An air capacitor of capacity $C = 10\,\mu F$ is connected to a constant voltage battery of $12\,V$. Now the space between the plates is filled with a liquid of dielectric constant $5$. The charge that flows now from battery to the capacitor is......$\mu C$

Two dielectric slab of dielectric constant $K_1$ and $K_2$ and of same thickness is inserted in parallel plats capacitor and $K_1 = 2K_2$ . Potential difference across slabs are $V_1$ and $V_2$ respectively then

While a capacitor remains connected to a battery and dielectric slab is applied between the plates, then

The capacity of a parallel plate condenser is $10\,\mu F$ without dielectric. Dielectric of constant $2$ is used to fill half the distance between the plates, the new capacitance in $\mu F$ is

The two metallic plates of radius $r$ are placed at a distance $d$ apart and its capacity is $C$. If a plate of radius $r/2$ and thickness $d$ of dielectric constant $6$ is placed between the plates of the condenser, then its capacity will be

Write the capacitance of parallel plate capacitor with medium of dielectric of dielectric constant $\mathrm{K}$.