Consider the situation shown in figure. The width of each plate is $b$. The capacitor plates are rigidly clamped in the laboratory and connected to a battery of emf  $V$. All surfaces are frictionless. The value of $M$ for which the dielectric slab will stay in equilibrium is

Suppose at some instant, the length of the slab, inside the cap is $x$
 the system is equivalent to 2 capacitor in parallel having capacitance  $C_1=\frac{{\epsilon }_0\left[b\right(\mathcal{l}-x\left)\right]}{d}$ and $C_2=\frac{{\epsilon }_0\left(bx\right)K}{d}$ .
 total capacitance  $\left(C_{eq}\right)=C_1+C_2=\frac{{\epsilon }_{0}b}{d}[\mathcal{l}+x(K-1\left)\right]$
 PE stored in system  $\left(U\right)=\frac{1}{2}\frac{{\epsilon }_{0}b{V}^2}{d}[\mathcal{l}+x(K-1\left)\right]$
 electric force on the slab, tending to pull it towards left  
$F=\frac{dU}{dx}=\frac{1}{2}\frac{{\epsilon }_{0}b{V}^2}{d}(K-1)$.
$Mg=\frac{{\epsilon }_{0}b{V}^2}{2d}(K-1)$