91Ó°ÊÓ

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

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

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.

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

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

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

Putting the values,

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

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

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

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 $9.109×{10}^{-31}\text{kg}$.

Putting the values,

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

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

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

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