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,$K_1=8$ and $K_2=4$ are the dielectric constants of the two dielectrics respectively).

$A$ 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 $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \,N m^2 C^{-2}$)

Consider a parallel plate capacitor of $10\,\mu F$ with air filled in the gap between the plates. Now,one half of the space between the plates is filled with a dielectric of dielectric constant $K = 4$,as shown in the figure. The capacity of the capacitor changes to.......$\mu F$.

Two parallel plates of area $A$ are separated by two different dielectrics as shown in the figure. Find the resultant capacitance.

When a dielectric substance is placed between the two plates of a capacitor,what happens to its capacitance,potential difference,and potential energy,respectively?

$A$ parallel plate capacitor has plates of length $l$,width $w$,and separation $d$. It is connected to a battery of emf $V$. $A$ dielectric slab of the same thickness $d$ and dielectric constant $k = 4$ is being inserted between the plates. At what length $x$ of the slab inside the plates will the energy stored in the capacitor be two times the initial energy stored?