$A$ parallel plate capacitor has plate area $40 \text{ cm}^2$ and plate separation $2 \text{ mm}$. The space between the plates is filled with a dielectric medium of thickness $1 \text{ mm}$ and dielectric constant $K = 5$. The capacitance of the system is:

$A$ parallel plate air capacitor has a capacitance $C$. When it is half filled with a dielectric of dielectric constant $K = 5$ as shown in the figure,the percentage increase in the capacitance will be......$\%$

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

$A$ parallel plate capacitor having a separation between the plates $d$,plate area $A$,and material with dielectric constant $K$ has capacitance $C_0$. Now,one-third of the material is replaced by another material with dielectric constant $2K$,such that effectively there are two capacitors: one with area $\frac{1}{3}A$,dielectric constant $2K$,and another with area $\frac{2}{3}A$ and dielectric constant $K$. If the capacitance of this new capacitor is $C$,then $\frac{C}{C_0}$ is:

$A$ capacitor is charged by using a battery which is then disconnected. $A$ dielectric slab is then inserted between the plates. What is the result?

$A$ 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 $\text{ pF}$)