91Ó°ÊÓ

Magnitude of the electric field,$E=1.40x{10}^6\text{N/C}$

The acceleration of the electron is given by Newton’s second law:,$\mathit{F}\mathbf{=}\mathit{m}\mathit{a}$where F is the electrostatic force.The magnitude of the force acting on the electron is,$\mathit{F}\mathbf{=}\mathit{q}\mathit{E}$, where E is the magnitude of the electric field at its location.We then use the concepts of kinematical equations.

Formulae:

Using Newton’s second law, the acceleration of the electron, $a=\frac{eE}{m}$ (i)

The first equation of kinematic motion, $v−v_0=at$ (ii)

The third equation of kinematic motion, $v^2−v_0^2=2ax$ (iii)

Using equation (i), we get the magnitude of the acceleration of the electron as:

$\begin{array}{c}a=\frac{(\text{1.60}×{\text{10}}^{\text{-19}}\text{C})}{(\text{9.11}×{\text{10}}^{\text{-31}}\text{kg})}(1.40×{10}^6\text{N/C})\\ =2.46×{10}^{17}{\text{m/s}}^{\text{2}}\end{array}$

Hence, the value of the acceleration is.$2.46×{10}^{17}{\text{m/s}}^{\text{2}}$

With the velocity of the electron given as:

$\begin{array}{c}v=c/10\\ =3.00x{10}^7\text{m/s}\end{array}$

The time taken by the electron using equation (ii) is given as:

1. $\begin{array}{c}t=\frac{(3.00×{10}^7\text{m/s}−0\text{m/s})}{(2.46×{10}^{17}{\text{m/s}}^{\text{2}})}\\ =1.22×{10}^{−10}\text{s}\end{array}$.

Hence, the value of the time taken is.$1.22×{10}^{−10}\text{s}$

Using kinematical equations we get the displacement using equation (iii) as follows:

$\begin{array}{c}\mathrm{Δ}x=\frac{{(3.00×{10}^7\text{m/s})}^2−{\left(0\text{m/s}\right)}^2}{2(2.46×{10}^{17}{\text{m/s}}^{\text{2}})}\\ =1.83×{10}^{−3}\text{m}\end{array}$

Hence, the distance travelled by the electron is.$1.83×{10}^{−3}\text{m}$