$A$ sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor

$A$ capacitor is charged by a battery and the energy stored is $U$. Now,the battery is removed and the distance between the plates is increased to four times its original value. The new energy stored becomes:

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

$A$ parallel plate capacitor is charged by connecting a $2 \ V$ battery across it. It is then disconnected from the battery and a glass slab is introduced between the plates. Which of the following pairs of quantities decrease?

The expression for the capacity of the capacitor formed by compound dielectric placed between the plates of a parallel plate capacitor as shown in the figure,will be (area of plate $= A$)

$A$ parallel plate capacitor having capacitance $C$ farad is connected to a battery of emf $V$ volt. Keeping the capacitor connected to the battery,a dielectric slab of dielectric constant $K$ is inserted between the plates. The dimensions of the slab are such that it fills the whole space between the capacitor plates. Then $:-$