Two large sheets of charge are places parallel to each other with a separation of 2cm between them.An electronstarting from rest near one of the sheets reaches the other plate in 2micro seconds .Find the surface charge density of the sheet of charge.

Dear student,
Distance travelled by the electron, d= 2 cm
Time taken to cross the region, t = 2×10^$-$6 s
Let the surface charge density at the conducting plates be σ.
Let the acceleration of the electron be a.
Applying the 2nd equation of motion, we get:
$$\begin{gathered} d=\frac{1}{2} a{t}^2 \\ \Rightarrow a=\frac{2d}{t^2} \end{gathered}$$
This acceleration is provided by the Coulombic force. So,

$$\begin{gathered} a=\frac{qE}{m}=\frac{2d}{t^2} \\ \Rightarrow E=\frac{2 md}{q{t}^2} \\ E=\frac{2\times (9.1\times {10}^{-31})\times (2\times {10}^{-2})}{(1.6\times {10}^{-19})\times (4\times {10}^{-12})} \\ E=5.6875\times {10}^{-2} \mathrm{N}/\mathrm{C} \end{gathered}$$

Also, we know that electric field due to a plate,
$$\begin{gathered} E=\frac{\sigma }{{\in }_0} \\ \Rightarrow \sigma = {\in }_{0}E \\ \Rightarrow \sigma =\left(8.85\times {10}^{-12}\right)\times \left(5.68\times {10}^{-2}\right) \mathrm{C}/{\mathrm{m}}^2 \\ \Rightarrow \sigma =50.33\times {10}^{-14} \mathrm{C}/{\mathrm{m}}^2=0.503\times {10}^{-12} \mathrm{C}/{\mathrm{m}}^2 \end{gathered}$$
Regards