$A$ parallel plate air capacitor, with plate separation $d$, has a capacitance of $9 \text{ pF}$. The space between the plates is now filled with two dielectrics, the first having $K_1=3$ and thickness $d_1=d/3$, while the second has $K_2=6$ and thickness $d_2=2d/3$. The capacitance of the new capacitor is: (in $\text{ pF}$)

The plates of a parallel plate capacitor are charged up to $200 \ V$. $A$ dielectric slab of thickness $4 \ mm$ is inserted between its plates. Then,to maintain the same potential difference between the plates of the capacitor,the distance between the plates is increased by $3.2 \ mm$. The dielectric constant of the dielectric slab is

$A$ capacitor stores $60 \mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,a charge of $120 \mu C$ flows through the battery. The dielectric constant of the material inserted is:

For changing the capacitance of a given parallel plate capacitor,a dielectric material of dielectric constant $K$ is used,which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4} d$,where $d$ is the separation between the plates of the parallel plate capacitor. The new capacitance $(C')$ in terms of original capacitance $(C_{0})$ is given by the following relation:

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

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......$V$