$A$ dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?

Two identical parallel plate capacitors are placed in series and connected to a constant voltage source of $V_0 \, V$. If one of the capacitors is completely immersed in a liquid with dielectric constant $K$,the potential difference between the plates of the other capacitor will change to:

$A$ capacitor has a capacitance of $5 \mu F$ when its parallel plates are separated by an air medium of thickness $d$. $A$ slab of material with a dielectric constant of $1.5$,having an area equal to that of the plates but a thickness of $\frac{d}{2}$,is inserted between the plates. The capacitance of the capacitor in the presence of the slab will be $..........\mu F$.

$A$ metal plate of thickness $2 \,mm$ and area $36 \pi \,mm^2$ is slid into a parallel plate capacitor of plate spacing $6 \,mm$ and area $36 \pi \,cm^2$. The metal plate is at a distance $3 \,mm$ from one of the plates. What is the capacitance of this arrangement (in $\,pF$)? (Let $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \,N m^2 C^{-2}$)

$A$ capacitor of $10 \mu F$ capacitance,whose plates are separated by $10 \text{ mm}$ through air and each plate has an area of $4 \text{ cm}^2$,is now filled equally with two dielectric media of $K_1=2$ and $K_2=3$ respectively,as shown in the figure. If the new force between the plates is $8 \text{ N}$,the supply voltage is . . . . . . $V$.

$A$ dielectric slab of thickness $d$ is inserted in a parallel plate capacitor whose negative plate is at $x = 0$ and positive plate is at $x = 3d$. The slab is equidistant from the plates. The capacitor is given some charge. As one goes from $x = 0$ to $x = 3d$: