When a dielectric material is introduced between the plates of a charged condenser,what happens to the electric field between the plates?

$A$ parallel plate capacitor is charged to a potential of $100 \ V$. $A$ dielectric slab of thickness $2 \ mm$ is inserted between the plates. To maintain the same potential difference,the distance between the plates is increased by $1.6 \ mm$. The dielectric constant of the slab is:

Assertion: The electrostatic force between the plates of a charged isolated capacitor decreases when a dielectric fills the whole space between the plates.
Reason: The electric field between the plates of a charged isolated capacitor increases when a dielectric fills the whole space between the plates.

$A$ voltmeter reads $4 \ V$ when connected to a parallel plate capacitor with air as a dielectric. When a dielectric slab is introduced between the plates for the same configuration,the voltmeter reads $2 \ V$. What is the dielectric constant of the material?

In a parallel plate air capacitor of plate separation $d$,a dielectric slab of thickness $t$ is introduced between the plates $(t < d)$. The capacitance becomes one-third of the original value. The dielectric constant of the slab will be

$A$ capacitor has some dielectric between its plates and the capacitor is connected to a $D.C.$ source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance,the energy stored in it,electric field,charge stored,and the voltage will increase,decrease,or remain constant.