If I have a capacitor charged to a 12V battery, the voltage difference is 12V across the plates. If I change the permitivity between the plates the pd must stay the same no and charge changes?

If this is so I am nut full convinced how may charge on the plates change because of permitivity? I know it has to do with the equation but mechanistically how does it affect charge?

* missing part of the question: and at the same time pd remains the same??

In terms of the mechanism for how the permittivity affects the charge, it has to do with the polarization of the molecules within the dielectric. The individual molecules or "cells" within the material have a small separation between the positive and negative charges. When you apply a voltage between the plates, the molecules in the dielectric reorient themselves because their negative sides are attracted to the positive plate, and their positive sides are attracted to the negative plate. Since the dielectric dipoles are oriented opposite of the field between the plates, this has the effect of canceling out some of the electric field. Thus, after increasing the permittivity, it takes more charge to get the same potential difference across the plates.

I hope that makes sense to you.

if u change the permitivity of space in b/w the plates by k, lets say.. then in the equations u may replace Eo (epsilon knot) by kEo...
the potential differecne b/w the plates remains the same as the plates are connected to battery and the capacitor contains the potential difference as it is present with the battery...
we know that Q=C x V where C is capacitance i.e. Eo x A/D... as permitivity is now KEo and the potential difference is same, the capacitance increases to kEo... putting the values, u get charge on the plates as kEo x V simple... so charge increases :)