$A$ parallel plate capacitor filled with oil of a dielectric constant $3$ between the plates has capacitance $C$. If the oil is removed,then the capacitance of the capacitor will be

Two metal plates each of area $A$ form a parallel plate capacitor with air in between the plates. The distance between the plates is $d$. $A$ metal plate of thickness $\frac{d}{2}$ and of same area $A$ is inserted between the plates to form two capacitors of capacitances $C_1$ and $C_2$ as shown in the figure. If the effective capacitance of the two capacitors is $C^{\prime}$ and the capacitance of the capacitor initially is $C$,then $\frac{C^{\prime}}{C}$ is

Consider a parallel plate capacitor of $10\,\mu F$ with air filled in the gap between the plates. Now,one half of the space between the plates is filled with a dielectric of dielectric constant $K = 4$,as shown in the figure. The capacity of the capacitor changes to.......$\mu F$.

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 $V$)

The variation of the electric field between the plates of a capacitor is shown in the figure. Find the relation between $K_1$,$K_2$,and $K_3$,if $2x = y$.

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