Parallel plates are boring!

There would be fringing fields as we take out the dielectric by a small angle say $\theta$ . We calculate the torque on the dielectric as follows.

Let an external agent displace it. Then, $W_{agent}+W_{battery}+W_{capacitor}=0$ .

So $\tau d\theta + dCV^2 - 1/2 d C V^2= 0$ .

So $\tau = - 1/2 V^2 \dfrac{dC}{d\theta}$ .

Now $C=\dfrac{(\theta+k(\pi-\theta))r^2\epsilon }{2d}$ .

Using this yields $\tau=1/(4d)×V^2r^2\epsilon (k-1)$ .

So in the absence of the external agent, the torque applied by the capacitor tries to bring the dielectric back to the initial position. Since $\tau$ is constant we have $\tau=I\alpha$ . And it's easy to conclude that $I=\psi Mr^2$ ( $\psi$ is a constant say here $\psi =1/2$ , $M$ is the mass of dielectric plate)

Time required is thus $t= √(2\theta /\alpha)$ . Plugging the values calculated, we get the time to be independent of the Radius. This was just a mathematical approach although one could simply conclude from physical interpretations. (say dimensional analysis), $t=√(8\theta \psi M×d/\epsilon V^2(k-1))$