Question

Consider two oppositely charged, parallel metal plates. The plates are square with sides L and carry charges Q and -Q. What is the magnitude of the electric field in the region between the plates

Answer 1

Answer:

 E = $\frac{Q}{L^2 \epsilon_o}$

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 = $q_{int}$ /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 = $\frac{Q}{L^2 \epsilon_o}$