The dielectric slab is introduced between the plates of a parallel plate charged capacitor. Which one of the following quantities will not change?

The distance between the plates of a charged parallel plate capacitor is $5 \ cm$ and the electric field inside the plates is $200 \ V/cm$. An uncharged metal bar of width $2 \ cm$ is fully immersed into the capacitor. The length of the metal bar is the same as that of the plate of the capacitor. The voltage across the capacitor after the immersion of the bar is......$V$

$A$ capacitor of capacitance $C = 10 \ \mu F$ is connected to a $12 \ V$ battery. When a dielectric slab of dielectric constant $K = 5$ is inserted between its plates,how much additional charge (in $\mu C$) will flow from the battery to the capacitor?

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.

$A$ parallel plate air capacitor has capacity $C$ and distance of separation between plates is $d$. If a conducting sheet of thickness $\frac{2d}{3}$ is inserted between the plates,the capacitance becomes $C_1$. The ratio of $\frac{C_1}{C}$ is (in $:1$)

$A$ parallel plate capacitor with plate area $A$ and plate separation $d$ is filled with a dielectric material of dielectric constant $K = 4$. The thickness of the dielectric material is $x$,where $x < d$.
Let $C_1$ and $C_2$ be the capacitance of the system for $x = \frac{d}{3}$ and $x = \frac{2d}{3}$,respectively. If $C_1 = 2 \mu F$,the value of $C_2$ is $........... \mu F$.