(c*) The nonconducting wall is replaced with a thick conducting wall with the same surface charge density on the right side of the conducting wall as was on the thin insulating layer. What is the electric field 6.45 cm in front of (just outside) the conducting wall if 6.45 cm is small compared with the dimensions of the wall

E = ½  σ₀/ 2ε₀  =  ½ E₀

Explanation:

The charge in the initial metal wall is in the right part, but in metals the electrons are mobile, so that the charge is divided into two parts of the wall, therefore the charge density is

                σ₂ = σ₀ / 2

the zero index is for the insulating wall and the index 2 for the metal wall

To find the field you can use Gauss's law,

            Ф = E. dA = /ε₀

If we use a cylinder as a Gaussian surface, the sides of the cylinder do not contribute to the flow and the bases give a scalar producer that is reduced to the algebraic product

            E A = q_{int} /ε₀

The surface charge density is

           σ = Q / A

          q_{int} = Q = σ A

          E A = σ₂ A /ε₀

            E = σ₂ /ε₀

As the electric field from the wall extends to both sides the value on one side is

            E = σ₂ / 2 ε₀

            E = σ₀/2   1/2ε₀

             E = ½  σ₀/ 2ε₀

             E = ½ E₀

The field is half of the field created by the insulating wall.

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle