91Ó°ÊÓ

• a)Initial velocity of electron,$V_1=40km/s$
• b)Electric field, strength,$\left|E\right|=50N/C$.
• c) Time interval,$\left(t\right)=1.5ns$

We combine the Equation for the magnitude of the electrostatic force on a point charge of magnitude q (given by F = qE, where E is the magnitude of the electric field at the location of the particle) with Newton’s second law. From this, we get the acceleration that is used to calculate the required final velocity and the distance of travel.

Formulae:

The force due to Newton’s second law of motion, F=ma (i)

The electrostatic force on a body, F = qE (ii)

The first equation of kinematic motion, $V_t-V_l=at$ (iii)

Due to the fact that the electron is negatively charged, then the field E pointing in the same direction as the velocity leads to deceleration. Thus, with $t=1.5×{10}^{-9}s$, we can find the speed of the electron, by substituting the values of equations (i) and (ii) in equation (iii) as follows:
$$\begin{gathered} V=V_O-\frac{eE}{m}t \\ =4.0×{10}^4\frac{m}{s}-\frac{\left(1.6×{10}^{-19}C\right)\left(50{\displaystyle \frac{N}{C}}\right)}{9.11×{10}^{-31}kg}\left(1.5×{10}^{-9}s\right) \\ =2.7×{10}^{4}m/s \end{gathered}$$

.

Hence, the speed of the electron is$2.7×{10}^{4}m/s$.

The displacement is equal to the distance since the electron does not change its direction of motion. The field is uniform, which implies the acceleration is constant. Thus, the distance covered by the electron is given as:

$$\begin{gathered} d=\frac{V+V_O}{2}t \\ =\frac{\left(2.7×{10}^{4}m/s+4×{10}^{4}m/s\right)1.5×{10}^{-9}s}{2} \\ =5.0×{10}^{-5}m \end{gathered}$$

Hence, the electron travels$5.0×{10}^{-5}mfar.$