$A$ parallel plate capacitor is connected to a battery. If a dielectric slab is inserted between the plates,which of the following quantities increases?

Capacitance of a parallel plate capacitor becomes $4/3$ times its original value if a dielectric slab of thickness $t = d/2$ is inserted between the plates ($d$ is the separation between the plates). The dielectric constant of the slab is

Two capacitors of capacities $2C$ and $C$ are joined in parallel and charged up to potential $V$. The battery is removed and the capacitor of capacity $C$ is filled completely with a medium of dielectric constant $K$. The potential difference across the capacitors will now be

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

$A$ parallel plate capacitor has plates of area $A$ separated by a distance $d$. If a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted between the plates,what is the new capacitance?

$A$ 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.