Answer:
E = [tex]\frac{Q}{L^2 \epsilon_o}[/tex]
Explanation:
For this exercise we use that the electric field is a vector, so the resulting field is
E_total = E₁ + E₂ (1)
since the field has the same direction in the space between the planes
Let's use Gauss's law for the electric field of each plate
Let's use a Gaussian surface that is a cylinder with the base parallel to the plate, therefore the normal to the surface and the field lines are parallel and the angle is zero so cos 0 = 1
Ф = ∫ .dA = [tex]q_{int}[/tex] /ε₀
if we assume that the charge is uniformly distributed on the plate we can define a charge density
σ = q_{int} A
as the field exists on both sides of the plate on the inside
E A = A σ / 2ε₀
E = σ / 2ε₀
we substitute in equation 1
E = σ /ε₀
for the complete plate
σ = Q / A = Q / L²
we substitute
E = [tex]\frac{Q}{L^2 \epsilon_o}[/tex]
