An air capacitor has a capacitance of $1 \mu F$. Now the space between the two plates of the capacitor is filled with two dielectrics as shown in the figure. The capacitance of the capacitor is: [$d=$ distance between two plates of the capacitor,$K_1$ and $K_2$ are dielectric constants of the first and second dielectric respectively] (in $\mu F$)

Two identical condensers $M$ and $N$ are connected in series with a battery. The space between the plates of $M$ is completely filled with a dielectric medium of dielectric constant $8$,and a copper plate of thickness $d/2$ is introduced between the plates of $N$ ($d$ is the distance between the plates). Then the potential differences across $M$ and $N$ are,respectively,in the ratio:

Assertion: The electrostatic force between the plates of a charged isolated capacitor decreases when a dielectric fills the whole space between the plates.
Reason: The electric field between the plates of a charged isolated capacitor increases when a dielectric fills the whole space between the plates.

Two condensers of capacities $2C$ and $C$ are joined in parallel and charged up to potential $V$. The battery is then disconnected and the condenser of capacity $C$ is filled completely with a medium of dielectric constant $K$. The potential difference across the capacitors in the second case is

When the air between the plates of a charged parallel plate capacitor is replaced by a dielectric medium,the intensity of the electric field:

$A$ 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