^0$

(e) It strikes the fluorescent screen S at $y=0.041m$

Step-by-step solution

Step 1:

Given data:

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

(a) As the charge is negative and the lower plate has a higher potential than the upper as well as the force of attraction with the lower and repulsive with the upper.

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

Step 2:

(b) Velocity is constant in the horizontal direction but it’s not in the case of the vertical 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.

(c) Time is taken to move the horizontal displacement as it is a projectile motion,

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$8.2\times {10}^{-3}\mathrm{m}$ .

Step 3:

(d) Angle

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

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$\theta =15.3^o$

Step 4:

(e) Time taken to move second horizontal displacement as it is a projectile motion,

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.