In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be (Given area of plate $=A$ )

A capacitor with plate separation $d$ is charged to $V$ volts. The battery is disconnected and a dielectric slab of thickness $\frac{d}{2}$ and dielectric constant ' $2$ ' is inserted between the plates. The potential difference across its terminals becomes

Separation between the plates of a parallel plate capacitor is $d$ and the area of each plate is $A$. When a slab of material of dielectric constant $k$ and thickness $t(t < d)$ is introduced between the plates, its capacitance becomes

If ${q}_{{f}}$ is the free charge on the capacitor plates and ${q}_{{b}}$ is the bound charge on the dielectric slab of dielectric constant $k$ placed between the capacitor plates, then bound charge $q_{b}$ can be expressed as 

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle $\theta$ with each other. When suspended in water the angle remains the same. If density of the material of the sphere is $1.5 \mathrm{~g} / \mathrm{cc}$, the dielectric constant of water will be

(Take density of water $=1 \mathrm{~g} / \mathrm{cc}$ )

A capacitor is kept connected to the battery and a dielectric slab is inserted between the plates. During this process