Consider two closely spaced and oppositely charged parallel metal plates. The plates are square with sides of length L and carry charges Q and -Q on their facing surfaces. What is the magnitude of the electric field in the region between the plates

  E_ {total} =

Explanation:

In this exercise you are asked to calculate the electric field between two plates, the electric field is a vector

         E_ {total} = E₁ + E₂

         E_ {total} = 2 E

where E₁ and E₂ are the fields of each plate, we have used that for the positively charged plate the field is outgoing and for the negatively charged plate the field is incoming, therefore in the space between the plates for a test charge the two fields point in the same direction

to calculate the field created by a plate let's use Gauss's law

          Ф = ∫ E . dA = q_{int} /ε₀

As a Gaussian surface we use a cylinder with the base parallel to the plate, therefore the direction of the electric field and the normal to the surface are parallel, therefore the scalar product is reduced to the algebraic product.

           E 2A = q_{int} / ε₀

where the 2 is due to the surface has two faces

indicate that the surface has a uniform charge for which we can define a surface density

           σ = q_{int} / A

           q_{int} = σ A

we substitute

           E 2A = σ A /ε₀

           E = σ / 2ε₀  

therefore the total field is

           E_ {total} = σ /ε₀

let's substitute the density for the charge of the whole plate

           σ= Q / L²

            E_ {total} =

the maxium height is 34m/s

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j