Question

Two large conducting plates are placed parallel to each other with a separation of 2⋅00 cm between them. An electron starting from rest near one of the plates reaches the other plate in 2⋅00 microseconds. Find the surface charge density on the inner surfaces.

Answer 1

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,

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

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