When a dielectric of constant $K = 2$ is filled in a capacitor,its capacitance becomes $C$. If the oil is removed,its capacitance will become ......

$A$ parallel plate capacitor has a capacity $80 \times 10^{-6} \,F$, when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant $20$. The capacitor is now connected to a battery of $30 \,V$ by wires. The dielectric slab is then removed. Then, the charge that passes now through the wire is

In a parallel plate capacitor, the radius of each circular plate is $12\,cm$ and the distance between the plates is $5\,mm$. A glass slab of $3\,mm$ thickness and radius $12\,cm$ with a dielectric constant of $6$ is placed between the plates. The capacitance of the capacitor will be:

$A$ parallel plate capacitor having cross-sectional area $A$ and separation $d$ has air in between the plates. Now,an insulating slab of same area but thickness $t = d/2$ is inserted between the plates as shown in the figure,having dielectric constant $K = 4$. The ratio of the new capacitance to its original capacitance will be:

The capacitance of an air capacitor is $15\,\mu F$ and the separation between the parallel plates is $6\,mm$. $A$ copper plate of $3\,mm$ thickness is introduced symmetrically between the plates. The capacitance now becomes.........$\mu F$.

The space between the plates of a parallel plate capacitor is filled with a dielectric whose dielectric constant varies with distance $x$ as per the relation: $K(x) = K_0 + \lambda x$ (where $\lambda$ is a constant). The capacitance $C$ of the capacitor would be related to its vacuum capacitance $C_0$ by the relation: