$A$ parallel plate capacitor has a capacitance of $50\,\mu F$ in air and $110\,\mu F$ when immersed in an oil. The dielectric constant $K$ of the oil is:

$A$ metal plate of thickness $2 \,mm$ and area $36 \pi \,mm^2$ is slid into a parallel plate capacitor of plate spacing $6 \,mm$ and area $36 \pi \,cm^2$. The metal plate is at a distance $3 \,mm$ from one of the plates. What is the capacitance of this arrangement (in $\,pF$)? (Let $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \,N m^2 C^{-2}$)

$A$ parallel plate capacitor with plate area $A$ and separation $d$ is filled with two dielectric materials as shown in the figure. The dielectric constants are $K_1$ and $K_2$ respectively. The capacitance will be:

Two parallel plate capacitors of capacity $C$ and $3C$ are connected in parallel and charged to a potential difference of $18V$. The battery is then disconnected,and the space between the plates of the capacitor of capacity $C$ is completely filled with a material of dielectric constant $K=9$. The final potential difference across the combination of capacitors will be $V^{\prime}$. Find $V^{\prime}$. (in $V$)

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:

The force between two point charges placed in a material medium of dielectric constant $K$ is $F$. If the material is removed,then the force between them becomes . . . . . . .