Two parallel plate air capacitors of same capacity $C$ are connected in parallel to a battery of e.m.f. $E$. Then,one of the capacitors is completely filled with a dielectric material of constant $K$. The change in the effective capacity of the parallel combination is:

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

$A$ parallel plate capacitor has a potential of $20 \, kV$ and a capacitance of $2 \times 10^{-4} \, \mu F$. If the area of the plates is $0.01 \, m^2$ and the distance between the plates is $2 \, mm$,find the dielectric constant of the medium.

$A$ parallel plate capacitor has plates with area $A$ and separation $d$. $A$ battery charges the plates to a potential difference $V_0$. The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced is

If the distance between parallel plates of a capacitor is halved and the dielectric constant is doubled,then the capacitance will become:

$A$ parallel plate capacitor has a capacity $80 \times 10^{-6} \ F$ when air is present between its plates. The space between the plates is filled with a dielectric slab of dielectric constant $20$. The capacitor is now connected to a battery of $30 \ V$. The dielectric slab is then removed. The charge passing through the wire is: