When a dielectric slab is inserted between the plates of a capacitor while it remains connected to a battery,which of the following occurs during this process?

In a parallel plate capacitor with air between the plates,each plate has an area of $6 \times 10^{-3} \, m^{2}$ and the distance between the plates is $3 \, mm$. The capacitance of the capacitor is $17.71 \, pF$. If this capacitor is connected to a $100 \, V$ supply,and a $3 \, mm$ thick mica sheet (of dielectric constant $k = 6$) is inserted between the plates,calculate the new capacitance,charge,and potential difference in the following cases:
$(a)$ While the voltage supply remains connected.
$(b)$ After the supply is disconnected.

The figure shows two identical parallel plate capacitors connected to a battery and a closed switch $S$. Now,the switch is opened and a dielectric material with dielectric constant $K = 3$ is inserted into the free space between the plates of both capacitors. What is the ratio of the total electrostatic energy stored in both capacitors before and after the insertion of the dielectric?

$A$ parallel plate capacitor has plates of length $l$,width $w$,and separation $d$. It is connected to a battery of emf $V$. $A$ dielectric slab of the same thickness $d$ and dielectric constant $k = 4$ is being inserted between the plates. At what length $x$ of the slab inside the plates will the energy stored in the capacitor be two times the initial energy stored?

$A$ parallel plate capacitor has two layers of dielectric as shown in the figure. This capacitor is connected across a battery. The graph between electric field $(E)$ and distance $(x)$ from the left plate will be:

$A$ parallel plate capacitor has a capacity of $80 \times 10^{-6} \ F$ when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant $K = 20$. The capacitor is connected to a battery of $30 \ V$. The dielectric slab is then removed while the capacitor remains connected to the battery. Calculate the charge that passes through the wire.