91Ó°ÊÓ

• a)Initial velocity of the electron,$V_I=0m/s$
• b)Distance between two plates,$X=0.02m$

c) Time interval,$t=1.5×{10}^{-8}S$

The magnitude of the electrostatic force on a point charge of magnitude q is given by F = qE, where E is the magnitude of the electric field at the location of the particle. And the concept of kinematics has been used.

Formulae:

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

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

The distance travelled between two plates,$∆x=V_{avg}torvt/2$(iii)

The second equation of kinematic motion,$∆x=(1/2)a{t}^2(\sin ce{v}_l=0m/s)$(iv)

Using equation (iii), the speed of the electron between the two plates is given as:

$$\begin{gathered} V=\frac{2\left(2.0×{10}^{-2}m\right)}{1.5×{10}^{-8}s} \\ =2.7×{10}^{6}m/s \end{gathered}$$

Hence, the speed of the electron is $2.7×{10}^{6}m/s$.

From equations (i), (ii) and (iv), the magnitude of the electric field can be given as:

$$\begin{gathered} E=\frac{ma}{e} \\ =\frac{2∆xm}{e{t}^2} \\ =\frac{2\left(2.0×{10}^{-2}m\right)\left(9.11×{10}^{-31}kg\right)}{\left(1.60×{10}^{-19}C\right){\left(1.5×{10}^{-8}s\right)}^2} \\ =1.0×{10}^{3}N/C \end{gathered}$$

Hence, the value of the electric field is$1.0×{10}^{3}N/C$.