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 gap between the plates of a parallel plate capacitor of area $A$ and distance between plates $d$ is filled with a dielectric whose permittivity varies linearly from $\varepsilon_1$ at one plate to $\varepsilon_2$ at the other. The capacitance of the capacitor is

If the distance between parallel plates of a capacitor is halved and the dielectric constant is doubled,then the capacitance will become:

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:

In the figure,a capacitor is filled with dielectrics of constants $K_1$,$K_2$,and $K_3$. The resultant capacitance is

If half of the space between the plates of a parallel plate capacitor is filled with a medium of dielectric constant $4$,the capacitance is $C_1$. If one third of the space between the plates of the capacitor is filled with the medium of dielectric constant $4$,the capacitance is $C_2$. If in both cases,the dielectric is placed parallel to the plates of the capacitor,then $C_1: C_2=$