Question

In a cathode-ray tube, an electron travels in a vacuum and enters a region between two deflection plates where there is an upward electric field of magnitude$\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{N}\mathbf{/}\mathbf{C}$(Figure 15.60).

(a) Sketch the trajectory of the electron, continuing on well past the deflection plates (the electron is going fast enough that it does not strike the plates). (b) Calculate the acceleration of the electron while it is between the deflection plates. (c) The deflection plates measure 12 cm by 3 cm, and the gap between them is 2.5 mm. The plates are charged equally and oppositely. What are the magnitude and sign of the charge on the upper plate?

Short answer

1. The trajectory curves downward when the electron is between the plates, and then it continues straight downward when outside the plates.
2. The acceleration of the electron is $1.75\times {10}^{16}\mathrm{m}/{\mathrm{s}}^2$.
3. The charge on the upper plate is negative and its value is$-3.19\times {10}^{-9}\mathrm{N}/\mathrm{C}$

Step-by-step solution

Identification of given data

The given data can be listed below,

• The upward electric field is,$E=1\times {10}^5\mathrm{N}/\mathrm{C}$
• The gap between deflection plates is,$d=2.5\mathrm{mm}$
• The length of the deflection plate is,$i=12\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)$
• The width of the plates is,$b=3\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)$

Concept/Significance of the acceleration.

The rate at which an object's velocity changes in relation to time is called acceleration. Newton's Second Law states that an object's acceleration equals the sum of all forces acting on it.

Determination of the trajectory of the electron, continuing on well past the deflection plates

The electron starts with an initial velocity parallel to the plates. The electric field accelerates it downward (because the electron has a negative charge), so the trajectory curves downward because we have two velocity components. One increases with time due to the electric force. After the plates, the electrons travel straight down because there is no longer a force, and it just maintains the velocity they had right before exiting the plates, and this velocity is pointed downward. It is shown in diagram below,

Thus,the trajectory curves downward when the electron is between the plates, and then it continues straight downward when outside the plates.

Step 4: the acceleration of the electron while it is between the deflection plates.

The force on the electron is given by,

$F=ma$ …(i)

Here, m is the mass of the electron whose value is and a is the acceleration

The force on the electron can also be given by,

$F=eE$ …(ii)

Here, e is the charge on the electron whose value is and E is the electric field.

From equation (i) and (ii).

$$\begin{gathered} ma=eE \\ a=\frac{eE}{m} \end{gathered}$$ …(iii)

Substitute all the values in the above

$$\begin{gathered} a=\frac{1.6\times {10}^{19}\mathrm{C}\left(1\times {10}^5\mathrm{N}/\mathrm{C}\right)}{9.1\times {10}^{-31}\mathrm{kg}} \\ =\frac{1.6\times {10}^{17}\mathrm{N}}{9.1\mathrm{kg}}\left(\frac{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right) \\ =1.75\times {10}^{16}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the acceleration of the electron is $1.75\times {10}^{16}\mathrm{m}/{\mathrm{s}}^2$.

Step 5: the magnitude and sign of the charge on the upper plate.

The charge on the plate is given by,

$$\begin{gathered} E=\frac{Q}{A{\epsilon }_0} \\ Q={\epsilon }_{0}AE \end{gathered}$$

Here, ${\epsilon }_0$is the permittivity of free space whose value is $8.85\times 10\mathrm{N}/\mathrm{m}.{\mathrm{C}}^2$, A is the area of plate, and E is the electric field on the plate.

Substitute all the values in the above,

$$\begin{gathered} -Q=\left(8.85\times {10}^{-12}\mathrm{N}/\mathrm{m}.{\mathrm{C}}^2\right)\left(0.12\mathrm{m}\times 0.03\mathrm{m}\right)\left(1\times {10}^5\mathrm{N}/\mathrm{C}\right) \\ =3.19\times {10}^{-9}\mathrm{N}/\mathrm{C} \end{gathered}$$

Thus, the charge on the upper plate is negative and its value is$-3.19\times {10}^{-9}\mathrm{N}/\mathrm{C}$