The charged plates of a parallel plate capacitor each have a charge density of sigma C/m^2. Using Gauss's law, compute the electric field between the plates.

Gauss' law says that

E*Area = sigma*Area/epsilon
or
E = sigma/epsilon

where epsilon is the permittitivy.

The answer will depend upon the dielectric constant of the material between the plates.

If there is a vacuum or air between the parallel plates,
epsilon = epsilonzero = 8.85*10^-12 farad/meter

To compute the electric field between the plates of a parallel plate capacitor using Gauss's law, we need to apply a cylindrical Gaussian surface between the plates. Gauss's law states that the electric flux through a closed surface is equal to the enclosed charge divided by the electric constant (ε₀).

Step 1: Determine the Gaussian surface
We choose a cylindrical Gaussian surface with its axis parallel to the plates, such that it extends between the plates and passes through them. The cylindrical surface is also symmetric with respect to its axis.

Step 2: Apply Gauss's law
The electric flux (Φₑ) through any closed surface is given by the equation:

Φₑ = ∮E⋅dA = (Qenclosed) / ε₀

Since the electric field (E) is constant and perpendicular to the cylindrical surface, we can take it out of the integral. The integral becomes a simple product of the electric field and the area of the cylindrical surface:

Φₑ = E ∮ dA = E * 2πrh

where r is the radius of the cylindrical surface and h is its height.

Step 3: Determine the charge enclosed
The charge enclosed (Qenclosed) is the charge density (σ) multiplied by the area of the cylindrical surface:

Qenclosed = σ * A

where A is the area of the cylindrical surface. Since the area of a cylinder is given by A = 2πrh, we can substitute this into the equation:

Qenclosed = σ * 2πrh

Step 4: Rearrange the equation
Now we can rearrange Gauss's law equation to solve for the electric field (E):

E * 2πrh = σ * 2πrh / ε₀

Dividing both sides of the equation by 2πrh, we get:

E = σ / ε₀

Therefore, the electric field (E) between the plates of a parallel plate capacitor with charge density sigma (σ) is given by E = σ / ε₀.

Remember to take into account the units of the variables to ensure consistent units for the electric field.