Solution

Both plates are infinite in the y and z directions. The total charge per unit area on each plate is \sigma = \rho h (positive on the first plate, negative on the second).

For the electric field, we use Gauss's law with a cylindrical surface whose axis is along x and cross-sectional area A. The total charge of the system is zero, so the field outside the plates (x < 0 and x > 3h) is zero.

For an arbitrary position x, we take a Gaussian surface with one cap at x < 0 (where the field is zero) and the other cap at x. The electric field \vec{E} = E(x)\hat{x} satisfies:

E(x) A = \frac{Q_{\text{enc}}(x)}{\varepsilon_0}

The enclosed charge Q_{\text{enc}}(x) is:

· 0 < x < h: Q_{\text{enc}} = \rho A x \rightarrow E(x) = \dfrac{\rho x}{\varepsilon_0}.
· h < x < 2h: Q_{\text{enc}} = \rho A h \rightarrow E(x) = \dfrac{\rho h}{\varepsilon_0} (constant).
· 2h < x < 3h: Q_{\text{enc}} = \rho A h - \rho A (x - 2h) = \rho A (3h - x) \rightarrow E(x) = \dfrac{\rho (3h - x)}{\varepsilon_0}.

The force on a volume element dV = A\,dx of the positive plate is dF = dq\, E(x), where E(x) is the total field at that point. Since the density is finite and the field is continuous, we can integrate directly:

F_+ = \int_{0}^{h} \rho \, E(x) \, A \, dx = \rho A \int_{0}^{h} \frac{\rho x}{\varepsilon_0} \, dx = \frac{\rho^2 A}{\varepsilon_0} \left[ \frac{x^2}{2} \right]_0^h = \frac{\rho^2 A h^2}{2\varepsilon_0}.

The force per unit area is then:

\boxed{\frac{F}{A} = \frac{\rho^2 h^2}{2\varepsilon_0}}

Since \sigma = \rho h, we can also write \displaystyle \frac{F}{A} = \frac{\sigma^2}{2\varepsilon_0}.

The result \frac{F}{A} = \frac{\sigma^2}{2\varepsilon_0} depends only on the charge per unit area \sigma, not on how the charge is distributed within the volume (thickness h). If we keep \sigma = \rho h constant while varying h, the force does not change.

This happens because:

· The electric field in the space between the plates is E = \sigma/\varepsilon_0 and depends only on \sigma.
· The net force on a thick plate is equivalent to that on a thin surface layer with the same total charge, since integrating the internal field averages the contributions from the plate itself and produces the same effect as if all the charge were concentrated on a surface.