Question

If we assume that the metallic plates are perfect conductors, the electric field in their interiors must vanish. given that the electric field e⃗ due to a charged sheet with surface charge +σ is given by e=σ2ϵ0, and that it points away from the plane of the sheet, how can the condition that the electric field in plate i vanishes be written?

Answer 1

here the charge density of metal plate is given as

$charge density = \sigma$

now the electric field is given Gauss law

$\int E. dA = \frac{q}{\epsilon_0}$

now here E = constant

so we will have

$E. \int dA = \frac{q}{\epsilon_0}$

Since total area on both sides of plate will be double and becomes 2A

$E. 2A  = \frac{q}{\epsilon_0}$

$E = \frac{q/A}{2\epsilon_0}$

$E = \frac{\sigma}{2\epsilon_0}$

Now if we will find the electric field inside the metal plate

Then as we know that total charge inside the plate will always be zero

so we have

$E = 0$