$A$ 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

$A$ parallel plate air-filled capacitor of capacitance $C_1$ has plate area $A$ and the distance between the plates $d$. When a metal sheet of thickness $\frac{d}{2}$ and of the same area $A$ is introduced between the plates,its capacitance becomes $C_2$. The ratio $C_2: C_1$ is

If a dielectric is inserted into a charged capacitor (battery removed),then the quantity that remains constant is

$A$ capacitor is charged by using a battery which is then disconnected. $A$ dielectric slab is then inserted between the plates. What is the result?

Four metallic plates $A, B, C$ and $D$ of the same size with the same separation between them are arranged as shown in the figure. Dielectric slabs of dielectric constant $K = 2$ are placed between $B, C$ and $C, D$ respectively. Plates $B$ and $D$ are connected together. The effective capacitance between $A$ and $C$ is (Assume capacitance of each pair of plates without dielectric is $C$):

$A$ parallel plate capacitor with oil between the plates (dielectric constant of oil $K = 2$) has a capacitance $C$. If the oil is removed,then the capacitance of the capacitor becomes: