Charging a capacitor and then introducing a dielectric
I unplug the wires and insert a dielectric inside. I know that
electrostatic energy between the capacitor plates change when I Do
this , however, where does this energy go / come from.


Charge remains the same,
Capacitance increases by X and
Voltage drops by X
Energy (\$CV^2/2\$) drops by X

Hence energy &quot;appears&quot; to have been lost but, energy is used when the dielectric enters the plates; it is attracted inwards between the plates and that apparent energy loss (due to work done in moving the dielectric) is retrieved when it is removed (using some small force). The dielectric material is pulled in by the charge on the plates.

how would the charges on the capacitor change at the point where I
remove the battery and also the point where I insert the dielectric?

They don't change; charge is conserved throughout unless the dielectric has conduction.

Also, finally how would I find the maximum amount of charge that I can
allocate onto the capacitor?

Well, this is limited by the applied voltage so if you can estimate the breakdown voltage between the plates then you are 90% there.