Question

15,000VIn the cathode ray tube found in old television sets, which contains a vacuum, electrons are boiled out of a very hot metal filament placed near a negative metal plate. These electrons start out nearly at rest and are accelerated toward a positive metal plate. They pass through a hole in the positive plate on their way toward the picture screen, as shown in the diagram in Figure 16.69. If the high-voltage supply in the television set maintains a potential difference of $\mathbf{15}\mathbf{,}\mathbf{000}\mathbf{}\mathit{V}$between the two plates, what speed do the electrons reach?

Short answer

The final speed of the electron is $\text{7.259}\times {10}^7\text{m/s}$.

Step-by-step solution

Identification of given data

The potential difference between the two plates is, $\Delta V=15000\text{V}$.

The distance between the two plates is, $L$.

Force acting on a moving charge

When a charge moves through a uniform electric field then an electric force acts on the charge.

The value of the electric force acting on the moving charge changes with the magnitude of the charge and the electric field.

Determining the velocity of the electron

The magnitude of the electric field acting between the two plates is given by,

$\begin{array}{rcl}E& =& \frac{\Delta V}{L}\\ -\frac{kq}{L^2}& =& \frac{\Delta V}{L}\end{array}$

Here, $k$is Coulomb’s constant and its value is $9\times {10}^9\text{N}·{\text{m}}^2{\text{/C}}^2$.

Putting the values,

$\begin{array}{rcl}-\frac{\left(9\times {10}^9\text{N}·{\text{m}}^2{\text{/C}}^2\right)\left(-1.6\times {10}^{-19}\text{C}\right)}{{\left(L\right)}^2}& =& \frac{15000\text{V}}{L}\\ \frac{1.44\times {10}^{-9}}{L^2}\text{N}·{\text{m}}^2\text{/C}& =& \frac{15000\text{V}}{L}\\ L& =& \frac{1.44\times {10}^{-9}\text{N}·{\text{m}}^2\text{/C}}{15000\text{V}}\\ L& =& 9.6\times {10}^{-14}\text{m}\end{array}$

The force acting on the electron due to electric field is given by,

$$\begin{gathered} F=eE \\ F=\left(1.6\times {10}^{-19}\text{C}\right)\left(\frac{14.4\times {10}^{-10}}{L^2}\text{N}·{\text{m}}^2\text{/C}\right) \\ F=\frac{23.04\times {10}^{-29}\text{N}·{\text{m}}^2}{{\left(9.6\times {10}^{-14}\text{m}\right)}^2} \\ F=0.025\text{N} \end{gathered}$$

Balancing the force using second law of motion,

$F=m_{e}a$

Here, $a$is the acceleration of the electron, and $m_e$is the mass of the electron, its value is $9.109\times {10}^{-31}\text{kg}$.

Putting the values,

$$\begin{gathered} 0.025\text{N}=\left(9.109\times {10}^{-31}\text{kg}\right)\times a \\ a=\frac{0.025\text{N}}{9.109\times {10}^{-31}\text{kg}}\times \frac{1\text{kg}·{\text{m/s}}^2}{1\text{N}} \\ a=2.75\times {10}^{28}{\text{m/s}}^2 \end{gathered}$$

Using the equation of motion, the final speed of the electron is given by,

$$\begin{gathered} v^2=u^2+2as \\ {v}^2={\left(0\text{m/s}\right)}^2+2\left(2.75\times {10}^{28}{\text{m/s}}^2\right)\left(9.6\times {10}^{-14}\text{m}\right) \\ {v}^2=5.269\times {10}^{15}{\text{m}}^2{\text{/s}}^2 \\ v=\text{7.259}\times {10}^7\text{m/s} \end{gathered}$$

Hence, the final speed of the electron is $\text{7.259}\times {10}^7\text{m/s}$.