Consider the situation depicted in the figure. Each plate has a...

Let $x$ be the length of the dielectric slab inside the capacitor plates.

The system can be modeled as two capacitors in parallel, one with air and one with the dielectric.

The equivalent capacitance is:

$C = \dfrac{\varepsilon_0 b (l - x)}{d} + \dfrac{K \varepsilon_0 b x}{d} = \dfrac{\varepsilon_0 b}{d} [l + x(K - 1)]$

Since the battery remains connected, the potential difference across the plates is constant at $\mathcal{E}$. The electric force pulling the dielectric slab into the capacitor is given by:

$F_e = \dfrac{1}{2} \mathcal{E}^2 \dfrac{dC}{dx}$

Differentiating $C$ with respect to $x$ gives:

$\dfrac{dC}{dx} = \dfrac{\varepsilon_0 b (K - 1)}{d}$

Substituting this into the force equation:

$F_e = \dfrac{\varepsilon_0 b \mathcal{E}^2 (K - 1)}{2d}$

For the dielectric slab to remain in equilibrium, the electric force must be balanced by the tension in the string, which is equal to the weight of the mass $M$.

$Mg = F_e$

$Mg = \dfrac{\varepsilon_0 b \mathcal{E}^2 (K - 1)}{2d}$

$M = \dfrac{\varepsilon_0 b \mathcal{E}^2 (K - 1)}{2dg}$