Step 1: Calculate Equivalent Capacitance

The capacitor is filled with three slabs, each of thickness $t = d/3$.

• Two dielectric slabs with dielectric constant $K = \varepsilon_r$.
• One metal slab. For a metal, the effective dielectric constant is infinite ($K \to \infty$).

The effective separation distance $d_{eff}$ for calculating capacitance is given by: $$ d_{eff} = \sum \frac{t_i}{K_i} = \frac{d/3}{\varepsilon_r} + \frac{d/3}{\infty} + \frac{d/3}{\varepsilon_r} $$ $$ d_{eff} = \frac{d}{3\varepsilon_r} + 0 + \frac{d}{3\varepsilon_r} = \frac{2d}{3\varepsilon_r} $$ The capacitance $C$ is: $$ C = \frac{\varepsilon_0 A}{d_{eff}} = \frac{\varepsilon_0 A}{\frac{2d}{3\varepsilon_r}} = \frac{3\varepsilon_r \varepsilon_0 A}{2d} $$

Step 2: Determine the Charge on the Plates

The potential difference $V$ is maintained by the battery. The charge $Q$ on the plates is: $$ Q = CV = \left( \frac{3\varepsilon_r \varepsilon_0 A}{2d} \right) V $$

Step 3: Calculate the Force on the Plates

The attractive force $F$ between the plates of a capacitor with charge $Q$ is given by the formula: $$ F = \frac{Q^2}{2\varepsilon_0 A} $$ Substituting the expression for $Q$: $$ F = \frac{1}{2\varepsilon_0 A} \left( \frac{3\varepsilon_r \varepsilon_0 A V}{2d} \right)^2 $$ $$ F = \frac{1}{2\varepsilon_0 A} \cdot \frac{9 \varepsilon_r^2 \varepsilon_0^2 A^2 V^2}{4d^2} $$ $$ F = \frac{9 \varepsilon_0 \varepsilon_r^2 A V^2}{8d^2} $$ Alternatively, this can be written as: $$ F = \frac{1}{2} \varepsilon_0 A \left( \frac{3 \varepsilon_r V}{2d} \right)^2 $$