When a slab of dielectric material is introduced between the parallel plates of a capacitor which remains connected to a battery,then the charge on the plates relative to the earlier charge:

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

Assertion : If the distance between parallel plates of a capacitor is halved and the dielectric constant is increased to three times its original value,then the capacitance becomes $6$ times.
Reason : The capacity of a capacitor does not depend upon the nature of the material between the plates.

If the distance between the plates of a parallel plate capacitor is halved and the dielectric constant of the dielectric is doubled,then its capacity will

$A$ parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant $\varepsilon_1$ and $\varepsilon_2$,as shown in the figures. The distance between the plates is $d$ and the area of each plate is $A$. If the capacitance in the first configuration and second configuration are $C_1$ and $C_2$ respectively,then $\frac{C_1}{C_2}$ is

Two square-shaped metal plates of side $1 \,m$, kept $0.01 \,m$ apart in air, form a parallel plate capacitor. It is connected to a battery of $500 \,V$. The plates of the capacitor are then immersed in an insulating oil by lowering the plates vertically with a speed of $0.001 \,m/s$. If the dielectric constant of the oil is $11$, then the current drawn from the battery during this process is: