#### 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.

#### Solution

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:

`"d" = 1/2 "a""t"^2`

`=> "a" = (2"d")/"t"^2`

This acceleration is provided by the Coulombic force. So,

`"a" = "qE"/m = "2d"/"t"^2`

`=> "E" = (2"md")/"qt"^2`

`"E" = (2 xx (9.1xx10^-31) xx (2 xx 10^-2))/((1.6 xx 10^-19) xx ( 4 xx 10^-12)`

E = 5.6875 × 10^{-2} N/C

Also, we know that electric field due to a plate,

`"E" = sigma/∈_0`

⇒ σ =∈_0 E

⇒ σ = ( 8.85 × 10^{-12}) × ( 5.68 × 10^{-2} ) C/m2

⇒ σ = 50.33 × 10^{-14} C/m2 = 0.503 × 10^{-12} C/m2